diff --git a/doc/pub/week37/html/._week37-bs000.html b/doc/pub/week37/html/._week37-bs000.html index eb476f9f4..d3cd07a53 100644 --- a/doc/pub/week37/html/._week37-bs000.html +++ b/doc/pub/week37/html/._week37-bs000.html @@ -37,6 +37,10 @@ doconce format html week37.do.txt --html_style=bootstrap --pygments_html_style=d
- -
We will now couple the discussions of ordinary least squares, Ridge -and Lasso regression with a statistical interpretation, that is we -move from a linear algebra analysis to a statistical analysis. In -particular, we will focus on what the regularization terms can result -in. We will amongst other things show that the regularization -parameter can reduce considerably the variance of the parameters -\( \beta \). -
- -The -advantage of doing linear regression is that we actually end up with -analytical expressions for several statistical quantities. -Standard least squares and Ridge regression allow us to -derive quantities like the variance and other expectation values in a -rather straightforward way. -
- -It is assumed that \( \varepsilon_i -\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are -independent, i.e.: -
-$$ -\begin{align*} -\mbox{Cov}(\varepsilon_{i_1}, -\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} -& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. -\end{align*} -$$ - -The randomness of \( \varepsilon_i \) implies that -\( \mathbf{y}_i \) is also a random variable. In particular, -\( \mathbf{y}_i \) is normally distributed, because \( \varepsilon_i \sim -\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \boldsymbol{\beta} \) is a -non-random scalar. To specify the parameters of the distribution of -\( \mathbf{y}_i \) we need to calculate its first two moments. -
- -Recall that \( \boldsymbol{X} \) is a matrix of dimensionality \( n\times p \). The -notation above \( \mathbf{X}_{i,\ast} \) means that we are looking at the -row number \( i \) and perform a sum over all values \( p \). -
+ +@@ -346,7 +309,7 @@ row number \( i \) and perform a sum over all values \( p \).
- -
The assumption we have made here can be summarized as (and this is going to be useful when we discuss the bias-variance trade off) -that there exists a function \( f(\boldsymbol{x}) \) and a normal distributed error \( \boldsymbol{\varepsilon}\sim \mathcal{N}(0, \sigma^2) \) -which describe our data +
We will now couple the discussions of ordinary least squares, Ridge +and Lasso regression with a statistical interpretation, that is we +move from a linear algebra analysis to a statistical analysis. In +particular, we will focus on what the regularization terms can result +in. We will amongst other things show that the regularization +parameter can reduce considerably the variance of the parameters +\( \beta \). +
+ +The +advantage of doing linear regression is that we actually end up with +analytical expressions for several statistical quantities. +Standard least squares and Ridge regression allow us to +derive quantities like the variance and other expectation values in a +rather straightforward way. +
+ +It is assumed that \( \varepsilon_i +\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are +independent, i.e.:
$$ -\boldsymbol{y} = f(\boldsymbol{x})+\boldsymbol{\varepsilon} +\begin{align*} +\mbox{Cov}(\varepsilon_{i_1}, +\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} +& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. +\end{align*} $$ -We approximate this function with our model from the solution of the linear regression equations, that is our -function \( f \) is approximated by \( \boldsymbol{\tilde{y}} \) where we want to minimize \( (\boldsymbol{y}-\boldsymbol{\tilde{y}})^2 \), our MSE, with +
The randomness of \( \varepsilon_i \) implies that +\( \mathbf{y}_i \) is also a random variable. In particular, +\( \mathbf{y}_i \) is normally distributed, because \( \varepsilon_i \sim +\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \boldsymbol{\beta} \) is a +non-random scalar. To specify the parameters of the distribution of +\( \mathbf{y}_i \) we need to calculate its first two moments.
-$$ -\boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta}. -$$ +Recall that \( \boldsymbol{X} \) is a matrix of dimensionality \( n\times p \). The +notation above \( \mathbf{X}_{i,\ast} \) means that we are looking at the +row number \( i \) and perform a sum over all values \( p \). +
@@ -321,7 +352,7 @@ $$
-
We can calculate the expectation value of \( \boldsymbol{y} \) for a given element \( i \)
-$$ -\begin{align*} -\mathbb{E}(y_i) & = -\mathbb{E}(\mathbf{X}_{i, \ast} \, \boldsymbol{\beta}) + \mathbb{E}(\varepsilon_i) -\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, -\end{align*} -$$ - -while -its variance is +
The assumption we have made here can be summarized as (and this is going to be useful when we discuss the bias-variance trade off) +that there exists a function \( f(\boldsymbol{x}) \) and a normal distributed error \( \boldsymbol{\varepsilon}\sim \mathcal{N}(0, \sigma^2) \) +which describe our data
$$ -\begin{align*} \mbox{Var}(y_i) & = \mathbb{E} \{ [y_i -- \mathbb{E}(y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( y_i^2 ) - -[\mathbb{E}(y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, -\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 \\ & -= \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 \varepsilon_i -\mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + \varepsilon_i^2 ] - ( \mathbf{X}_{i, -\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 -\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + -\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 -\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, -\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. -\end{align*} +\boldsymbol{y} = f(\boldsymbol{x})+\boldsymbol{\varepsilon} $$ -Hence, \( y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta}, \sigma^2) \), that is \( \boldsymbol{y} \) follows a normal distribution with -mean value \( \boldsymbol{X}\boldsymbol{\beta} \) and variance \( \sigma^2 \) (not be confused with the singular values of the SVD). +
We approximate this function with our model from the solution of the linear regression equations, that is our +function \( f \) is approximated by \( \boldsymbol{\tilde{y}} \) where we want to minimize \( (\boldsymbol{y}-\boldsymbol{\tilde{y}})^2 \), our MSE, with
+$$ +\boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta}. +$$ +@@ -337,7 +327,7 @@ mean value \( \boldsymbol{X}\boldsymbol{\beta} \) and variance \( \sigma^2 \) (n
-
With the OLS expressions for the optimal parameters \( \boldsymbol{\hat{\beta}} \) we can evaluate the expectation value
+We can calculate the expectation value of \( \boldsymbol{y} \) for a given element \( i \)
$$ -\mathbb{E}(\boldsymbol{\hat{\beta}}) = \mathbb{E}[ (\mathbf{X}^{\top} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbb{E}[ \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1} \mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta}=\boldsymbol{\beta}. +\begin{align*} +\mathbb{E}(y_i) & = +\mathbb{E}(\mathbf{X}_{i, \ast} \, \boldsymbol{\beta}) + \mathbb{E}(\varepsilon_i) +\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, +\end{align*} $$ -This means that the estimator of the regression parameters is unbiased.
- -We can also calculate the variance
- -The variance of the optimal value \( \boldsymbol{\hat{\beta}} \) is
-$$ -\begin{eqnarray*} -\mbox{Var}(\boldsymbol{\hat{\beta}}) & = & \mathbb{E} \{ [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})] [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})]^{T} \} -\\ -& = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}]^{T} \} -\\ -% & = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}]^{T} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -% \\ -% & = & \mathbb{E} \{ (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} \, \mathbf{Y}^{T} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -% \\ -& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{T} \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -\\ -& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \{ \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + \sigma^2 \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -% \\ -% & = & (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^T \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1} -% \\ -% & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \boldsymbol{\beta} \boldsymbol{\beta}^T -\\ -& = & \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} + \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} -\, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1}, -\end{eqnarray*} -$$ - -where we have used that \( \mathbb{E} (\mathbf{Y} \mathbf{Y}^{T}) = -\mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + -\sigma^2 \, \mathbf{I}_{nn} \). From \( \mbox{Var}(\boldsymbol{\beta}) = \sigma^2 -\, (\mathbf{X}^{T} \mathbf{X})^{-1} \), one obtains an estimate of the -variance of the estimate of the \( j \)-th regression coefficient: -\( \boldsymbol{\sigma}^2 (\boldsymbol{\beta}_j ) = \boldsymbol{\sigma}^2 [(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} \). This may be used to -construct a confidence interval for the estimates. +
while +its variance is
- -In a similar way, we can obtain analytical expressions for say the -expectation values of the parameters \( \boldsymbol{\beta} \) and their variance -when we employ Ridge regression, allowing us again to define a confidence interval. -
- -It is rather straightforward to show that
$$ -\mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big]=(\mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X})\boldsymbol{\beta}^{\mathrm{OLS}}. +\begin{align*} \mbox{Var}(y_i) & = \mathbb{E} \{ [y_i +- \mathbb{E}(y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( y_i^2 ) - +[\mathbb{E}(y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, +\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 \\ & += \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 \varepsilon_i +\mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + \varepsilon_i^2 ] - ( \mathbf{X}_{i, +\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 + 2 +\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \boldsymbol{\beta} + +\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta})^2 +\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, +\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. +\end{align*} $$ -We see clearly that -\( \mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big] \not= \boldsymbol{\beta}^{\mathrm{OLS}} \) for any \( \lambda > 0 \). We say then that the ridge estimator is biased. -
- -We can also compute the variance as
- -$$ -\mbox{Var}[\boldsymbol{\beta}^{\mathrm{Ridge}}]=\sigma^2[ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1} \mathbf{X}^{T} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}, -$$ - -and it is easy to see that if the parameter \( \lambda \) goes to infinity then the variance of Ridge parameters \( \boldsymbol{\beta} \) goes to zero.
- -With this, we can compute the difference
- -$$ -\mbox{Var}[\boldsymbol{\beta}^{\mathrm{OLS}}]-\mbox{Var}(\boldsymbol{\beta}^{\mathrm{Ridge}})=\sigma^2 [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}[ 2\lambda\mathbf{I} + \lambda^2 (\mathbf{X}^{T} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}. -$$ - -The difference is non-negative definite since each component of the -matrix product is non-negative definite. -This means the variance we obtain with the standard OLS will always for \( \lambda > 0 \) be larger than the variance of \( \boldsymbol{\beta} \) obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below. +
Hence, \( y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta}, \sigma^2) \), that is \( \boldsymbol{y} \) follows a normal distribution with +mean value \( \boldsymbol{X}\boldsymbol{\beta} \) and variance \( \sigma^2 \) (not be confused with the singular values of the SVD).
@@ -382,7 +343,7 @@ This means the variance we obtain with the standard OLS will always for \( \lamb
-
With the OLS expressions for the optimal parameters \( \boldsymbol{\hat{\beta}} \) we can evaluate the expectation value
+$$ +\mathbb{E}(\boldsymbol{\hat{\beta}}) = \mathbb{E}[ (\mathbf{X}^{\top} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbb{E}[ \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1} \mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta}=\boldsymbol{\beta}. +$$ + +This means that the estimator of the regression parameters is unbiased.
+ +We can also calculate the variance
+ +The variance of the optimal value \( \boldsymbol{\hat{\beta}} \) is
+$$ +\begin{eqnarray*} +\mbox{Var}(\boldsymbol{\hat{\beta}}) & = & \mathbb{E} \{ [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})] [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})]^{T} \} +\\ +& = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}]^{T} \} +\\ +% & = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}]^{T} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +% \\ +% & = & \mathbb{E} \{ (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} \, \mathbf{Y}^{T} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +% \\ +& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{T} \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +\\ +& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \{ \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + \sigma^2 \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +% \\ +% & = & (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^T \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1} +% \\ +% & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \boldsymbol{\beta} \boldsymbol{\beta}^T +\\ +& = & \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} + \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} +\, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1}, +\end{eqnarray*} +$$ + +where we have used that \( \mathbb{E} (\mathbf{Y} \mathbf{Y}^{T}) = +\mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \, \mathbf{X}^{T} + +\sigma^2 \, \mathbf{I}_{nn} \). From \( \mbox{Var}(\boldsymbol{\beta}) = \sigma^2 +\, (\mathbf{X}^{T} \mathbf{X})^{-1} \), one obtains an estimate of the +variance of the estimate of the \( j \)-th regression coefficient: +\( \boldsymbol{\sigma}^2 (\boldsymbol{\beta}_j ) = \boldsymbol{\sigma}^2 [(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} \). This may be used to +construct a confidence interval for the estimates. +
+ +In a similar way, we can obtain analytical expressions for say the +expectation values of the parameters \( \boldsymbol{\beta} \) and their variance +when we employ Ridge regression, allowing us again to define a confidence interval. +
+ +It is rather straightforward to show that
+$$ +\mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big]=(\mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X})\boldsymbol{\beta}^{\mathrm{OLS}}. +$$ + +We see clearly that +\( \mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big] \not= \boldsymbol{\beta}^{\mathrm{OLS}} \) for any \( \lambda > 0 \). We say then that the ridge estimator is biased. +
+ +We can also compute the variance as
+ +$$ +\mbox{Var}[\boldsymbol{\beta}^{\mathrm{Ridge}}]=\sigma^2[ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1} \mathbf{X}^{T} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}, +$$ + +and it is easy to see that if the parameter \( \lambda \) goes to infinity then the variance of Ridge parameters \( \boldsymbol{\beta} \) goes to zero.
+ +With this, we can compute the difference
+ +$$ +\mbox{Var}[\boldsymbol{\beta}^{\mathrm{OLS}}]-\mbox{Var}(\boldsymbol{\beta}^{\mathrm{Ridge}})=\sigma^2 [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}[ 2\lambda\mathbf{I} + \lambda^2 (\mathbf{X}^{T} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}. +$$ + +The difference is non-negative definite since each component of the +matrix product is non-negative definite. +This means the variance we obtain with the standard OLS will always for \( \lambda > 0 \) be larger than the variance of \( \boldsymbol{\beta} \) obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below. +
+ +For more discussions of Ridge regression and calculation of averages, Wessel van Wieringen's article is highly recommended.
@@ -308,7 +390,7 @@ MathJax.Hub.Config({
-
Our basic assumption when we derived the OLS equations was to assume -that our output is determined by a given continuous function -\( f(\boldsymbol{x}) \) and a random noise \( \boldsymbol{\epsilon} \) given by the normal -distribution with zero mean value and an undetermined variance -\( \sigma^2 \). -
- -We found above that the outputs \( \boldsymbol{y} \) have a mean value given by -\( \boldsymbol{X}\hat{\boldsymbol{\beta}} \) and variance \( \sigma^2 \). Since the entries to -the design matrix are not stochastic variables, we can assume that the -probability distribution of our targets is also a normal distribution -but now with mean value \( \boldsymbol{X}\hat{\boldsymbol{\beta}} \). This means that a -single output \( y_i \) is given by the Gaussian distribution -
- -$$ -y_i\sim \mathcal{N}(\boldsymbol{X}_{i,*}\boldsymbol{\beta}, \sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. -$$ - +@@ -329,7 +314,7 @@ $$
-
We assume now that the various \( y_i \) values are stochastically distributed according to the above Gaussian distribution. -We define this distribution as +
Our basic assumption when we derived the OLS equations was to assume +that our output is determined by a given continuous function +\( f(\boldsymbol{x}) \) and a random noise \( \boldsymbol{\epsilon} \) given by the normal +distribution with zero mean value and an undetermined variance +\( \sigma^2 \).
-$$ -p(y_i, \boldsymbol{X}\vert\boldsymbol{\beta})=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}, -$$ -which reads as finding the likelihood of an event \( y_i \) with the input variables \( \boldsymbol{X} \) given the parameters (to be determined) \( \boldsymbol{\beta} \).
- -Since these events are assumed to be independent and identicall distributed we can build the probability distribution function (PDF) for all possible event \( \boldsymbol{y} \) as the product of the single events, that is we have
- -$$ -p(\boldsymbol{y},\boldsymbol{X}\vert\boldsymbol{\beta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}=\prod_{i=0}^{n-1}p(y_i,\boldsymbol{X}\vert\boldsymbol{\beta}). -$$ - -We will write this in a more compact form reserving \( \boldsymbol{D} \) for the domain of events, including the ouputs (targets) and the inputs. That is -in case we have a simple one-dimensional input and output case +
We found above that the outputs \( \boldsymbol{y} \) have a mean value given by +\( \boldsymbol{X}\hat{\boldsymbol{\beta}} \) and variance \( \sigma^2 \). Since the entries to +the design matrix are not stochastic variables, we can assume that the +probability distribution of our targets is also a normal distribution +but now with mean value \( \boldsymbol{X}\hat{\boldsymbol{\beta}} \). This means that a +single output \( y_i \) is given by the Gaussian distribution
+ $$ -\boldsymbol{D}=[(x_0,y_0), (x_1,y_1),\dots, (x_{n-1},y_{n-1})]. +y_i\sim \mathcal{N}(\boldsymbol{X}_{i,*}\boldsymbol{\beta}, \sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. $$ -In the more general case the various inputs should be replaced by the possible features represented by the input data set \( \boldsymbol{X} \). -We can now rewrite the above probability as -
-$$ -p(\boldsymbol{D}\vert\boldsymbol{\beta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. -$$ - -It is a conditional probability (see below) and reads as the likelihood of a domain of events \( \boldsymbol{D} \) given a set of parameters \( \boldsymbol{\beta} \).
@@ -341,7 +335,7 @@ $$
-
In statistics, maximum likelihood estimation (MLE) is a method of -estimating the parameters of an assumed probability distribution, -given some observed data. This is achieved by maximizing a likelihood -function so that, under the assumed statistical model, the observed -data is the most probable. +
We assume now that the various \( y_i \) values are stochastically distributed according to the above Gaussian distribution. +We define this distribution as
+$$ +p(y_i, \boldsymbol{X}\vert\boldsymbol{\beta})=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}, +$$ -We will assume here that our events are given by the above Gaussian -distribution and we will determine the optimal parameters \( \beta \) by -maximizing the above PDF. However, computing the derivatives of a -product function is cumbersome and can easily lead to overflow and/or -underflowproblems, with potentials for loss of numerical precision. -
+which reads as finding the likelihood of an event \( y_i \) with the input variables \( \boldsymbol{X} \) given the parameters (to be determined) \( \boldsymbol{\beta} \).
-In practice, it is more convenient to maximize the logarithm of the -PDF because it is a monotonically increasing function of the argument. -Alternatively, and this will be our option, we will minimize the -negative of the logarithm since this is a monotonically decreasing -function. -
+Since these events are assumed to be independent and identicall distributed we can build the probability distribution function (PDF) for all possible event \( \boldsymbol{y} \) as the product of the single events, that is we have
-Note also that maximization/minimization of the logarithm of the PDF -is equivalent to the maximization/minimization of the function itself. +$$ +p(\boldsymbol{y},\boldsymbol{X}\vert\boldsymbol{\beta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}=\prod_{i=0}^{n-1}p(y_i,\boldsymbol{X}\vert\boldsymbol{\beta}). +$$ + +
We will write this in a more compact form reserving \( \boldsymbol{D} \) for the domain of events, including the ouputs (targets) and the inputs. That is +in case we have a simple one-dimensional input and output case
+$$ +\boldsymbol{D}=[(x_0,y_0), (x_1,y_1),\dots, (x_{n-1},y_{n-1})]. +$$ + +In the more general case the various inputs should be replaced by the possible features represented by the input data set \( \boldsymbol{X} \). +We can now rewrite the above probability as +
+$$ +p(\boldsymbol{D}\vert\boldsymbol{\beta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. +$$ + +It is a conditional probability (see below) and reads as the likelihood of a domain of events \( \boldsymbol{D} \) given a set of parameters \( \boldsymbol{\beta} \).
@@ -336,7 +347,7 @@ is equivalent to the maximization/minimization of the function itself.
-
We could now define a new cost function to minimize, namely the negative logarithm of the above PDF
+In statistics, maximum likelihood estimation (MLE) is a method of +estimating the parameters of an assumed probability distribution, +given some observed data. This is achieved by maximizing a likelihood +function so that, under the assumed statistical model, the observed +data is the most probable. +
-$$ -C(\boldsymbol{\beta}=-\log{\prod_{i=0}^{n-1}p(y_i,\boldsymbol{X}\vert\boldsymbol{\beta})}=-\sum_{i=0}^{n-1}\log{p(y_i,\boldsymbol{X}\vert\boldsymbol{\beta})}, -$$ +We will assume here that our events are given by the above Gaussian +distribution and we will determine the optimal parameters \( \beta \) by +maximizing the above PDF. However, computing the derivatives of a +product function is cumbersome and can easily lead to overflow and/or +underflowproblems, with potentials for loss of numerical precision. +
-which becomes
-$$ -C(\boldsymbol{\beta}=\frac{n}{2}\log{2\pi\sigma^2}+\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}. -$$ +In practice, it is more convenient to maximize the logarithm of the +PDF because it is a monotonically increasing function of the argument. +Alternatively, and this will be our option, we will minimize the +negative of the logarithm since this is a monotonically decreasing +function. +
-Taking the derivative of the new cost function with respect to the parameters \( \beta \) we recognize our familiar OLS equation, namely
- -$$ -\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right) =0, -$$ - -which leads to the well-known OLS equation for the optimal paramters \( \beta \)
-$$ -\hat{\boldsymbol{\beta}}^{\mathrm{OLS}}=\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}! -$$ - -Before we make a similar analysis for Ridge and Lasso regression, we need a short reminder on statistics.
+Note also that maximization/minimization of the logarithm of the PDF +is equivalent to the maximization/minimization of the function itself. +
@@ -336,7 +342,7 @@ $$
-
A central theorem in statistics is Bayes' theorem. This theorem plays a similar role as the good old Pythagoras' theorem in geometry. -Bayes' theorem is extremely simple to derive. But to do so we need some basic axioms from statistics. -
+We could now define a new cost function to minimize, namely the negative logarithm of the above PDF
-Assume we have two domains of events \( X=[x_0,x_1,\dots,x_{n-1}] \) and \( Y=[y_0,y_1,\dots,y_{n-1}] \).
- -We define also the likelihood for \( X \) and \( Y \) as \( p(X) \) and \( p(Y) \) respectively. -The likelihood of a specific event \( x_i \) (or \( y_i \)) is then written as \( p(X=x_i) \) or just \( p(x_i)=p_i \). -
- -where we read \( p(X\vert Y) \) as the likelihood of obtaining \( X \) given \( Y \).
-which becomes
+$$ +C(\boldsymbol{\beta}=\frac{n}{2}\log{2\pi\sigma^2}+\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}. +$$ +Taking the derivative of the new cost function with respect to the parameters \( \beta \) we recognize our familiar OLS equation, namely
-If we have independent events then \( p(X,Y)=p(X)p(Y) \).
+$$ +\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right) =0, +$$ + +which leads to the well-known OLS equation for the optimal paramters \( \beta \)
+$$ +\hat{\boldsymbol{\beta}}^{\mathrm{OLS}}=\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}! +$$ + +Before we make a similar analysis for Ridge and Lasso regression, we need a short reminder on statistics.
@@ -346,7 +341,7 @@ $$
-
A central theorem in statistics is Bayes' theorem. This theorem plays a similar role as the good old Pythagoras' theorem in geometry. +Bayes' theorem is extremely simple to derive. But to do so we need some basic axioms from statistics. +
+ +Assume we have two domains of events \( X=[x_0,x_1,\dots,x_{n-1}] \) and \( Y=[y_0,y_1,\dots,y_{n-1}] \).
+ +We define also the likelihood for \( X \) and \( Y \) as \( p(X) \) and \( p(Y) \) respectively. +The likelihood of a specific event \( x_i \) (or \( y_i \)) is then written as \( p(X=x_i) \) or just \( p(x_i)=p_i \). +
-The marginal probability is defined in terms of only one of the set of variables \( X,Y \). For a discrete probability we have
where we read \( p(X\vert Y) \) as the likelihood of obtaining \( X \) given \( Y \).
+If we have independent events then \( p(X,Y)=p(X)p(Y) \).
+diff --git a/doc/pub/week37/html/._week37-bs013.html b/doc/pub/week37/html/._week37-bs013.html index 3df2205ab..cd1a20a82 100644 --- a/doc/pub/week37/html/._week37-bs013.html +++ b/doc/pub/week37/html/._week37-bs013.html @@ -37,6 +37,10 @@ doconce format html week37.do.txt --html_style=bootstrap --pygments_html_style=d
-
The conditional probability, if \( p(Y) > 0 \), is
+The marginal probability is defined in terms of only one of the set of variables \( X,Y \). For a discrete probability we have
-
If we combine the conditional probability with the marginal probability and the standard product rule, we have
+The conditional probability, if \( p(Y) > 0 \), is
+which we can rewrite as
- -$$ -p(X\vert Y)= \frac{p(X,Y)}{\sum_{i=0}^{n-1}p(Y\vert X=x_i)p(x_i)}=\frac{p(Y\vert X)p(X)}{\sum_{i=0}^{n-1}p(Y\vert X=x_i)p(x_i)}, -$$ - -which is Bayes' theorem. It allows us to evaluate the uncertainty in in \( X \) after we have observed \( Y \). We can easily interchange \( X \) with \( Y \).
@@ -325,7 +328,7 @@ $$
-
The quantity \( p(Y\vert X) \) on the right-hand side of the theorem is -evaluated for the observed data \( Y \) and can be viewed as a function of -the parameter space represented by \( X \). This function is not -necesseraly normalized and is normally called the likelihood function. -
+If we combine the conditional probability with the marginal probability and the standard product rule, we have
+$$ +p(X\vert Y)= \frac{p(X,Y)}{p(Y)}, +$$ -The function \( p(X) \) on the right hand side is called the prior while the function on the left hand side is the called the posterior probability. The denominator on the right hand side serves as a normalization factor for the posterior distribution.
+which we can rewrite as
-Let us try to illustrate Bayes' theorem through an example.
+$$ +p(X\vert Y)= \frac{p(X,Y)}{\sum_{i=0}^{n-1}p(Y\vert X=x_i)p(x_i)}=\frac{p(Y\vert X)p(X)}{\sum_{i=0}^{n-1}p(Y\vert X=x_i)p(x_i)}, +$$ + +which is Bayes' theorem. It allows us to evaluate the uncertainty in in \( X \) after we have observed \( Y \). We can easily interchange \( X \) with \( Y \).
@@ -322,7 +330,7 @@ necesseraly normalized and is normally called the likelihood function.
-
Let us suppose that you are undergoing a series of mammography scans in -order to rule out possible breast cancer cases. We define the -sensitivity for a positive event by the variable \( X \). It takes binary -values with \( X=1 \) representing a positive event and \( X=0 \) being a -negative event. We reserve \( Y \) as a classification parameter for -either a negative or a positive breast cancer confirmation. (Short note on wordings: positive here means having breast cancer, although none of us would consider this being a positive thing). +
The quantity \( p(Y\vert X) \) on the right-hand side of the theorem is +evaluated for the observed data \( Y \) and can be viewed as a function of +the parameter space represented by \( X \). This function is not +necesseraly normalized and is normally called the likelihood function.
-We let \( Y=1 \) represent the the case of having breast cancer and \( Y=0 \) as not.
+The function \( p(X) \) on the right hand side is called the prior while the function on the left hand side is the called the posterior probability. The denominator on the right hand side serves as a normalization factor for the posterior distribution.
-Let us assume that if you have breast cancer, the test will be positive with a probability of \( 0.8 \), that is we have
- -$$ -p(X=1\vert Y=1) =0.8. -$$ - -This obviously sounds scary since many would conclude that if the test is positive, there is a likelihood of \( 80\% \) for having cancer. -It is however not correct, as the following Bayesian analysis shows. -
+Let us try to illustrate Bayes' theorem through an example.
@@ -332,7 +327,7 @@ It is however not correct, as the following Bayesian analysis shows.
-
If we look at various national surveys on breast cancer, the general likelihood of developing breast cancer is a very small number. -Let us assume that the prior probability in the population as a whole is +
Let us suppose that you are undergoing a series of mammography scans in +order to rule out possible breast cancer cases. We define the +sensitivity for a positive event by the variable \( X \). It takes binary +values with \( X=1 \) representing a positive event and \( X=0 \) being a +negative event. We reserve \( Y \) as a classification parameter for +either a negative or a positive breast cancer confirmation. (Short note on wordings: positive here means having breast cancer, although none of us would consider this being a positive thing).
-$$ -p(Y=1) =0.004. -$$ +We let \( Y=1 \) represent the the case of having breast cancer and \( Y=0 \) as not.
-We need also to account for the fact that the test may produce a false positive result (false alarm). Let us here assume that we have
-$$ -p(X=1\vert Y=0) =0.1. -$$ - -Using Bayes' theorem we can then find the posterior probability that the person has breast cancer in case of a positive test, that is we can compute
+Let us assume that if you have breast cancer, the test will be positive with a probability of \( 0.8 \), that is we have
$$ -p(Y=1\vert X=1)=\frac{p(X=1\vert Y=1)p(Y=1)}{p(X=1\vert Y=1)p(Y=1)+p(X=1\vert Y=0)p(Y=0)}=\frac{0.8\times 0.004}{0.8\times 0.004+0.1\times 0.996}=0.031. +p(X=1\vert Y=1) =0.8. $$ -That is, in case of a positive test, there is only a \( 3\% \) chance of having breast cancer!
+This obviously sounds scary since many would conclude that if the test is positive, there is a likelihood of \( 80\% \) for having cancer. +It is however not correct, as the following Bayesian analysis shows. +
@@ -333,7 +337,7 @@ $$
-
Using Bayes' theorem we can gain a better intuition about Ridge and Lasso regression.
- -For ordinary least squares we postulated that the maximum likelihood for the doamin of events \( \boldsymbol{D} \) (one-dimensional case)
-$$ -\boldsymbol{D}=[(x_0,y_0), (x_1,y_1),\dots, (x_{n-1},y_{n-1})], -$$ - -is given by
-$$ -p(\boldsymbol{D}\vert\boldsymbol{\beta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. -$$ - -In Bayes' theorem this function plays the role of the so-called likelihood. We could now ask the question what is the posterior probability of a parameter set \( \boldsymbol{\beta} \) given a domain of events \( \boldsymbol{D} \)? That is, how can we define the posterior probability
+If we look at various national surveys on breast cancer, the general likelihood of developing breast cancer is a very small number. +Let us assume that the prior probability in the population as a whole is +
$$ -p(\boldsymbol{\beta}\vert\boldsymbol{D}). +p(Y=1) =0.004. $$ -Bayes' theorem comes to our rescue here since (omitting the normalization constant)
+We need also to account for the fact that the test may produce a false positive result (false alarm). Let us here assume that we have
$$ -p(\boldsymbol{\beta}\vert\boldsymbol{D})\propto p(\boldsymbol{D}\vert\boldsymbol{\beta})p(\boldsymbol{\beta}). +p(X=1\vert Y=0) =0.1. $$ -We have a model for \( p(\boldsymbol{D}\vert\boldsymbol{\beta}) \) but need one for the prior \( p(\boldsymbol{\beta} \)!
+Using Bayes' theorem we can then find the posterior probability that the person has breast cancer in case of a positive test, that is we can compute
+ +$$ +p(Y=1\vert X=1)=\frac{p(X=1\vert Y=1)p(Y=1)}{p(X=1\vert Y=1)p(Y=1)+p(X=1\vert Y=0)p(Y=0)}=\frac{0.8\times 0.004}{0.8\times 0.004+0.1\times 0.996}=0.031. +$$ + +That is, in case of a positive test, there is only a \( 3\% \) chance of having breast cancer!
@@ -337,7 +338,7 @@ $$
-
With the posterior probability defined by a likelihood which we have -already modeled and an unknown prior, we are now ready to make -additional models for the prior. -
- -We can, based on our discussions of the variance of \( \boldsymbol{\beta} \) and the mean value, assume that the prior for the values \( \boldsymbol{\beta} \) is given by a Gaussian with mean value zero and variance \( \tau^2 \), that is
+Using Bayes' theorem we can gain a better intuition about Ridge and Lasso regression.
+For ordinary least squares we postulated that the maximum likelihood for the doamin of events \( \boldsymbol{D} \) (one-dimensional case)
$$ -p(\boldsymbol{\beta})=\prod_{j=0}^{p-1}\exp{\left(-\frac{\beta_j^2}{2\tau^2}\right)}. +\boldsymbol{D}=[(x_0,y_0), (x_1,y_1),\dots, (x_{n-1},y_{n-1})], $$ -Our posterior probability becomes then (omitting the normalization factor which is just a constant)
+is given by
$$ -p(\boldsymbol{\beta\vert\boldsymbol{D})}=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}\prod_{j=0}^{p-1}\exp{\left(-\frac{\beta_j^2}{2\tau^2}\right)}. +p(\boldsymbol{D}\vert\boldsymbol{\beta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. $$ -We can now optimize this quantity with respect to \( \boldsymbol{\beta} \). As we -did for OLS, this is most conveniently done by taking the negative -logarithm of the posterior probability. Doing so and leaving out the -constants terms that do not depend on \( \beta \), we have -
+In Bayes' theorem this function plays the role of the so-called likelihood. We could now ask the question what is the posterior probability of a parameter set \( \boldsymbol{\beta} \) given a domain of events \( \boldsymbol{D} \)? That is, how can we define the posterior probability
$$ -C(\boldsymbol{\beta})=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\frac{1}{2\tau^2}\vert\vert\boldsymbol{\beta}\vert\vert_2^2, +p(\boldsymbol{\beta}\vert\boldsymbol{D}). $$ -and replacing \( 1/2\tau^2 \) with \( \lambda \) we have
- +Bayes' theorem comes to our rescue here since (omitting the normalization constant)
$$ -C(\boldsymbol{\beta})=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_2^2, +p(\boldsymbol{\beta}\vert\boldsymbol{D})\propto p(\boldsymbol{D}\vert\boldsymbol{\beta})p(\boldsymbol{\beta}). $$ -which is our Ridge cost function! Nice, isn't it?
+We have a model for \( p(\boldsymbol{D}\vert\boldsymbol{\beta}) \) but need one for the prior \( p(\boldsymbol{\beta} \)!
@@ -346,7 +342,7 @@ $$
-
To derive the Lasso cost function, we simply replace the Gaussian prior with an exponential distribution (Laplace in this case) with zero mean value, that is
+With the posterior probability defined by a likelihood which we have +already modeled and an unknown prior, we are now ready to make +additional models for the prior. +
+ +We can, based on our discussions of the variance of \( \boldsymbol{\beta} \) and the mean value, assume that the prior for the values \( \boldsymbol{\beta} \) is given by a Gaussian with mean value zero and variance \( \tau^2 \), that is
$$ -p(\boldsymbol{\beta})=\prod_{j=0}^{p-1}\exp{\left(-\frac{\vert\beta_j\vert}{\tau}\right)}. +p(\boldsymbol{\beta})=\prod_{j=0}^{p-1}\exp{\left(-\frac{\beta_j^2}{2\tau^2}\right)}. $$Our posterior probability becomes then (omitting the normalization factor which is just a constant)
$$ -p(\boldsymbol{\beta}\vert\boldsymbol{D})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}\prod_{j=0}^{p-1}\exp{\left(-\frac{\vert\beta_j\vert}{\tau}\right)}. +p(\boldsymbol{\beta\vert\boldsymbol{D})}=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}\prod_{j=0}^{p-1}\exp{\left(-\frac{\beta_j^2}{2\tau^2}\right)}. $$ -Taking the negative -logarithm of the posterior probability and leaving out the +
We can now optimize this quantity with respect to \( \boldsymbol{\beta} \). As we +did for OLS, this is most conveniently done by taking the negative +logarithm of the posterior probability. Doing so and leaving out the constants terms that do not depend on \( \beta \), we have
$$ -C(\boldsymbol{\beta}=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\frac{1}{\tau}\vert\vert\boldsymbol{\beta}\vert\vert_1, +C(\boldsymbol{\beta})=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\frac{1}{2\tau^2}\vert\vert\boldsymbol{\beta}\vert\vert_2^2, $$ -and replacing \( 1/\tau \) with \( \lambda \) we have
+and replacing \( 1/2\tau^2 \) with \( \lambda \) we have
$$ -C(\boldsymbol{\beta}=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_1, +C(\boldsymbol{\beta})=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_2^2, $$ -which is our Lasso cost function!
+which is our Ridge cost function! Nice, isn't it?
@@ -340,7 +351,7 @@ $$
-
We will play around with a study of the values for the optimal -parameters \( \boldsymbol{\beta} \) using OLS, Ridge and Lasso regression. For -OLS, you will notice as function of the noise and polynomial degree, -that the parameters \( \beta \) will fluctuate from order to order in the -polynomial fit and that for larger and larger polynomial degrees of freedom, the parameters will tend to increase in value for OLS. +
To derive the Lasso cost function, we simply replace the Gaussian prior with an exponential distribution (Laplace in this case) with zero mean value, that is
+ +$$ +p(\boldsymbol{\beta})=\prod_{j=0}^{p-1}\exp{\left(-\frac{\vert\beta_j\vert}{\tau}\right)}. +$$ + +Our posterior probability becomes then (omitting the normalization factor which is just a constant)
+$$ +p(\boldsymbol{\beta}\vert\boldsymbol{D})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}\prod_{j=0}^{p-1}\exp{\left(-\frac{\vert\beta_j\vert}{\tau}\right)}. +$$ + +Taking the negative +logarithm of the posterior probability and leaving out the +constants terms that do not depend on \( \beta \), we have
-For Ridge and Lasso regression, the higher order parameters will typically be reduced, providing thereby less fluctuations from one order to another one.
+$$ +C(\boldsymbol{\beta}=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\frac{1}{\tau}\vert\vert\boldsymbol{\beta}\vert\vert_1, +$$ +and replacing \( 1/\tau \) with \( \lambda \) we have
- -import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.model_selection import train_test_split
-from sklearn import linear_model
+$$
+C(\boldsymbol{\beta}=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_1,
+$$
-def R2(y_data, y_model):
- return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
-def MSE(y_data,y_model):
- n = np.size(y_model)
- return np.sum((y_data-y_model)**2)/n
-
-# Make data set.
-n = 10000
-x = np.random.rand(n)
-y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.randn(n)
-
-Maxpolydegree = 5
-X = np.zeros((len(x),Maxpolydegree))
-X[:,0] = 1.0
-
-for polydegree in range(1, Maxpolydegree+1):
- for degree in range(polydegree):
- X[:,degree] = x**(degree)
-
-
-# We split the data in test and training data
-X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
-
-# matrix inversion to find beta
-OLSbeta = np.linalg.pinv(X_train.T @ X_train) @ X_train.T @ y_train
-print(OLSbeta)
-ypredictOLS = X_test @ OLSbeta
-print("Test MSE OLS")
-print(MSE(y_test,ypredictOLS))
-# Repeat now for Lasso and Ridge regression and various values of the regularization parameter using Scikit-Learn
-# Decide which values of lambda to use
-nlambdas = 4
-MSERidgePredict = np.zeros(nlambdas)
-MSELassoPredict = np.zeros(nlambdas)
-lambdas = np.logspace(-3, 1, nlambdas)
-for i in range(nlambdas):
- lmb = lambdas[i]
- # Make the fit using Ridge and Lasso
- RegRidge = linear_model.Ridge(lmb,fit_intercept=False)
- RegRidge.fit(X_train,y_train)
- RegLasso = linear_model.Lasso(lmb,fit_intercept=False)
- RegLasso.fit(X_train,y_train)
- # and then make the prediction
- ypredictRidge = RegRidge.predict(X_test)
- ypredictLasso = RegLasso.predict(X_test)
- # Compute the MSE and print it
- MSERidgePredict[i] = MSE(y_test,ypredictRidge)
- MSELassoPredict[i] = MSE(y_test,ypredictLasso)
- print(lmb,RegRidge.coef_)
- print(lmb,RegLasso.coef_)
-# Now plot the results
-plt.figure()
-plt.plot(np.log10(lambdas), MSERidgePredict, 'b', label = 'MSE Ridge Test')
-plt.plot(np.log10(lambdas), MSELassoPredict, 'r', label = 'MSE Lasso Test')
-plt.xlabel('log10(lambda)')
-plt.ylabel('MSE')
-plt.legend()
-plt.show()
-
-How can we understand this?
+which is our Lasso cost function!
@@ -408,7 +345,7 @@ plt.show()
-
Let us write out the values of the coefficients \( \beta_i \) as functions -of the polynomial degree and noise. We will focus only on the Ridge -results and some few selected values of the hyperparameter \( \lambda \). +
We will play around with a study of the values for the optimal +parameters \( \boldsymbol{\beta} \) using OLS, Ridge and Lasso regression. For +OLS, you will notice as function of the noise and polynomial degree, +that the parameters \( \beta \) will fluctuate from order to order in the +polynomial fit and that for larger and larger polynomial degrees of freedom, the parameters will tend to increase in value for OLS.
-If we don't include any noise and run this code for different values -of the polynomial degree, we notice that the results for \( \beta_i \) do -not show great changes from one order to the next. This is an -indication that for higher polynomial orders, our parameters become -less important. -
- -If we however add noise, what happens is that the polynomial fit is -trying to adjust the fit to traverse in the best possible way all data -points. This can lead to large fluctuations in the parameters -\( \beta_i \) as functions of polynomial order. It will also be reflected -in a larger value of the variance of each parameter \( \beta_i \). What -Ridge regression (and Lasso as well) are doing then is to try to -quench the fluctuations in the parameters of \( \beta_i \) which have a -large variance (normally for higher orders in the polynomial). -
+For Ridge and Lasso regression, the higher order parameters will typically be reduced, providing thereby less fluctuations from one order to another one.
@@ -317,14 +309,18 @@ large variance (normally for higher orders in the polynomial).import numpy as np
-import pandas as pd
-from IPython.display import display
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn import linear_model
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
# Make data set.
-n = 1000
+n = 10000
x = np.random.rand(n)
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.randn(n)
@@ -340,20 +336,41 @@ X[:,0] =
# We split the data in test and training data
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+# matrix inversion to find beta
+OLSbeta = np.linalg.pinv(X_train.T @ X_train) @ X_train.T @ y_train
+print(OLSbeta)
+ypredictOLS = X_test @ OLSbeta
+print("Test MSE OLS")
+print(MSE(y_test,ypredictOLS))
+# Repeat now for Lasso and Ridge regression and various values of the regularization parameter using Scikit-Learn
# Decide which values of lambda to use
-nlambdas = 5
-lambdas = np.logspace(-3, 2, nlambdas)
+nlambdas = 4
+MSERidgePredict = np.zeros(nlambdas)
+MSELassoPredict = np.zeros(nlambdas)
+lambdas = np.logspace(-3, 1, nlambdas)
for i in range(nlambdas):
lmb = lambdas[i]
- # Make the fit using Ridge only
+ # Make the fit using Ridge and Lasso
RegRidge = linear_model.Ridge(lmb,fit_intercept=False)
RegRidge.fit(X_train,y_train)
+ RegLasso = linear_model.Lasso(lmb,fit_intercept=False)
+ RegLasso.fit(X_train,y_train)
# and then make the prediction
ypredictRidge = RegRidge.predict(X_test)
- Coeffs = np.array(RegRidge.coef_)
- BetaValues = pd.DataFrame(Coeffs)
- BetaValues.columns = ['beta']
- display(BetaValues)
+ ypredictLasso = RegLasso.predict(X_test)
+ # Compute the MSE and print it
+ MSERidgePredict[i] = MSE(y_test,ypredictRidge)
+ MSELassoPredict[i] = MSE(y_test,ypredictLasso)
+ print(lmb,RegRidge.coef_)
+ print(lmb,RegLasso.coef_)
+# Now plot the results
+plt.figure()
+plt.plot(np.log10(lambdas), MSERidgePredict, 'b', label = 'MSE Ridge Test')
+plt.plot(np.log10(lambdas), MSELassoPredict, 'r', label = 'MSE Lasso Test')
+plt.xlabel('log10(lambda)')
+plt.ylabel('MSE')
+plt.legend()
+plt.show()
How can we understand this?
@@ -395,7 +413,7 @@ lambdas = np.
Before we proceed, we need to rethink what we have been doing. In our
-eager to fit the data, we have omitted several important elements in
-our regression analysis. In what follows we will
+ Let us write out the values of the coefficients \( \beta_i \) as functions
+of the polynomial degree and noise. We will focus only on the Ridge
+results and some few selected values of the hyperparameter \( \lambda \).
and discuss how to select a given model (one of the difficult parts in machine learning). If we don't include any noise and run this code for different values
+of the polynomial degree, we notice that the results for \( \beta_i \) do
+not show great changes from one order to the next. This is an
+indication that for higher polynomial orders, our parameters become
+less important.
+ If we however add noise, what happens is that the polynomial fit is
+trying to adjust the fit to traverse in the best possible way all data
+points. This can lead to large fluctuations in the parameters
+\( \beta_i \) as functions of polynomial order. It will also be reflected
+in a larger value of the variance of each parameter \( \beta_i \). What
+Ridge regression (and Lasso as well) are doing then is to try to
+quench the fluctuations in the parameters of \( \beta_i \) which have a
+large variance (normally for higher orders in the polynomial).
+
@@ -322,7 +400,7 @@ our regression analysis. In what follows we will
Resampling methods are an indispensable tool in modern
-statistics. They involve repeatedly drawing samples from a training
-set and refitting a model of interest on each sample in order to
-obtain additional information about the fitted model. For example, in
-order to estimate the variability of a linear regression fit, we can
-repeatedly draw different samples from the training data, fit a linear
-regression to each new sample, and then examine the extent to which
-the resulting fits differ. Such an approach may allow us to obtain
-information that would not be available from fitting the model only
-once using the original training sample.
- Two resampling methods are often used in Machine Learning analyses, Before we proceed, we need to rethink what we have been doing. In our
+eager to fit the data, we have omitted several important elements in
+our regression analysis. In what follows we will
+ In addition there are several other methods such as the Jackknife and the Blocking methods. We will discuss in particular
-cross-validation and the bootstrap method.
- and discuss how to select a given model (one of the difficult parts in machine learning).
@@ -338,7 +327,7 @@ cross-validation and the bootstrap method.
Resampling methods are an indispensable tool in modern
+statistics. They involve repeatedly drawing samples from a training
+set and refitting a model of interest on each sample in order to
+obtain additional information about the fitted model. For example, in
+order to estimate the variability of a linear regression fit, we can
+repeatedly draw different samples from the training data, fit a linear
+regression to each new sample, and then examine the extent to which
+the resulting fits differ. Such an approach may allow us to obtain
+information that would not be available from fitting the model only
+once using the original training sample.
+ Resampling approaches can be computationally expensive, because they
-involve fitting the same statistical method multiple times using
-different subsets of the training data. However, due to recent
-advances in computing power, the computational requirements of
-resampling methods generally are not prohibitive. In this chapter, we
-discuss two of the most commonly used resampling methods,
-cross-validation and the bootstrap. Both methods are important tools
-in the practical application of many statistical learning
-procedures. For example, cross-validation can be used to estimate the
-test error associated with a given statistical learning method in
-order to evaluate its performance, or to select the appropriate level
-of flexibility. The process of evaluating a model’s performance is
-known as model assessment, whereas the process of selecting the proper
-level of flexibility for a model is known as model selection. The
-bootstrap is widely used.
+ Two resampling methods are often used in Machine Learning analyses, In addition there are several other methods such as the Jackknife and the Blocking methods. We will discuss in particular
+cross-validation and the bootstrap method.
Resampling approaches can be computationally expensive, because they
+involve fitting the same statistical method multiple times using
+different subsets of the training data. However, due to recent
+advances in computing power, the computational requirements of
+resampling methods generally are not prohibitive. In this chapter, we
+discuss two of the most commonly used resampling methods,
+cross-validation and the bootstrap. Both methods are important tools
+in the practical application of many statistical learning
+procedures. For example, cross-validation can be used to estimate the
+test error associated with a given statistical learning method in
+order to evaluate its performance, or to select the appropriate level
+of flexibility. The process of evaluating a model’s performance is
+known as model assessment, whereas the process of selecting the proper
+level of flexibility for a model is known as model selection. The
+bootstrap is widely used.
+
@@ -324,7 +340,7 @@ MathJax.Hub.Config({
With all these analytical equations for both the OLS and Ridge
-regression, we will now outline how to assess a given model. This will
-lead to a discussion of the so-called bias-variance tradeoff (see
-below) and so-called resampling methods.
- One of the quantities we have discussed as a way to measure errors is
-the mean-squared error (MSE), mainly used for fitting of continuous
-functions. Another choice is the absolute error.
- In the discussions below we will focus on the MSE and in particular since we will split the data into test and training data,
-we discuss the
- As our model becomes more and more complex, more of the training data tends to used. The training may thence adapt to more complicated structures in the data. This may lead to a decrease in the bias (see below for code example) and a slight increase of the variance for the test error.
-For a certain level of complexity the test error will reach minimum, before starting to increase again. The
-training error reaches a saturation.
-
@@ -335,7 +333,7 @@ training error reaches a saturation.
Bootstrapping is a non-parametric approach to statistical inference
-that substitutes computation for more traditional distributional
-assumptions and asymptotic results. Bootstrapping offers a number of
-advantages:
+ With all these analytical equations for both the OLS and Ridge
+regression, we will now outline how to assess a given model. This will
+lead to a discussion of the so-called bias-variance tradeoff (see
+below) and so-called resampling methods.
+ One of the quantities we have discussed as a way to measure errors is
+the mean-squared error (MSE), mainly used for fitting of continuous
+functions. Another choice is the absolute error.
+ In the discussions below we will focus on the MSE and in particular since we will split the data into test and training data,
+we discuss the
The textbook by Davison on the Bootstrap Methods and their Applications provides many more insights and proofs. In this course we will take a more practical approach and use the results and theorems provided in the literature. For those interested in reading more about the bootstrap methods, we recommend the above text and the one by Efron and Tibshirani. Before we proceed however, we need to remind ourselves about a central theorem in statistics, namely the so-called central limit theorem. As our model becomes more and more complex, more of the training data tends to used. The training may thence adapt to more complicated structures in the data. This may lead to a decrease in the bias (see below for code example) and a slight increase of the variance for the test error.
+For a certain level of complexity the test error will reach minimum, before starting to increase again. The
+training error reaches a saturation.
+
@@ -333,7 +340,7 @@ advantages:
Suppose we have a PDF \( p(x) \) from which we generate a series \( N \)
-of averages \( \mathbb{E}[x_i] \). Each mean value \( \mathbb{E}[x_i] \)
-is viewed as the average of a specific measurement, e.g., throwing
-dice 100 times and then taking the average value, or producing a certain
-amount of random numbers.
-For notational ease, we set \( \mathbb{E}[x_i]=x_i \) in the discussion
-which follows. We do the same for \( \mathbb{E}[z]=z \).
+ Bootstrapping is a non-parametric approach to statistical inference
+that substitutes computation for more traditional distributional
+assumptions and asymptotic results. Bootstrapping offers a number of
+advantages:
If we compute the mean \( z \) of \( m \) such mean values \( x_i \) the question we pose is which is the PDF of the new variable \( z \). The textbook by Davison on the Bootstrap Methods and their Applications provides many more insights and proofs. In this course we will take a more practical approach and use the results and theorems provided in the literature. For those interested in reading more about the bootstrap methods, we recommend the above text and the one by Efron and Tibshirani. Before we proceed however, we need to remind ourselves about a central theorem in statistics, namely the so-called central limit theorem.
@@ -328,7 +338,7 @@ $$
The probability of obtaining an average value \( z \) is the product of the
-probabilities of obtaining arbitrary individual mean values \( x_i \),
-but with the constraint that the average is \( z \). We can express this through
-the following expression
+ Suppose we have a PDF \( p(x) \) from which we generate a series \( N \)
+of averages \( \mathbb{E}[x_i] \). Each mean value \( \mathbb{E}[x_i] \)
+is viewed as the average of a specific measurement, e.g., throwing
+dice 100 times and then taking the average value, or producing a certain
+amount of random numbers.
+For notational ease, we set \( \mathbb{E}[x_i]=x_i \) in the discussion
+which follows. We do the same for \( \mathbb{E}[z]=z \).
If we compute the mean \( z \) of \( m \) such mean values \( x_i \) where the \( \delta \)-function enbodies the constraint that the mean is \( z \).
-All measurements that lead to each individual \( x_i \) are expected to
-be independent, which in turn means that we can express \( \tilde{p} \) as the
-product of individual \( p(x_i) \). The independence assumption is important in the derivation of the central limit theorem.
- the question we pose is which is the PDF of the new variable \( z \).
@@ -328,7 +333,7 @@ product of individual \( p(x_i) \). The independence assumption is important in
If we use the integral expression for the \( \delta \)-function and inserting \( e^{i\mu q-i\mu q} \) where \( \mu \) is the mean value
-we arrive at
+ The probability of obtaining an average value \( z \) is the product of the
+probabilities of obtaining arbitrary individual mean values \( x_i \),
+but with the constraint that the average is \( z \). We can express this through
+the following expression
with the integral over \( x \) resulting in where the \( \delta \)-function enbodies the constraint that the mean is \( z \).
+All measurements that lead to each individual \( x_i \) are expected to
+be independent, which in turn means that we can express \( \tilde{p} \) as the
+product of individual \( p(x_i) \). The independence assumption is important in the derivation of the central limit theorem.
+
@@ -337,7 +333,7 @@ $$
The second term on the rhs disappears since this is just the mean and
-employing the definition of \( \sigma^2 \) we have
+ If we use the integral expression for the \( \delta \)-function and inserting \( e^{i\mu q-i\mu q} \) where \( \mu \) is the mean value
+we arrive at
resulting in with the integral over \( x \) resulting in and in the limit \( m\rightarrow \infty \) we obtain which is the normal distribution with variance
-\( \sigma^2_m=\sigma^2/m \), where \( \sigma \) is the variance of the PDF \( p(x) \)
-and \( \mu \) is also the mean of the PDF \( p(x) \).
-
@@ -339,7 +342,7 @@ and \( \mu \) is also the mean of the PDF \( p(x) \).
Thus, the central limit theorem states that the PDF \( \tilde{p}(z) \) of
-the average of \( m \) random values corresponding to a PDF \( p(x) \)
-is a normal distribution whose mean is the
-mean value of the PDF \( p(x) \) and whose variance is the variance
-of the PDF \( p(x) \) divided by \( m \), the number of values used to compute \( z \).
- The central limit theorem leads to the well-known expression for the
-standard deviation, given by
- The latter is true only if the average value is known exactly. This is obtained in the limit
-\( m\rightarrow \infty \) only. Because the mean and the variance are measured quantities we obtain
-the familiar expression in statistics (the so-called Bessel correction)
+ The second term on the rhs disappears since this is just the mean and
+employing the definition of \( \sigma^2 \) we have
In many cases however the above estimate for the standard deviation,
-in particular if correlations are strong, may be too simplistic. Keep
-in mind that we have assumed that the variables \( x \) are independent
-and identically distributed. This is obviously not always the
-case. For example, the random numbers (or better pseudorandom numbers)
-we generate in various calculations do always exhibit some
-correlations.
- resulting in The theorem is satisfied by a large class of PDFs. Note however that for a
-finite \( m \), it is not always possible to find a closed form /analytic expression for
-\( \tilde{p}(x) \).
+$$
+ \left[\int_{-\infty}^{\infty}dxp(x)\exp{\left(iq(\mu-x)/m\right)}\right]^m\approx
+ \left[1-\frac{q^2\sigma^2}{2m^2}+\dots \right]^m,
+$$
+
+ and in the limit \( m\rightarrow \infty \) we obtain which is the normal distribution with variance
+\( \sigma^2_m=\sigma^2/m \), where \( \sigma \) is the variance of the PDF \( p(x) \)
+and \( \mu \) is also the mean of the PDF \( p(x) \).
@@ -351,7 +344,7 @@ finite \( m \), it is not always possible to find a closed form /analytic expres
Confidence intervals are used in statistics and represent a type of estimate
-computed from the observed data. This gives a range of values for an
-unknown parameter such as the parameters \( \boldsymbol{\beta} \) from linear regression.
+ Thus, the central limit theorem states that the PDF \( \tilde{p}(z) \) of
+the average of \( m \) random values corresponding to a PDF \( p(x) \)
+is a normal distribution whose mean is the
+mean value of the PDF \( p(x) \) and whose variance is the variance
+of the PDF \( p(x) \) divided by \( m \), the number of values used to compute \( z \).
With the OLS expressions for the parameters \( \boldsymbol{\beta} \) we found
-\( \mathbb{E}(\boldsymbol{\beta}) = \boldsymbol{\beta} \), which means that the estimator of the regression parameters is unbiased.
+ The central limit theorem leads to the well-known expression for the
+standard deviation, given by
We found also that the variance of the estimate of the \( j \)-th regression coefficient is
-\( \boldsymbol{\sigma}^2 (\boldsymbol{\beta}_j ) = \boldsymbol{\sigma}^2 [(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} \).
+$$
+ \sigma_m=
+\frac{\sigma}{\sqrt{m}}.
+$$
+
+ The latter is true only if the average value is known exactly. This is obtained in the limit
+\( m\rightarrow \infty \) only. Because the mean and the variance are measured quantities we obtain
+the familiar expression in statistics (the so-called Bessel correction)
+ In many cases however the above estimate for the standard deviation,
+in particular if correlations are strong, may be too simplistic. Keep
+in mind that we have assumed that the variables \( x \) are independent
+and identically distributed. This is obviously not always the
+case. For example, the random numbers (or better pseudorandom numbers)
+we generate in various calculations do always exhibit some
+correlations.
This quantity will be used to
-construct a confidence interval for the estimates.
+ The theorem is satisfied by a large class of PDFs. Note however that for a
+finite \( m \), it is not always possible to find a closed form /analytic expression for
+\( \tilde{p}(x) \).
@@ -329,7 +356,7 @@ construct a confidence interval for the estimates.
We will assume that the parameters \( \beta \) follow a normal
-distribution. We can then define the confidence interval. Here we will be using as
-shorthands \( \mu_{\beta} \) for the above mean value and \( \sigma_{\beta} \)
-for the standard deviation. We have then a confidence interval
+ Confidence intervals are used in statistics and represent a type of estimate
+computed from the observed data. This gives a range of values for an
+unknown parameter such as the parameters \( \boldsymbol{\beta} \) from linear regression.
where \( z \) defines the level of certainty (or confidence). For a normal
-distribution typical parameters are \( z=2.576 \) which corresponds to a
-confidence of \( 99\% \) while \( z=1.96 \) corresponds to a confidence of
-\( 95\% \). A confidence level of \( 95\% \) is commonly used and it is
-normally referred to as a two-sigmas confidence level, that is we
-approximate \( z\approx 2 \).
+ With the OLS expressions for the parameters \( \boldsymbol{\beta} \) we found
+\( \mathbb{E}(\boldsymbol{\beta}) = \boldsymbol{\beta} \), which means that the estimator of the regression parameters is unbiased.
For more discussions of confidence intervals (and in particular linked with a discussion of the bootstrap method), see chapter 5 of the textbook by Davison on the Bootstrap Methods and their Applications We found also that the variance of the estimate of the \( j \)-th regression coefficient is
+\( \boldsymbol{\sigma}^2 (\boldsymbol{\beta}_j ) = \boldsymbol{\sigma}^2 [(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} \).
+ In this text you will also find an in-depth discussion of the
-Bootstrap method, why it works and various theorems related to it.
+ This quantity will be used to
+construct a confidence interval for the estimates.
@@ -336,7 +334,7 @@ Bootstrap method, why it works and various theorems related to it.
Since \( \widehat{\beta} = \widehat{\beta}(\boldsymbol{X}) \) is a function of random variables,
-\( \widehat{\beta} \) itself must be a random variable. Thus it has
-a pdf, call this function \( p(\boldsymbol{t}) \). The aim of the bootstrap is to
-estimate \( p(\boldsymbol{t}) \) by the relative frequency of
-\( \widehat{\beta} \). You can think of this as using a histogram
-in the place of \( p(\boldsymbol{t}) \). If the relative frequency closely
-resembles \( p(\vec{t}) \), then using numerics, it is straight forward to
-estimate all the interesting parameters of \( p(\boldsymbol{t}) \) using point
-estimators.
+ We will assume that the parameters \( \beta \) follow a normal
+distribution. We can then define the confidence interval. Here we will be using as
+shorthands \( \mu_{\beta} \) for the above mean value and \( \sigma_{\beta} \)
+for the standard deviation. We have then a confidence interval
+ where \( z \) defines the level of certainty (or confidence). For a normal
+distribution typical parameters are \( z=2.576 \) which corresponds to a
+confidence of \( 99\% \) while \( z=1.96 \) corresponds to a confidence of
+\( 95\% \). A confidence level of \( 95\% \) is commonly used and it is
+normally referred to as a two-sigmas confidence level, that is we
+approximate \( z\approx 2 \).
+ For more discussions of confidence intervals (and in particular linked with a discussion of the bootstrap method), see chapter 5 of the textbook by Davison on the Bootstrap Methods and their Applications In this text you will also find an in-depth discussion of the
+Bootstrap method, why it works and various theorems related to it.
@@ -323,7 +341,7 @@ estimators.
In the case that \( \widehat{\beta} \) has
-more than one component, and the components are independent, we use the
-same estimator on each component separately. If the probability
-density function of \( X_i \), \( p(x) \), had been known, then it would have
-been straightforward to do this by:
- By repeated use of the above two points, many
-estimates of \( \widehat{\beta} \) can be obtained. The
-idea is to use the relative frequency of \( \widehat{\beta}^* \)
-(think of a histogram) as an estimate of \( p(\boldsymbol{t}) \).
+ Since \( \widehat{\beta} = \widehat{\beta}(\boldsymbol{X}) \) is a function of random variables,
+\( \widehat{\beta} \) itself must be a random variable. Thus it has
+a pdf, call this function \( p(\boldsymbol{t}) \). The aim of the bootstrap is to
+estimate \( p(\boldsymbol{t}) \) by the relative frequency of
+\( \widehat{\beta} \). You can think of this as using a histogram
+in the place of \( p(\boldsymbol{t}) \). If the relative frequency closely
+resembles \( p(\vec{t}) \), then using numerics, it is straight forward to
+estimate all the interesting parameters of \( p(\boldsymbol{t}) \) using point
+estimators.
@@ -328,7 +328,7 @@ idea is to use the relative frequency of \( \widehat{\beta}^* \)
But
-unless there is enough information available about the process that
-generated \( X_1,X_2,\cdots,X_n \), \( p(x) \) is in general
-unknown. Therefore, Efron in 1979 asked the
-question: What if we replace \( p(x) \) by the relative frequency
-of the observation \( X_i \)?
+ In the case that \( \widehat{\beta} \) has
+more than one component, and the components are independent, we use the
+same estimator on each component separately. If the probability
+density function of \( X_i \), \( p(x) \), had been known, then it would have
+been straightforward to do this by:
If we draw observations in accordance with
-the relative frequency of the observations, will we obtain the same
-result in some asymptotic sense? The answer is yes.
+ By repeated use of the above two points, many
+estimates of \( \widehat{\beta} \) can be obtained. The
+idea is to use the relative frequency of \( \widehat{\beta}^* \)
+(think of a histogram) as an estimate of \( p(\boldsymbol{t}) \).
@@ -325,7 +333,7 @@ result in some asymptotic sense? The answer is yes.
The independent bootstrap works like this: But
+unless there is enough information available about the process that
+generated \( X_1,X_2,\cdots,X_n \), \( p(x) \) is in general
+unknown. Therefore, Efron in 1979 asked the
+question: What if we replace \( p(x) \) by the relative frequency
+of the observation \( X_i \)?
+ When you are done, you can draw a histogram of the relative frequency
-of \( \widehat \beta^* \). This is your estimate of the probability
-distribution \( p(t) \). Using this probability distribution you can
-estimate any statistics thereof. In principle you never draw the
-histogram of the relative frequency of \( \widehat{\beta}^* \). Instead
-you use the estimators corresponding to the statistic of interest. For
-example, if you are interested in estimating the variance of \( \widehat
-\beta \), apply the etsimator \( \widehat \sigma^2 \) to the values
-\( \widehat \beta^* \).
+ If we draw observations in accordance with
+the relative frequency of the observations, will we obtain the same
+result in some asymptotic sense? The answer is yes.
@@ -331,7 +330,7 @@ example, if you are interested in estimating the variance of \( \widehat
The following code starts with a Gaussian distribution with mean value
-\( \mu =100 \) and variance \( \sigma=15 \). We use this to generate the data
-used in the bootstrap analysis. The bootstrap analysis returns a data
-set after a given number of bootstrap operations (as many as we have
-data points). This data set consists of estimated mean values for each
-bootstrap operation. The histogram generated by the bootstrap method
-shows that the distribution for these mean values is also a Gaussian,
-centered around the mean value \( \mu=100 \) but with standard deviation
-\( \sigma/\sqrt{n} \), where \( n \) is the number of bootstrap samples (in
-this case the same as the number of original data points). The value
-of the standard deviation is what we expect from the central limit
-theorem.
+ The independent bootstrap works like this: When you are done, you can draw a histogram of the relative frequency
+of \( \widehat \beta^* \). This is your estimate of the probability
+distribution \( p(t) \). Using this probability distribution you can
+estimate any statistics thereof. In principle you never draw the
+histogram of the relative frequency of \( \widehat{\beta}^* \). Instead
+you use the estimators corresponding to the statistic of interest. For
+example, if you are interested in estimating the variance of \( \widehat
+\beta \), apply the etsimator \( \widehat \sigma^2 \) to the values
+\( \widehat \beta^* \).
We see that our new variance and from that the standard deviation, agrees with the central limit theorem.
The following code starts with a Gaussian distribution with mean value
+\( \mu =100 \) and variance \( \sigma=15 \). We use this to generate the data
+used in the bootstrap analysis. The bootstrap analysis returns a data
+set after a given number of bootstrap operations (as many as we have
+data points). This data set consists of estimated mean values for each
+bootstrap operation. The histogram generated by the bootstrap method
+shows that the distribution for these mean values is also a Gaussian,
+centered around the mean value \( \mu=100 \) but with standard deviation
+\( \sigma/\sqrt{n} \), where \( n \) is the number of bootstrap samples (in
+this case the same as the number of original data points). The value
+of the standard deviation is what we expect from the central limit
+theorem.
+ We see that our new variance and from that the standard deviation, agrees with the central limit theorem.
@@ -343,7 +381,7 @@ plt.show()
We will discuss the bias-variance tradeoff in the context of
-continuous predictions such as regression. However, many of the
-intuitions and ideas discussed here also carry over to classification
-tasks. Consider a dataset \( \mathcal{D} \) consisting of the data
-\( \mathbf{X}_\mathcal{D}=\{(y_j, \boldsymbol{x}_j), j=0\ldots n-1\} \).
- Let us assume that the true data is generated from a noisy model where \( \epsilon \) is normally distributed with mean zero and standard deviation \( \sigma^2 \). In our derivation of the ordinary least squares method we defined then
-an approximation to the function \( f \) in terms of the parameters
-\( \boldsymbol{\beta} \) and the design matrix \( \boldsymbol{X} \) which embody our model,
-that is \( \boldsymbol{\tilde{y}}=\boldsymbol{X}\boldsymbol{\beta} \).
- Thereafter we found the parameters \( \boldsymbol{\beta} \) by optimizing the means squared error via the so-called cost function We can rewrite this as The three terms represent the square of the bias of the learning
-method, which can be thought of as the error caused by the simplifying
-assumptions built into the method. The second term represents the
-variance of the chosen model and finally the last terms is variance of
-the error \( \boldsymbol{\epsilon} \).
- To derive this equation, we need to recall that the variance of \( \boldsymbol{y} \) and \( \boldsymbol{\epsilon} \) are both equal to \( \sigma^2 \). The mean value of \( \boldsymbol{\epsilon} \) is by definition equal to zero. Furthermore, the function \( f \) is not a stochastics variable, idem for \( \boldsymbol{\tilde{y}} \).
-We use a more compact notation in terms of the expectation value
- and adding and subtracting \( \mathbb{E}\left[\boldsymbol{\tilde{y}}\right] \) we get which, using the abovementioned expectation values can be rewritten as that is the rewriting in terms of the so-called bias, the variance of the model \( \boldsymbol{\tilde{y}} \) and the variance of \( \boldsymbol{\epsilon} \).
@@ -369,7 +348,7 @@ $$
We will discuss the bias-variance tradeoff in the context of
+continuous predictions such as regression. However, many of the
+intuitions and ideas discussed here also carry over to classification
+tasks. Consider a dataset \( \mathcal{D} \) consisting of the data
+\( \mathbf{X}_\mathcal{D}=\{(y_j, \boldsymbol{x}_j), j=0\ldots n-1\} \).
+ Let us assume that the true data is generated from a noisy model where \( \epsilon \) is normally distributed with mean zero and standard deviation \( \sigma^2 \). In our derivation of the ordinary least squares method we defined then
+an approximation to the function \( f \) in terms of the parameters
+\( \boldsymbol{\beta} \) and the design matrix \( \boldsymbol{X} \) which embody our model,
+that is \( \boldsymbol{\tilde{y}}=\boldsymbol{X}\boldsymbol{\beta} \).
+ Thereafter we found the parameters \( \boldsymbol{\beta} \) by optimizing the means squared error via the so-called cost function We can rewrite this as The three terms represent the square of the bias of the learning
+method, which can be thought of as the error caused by the simplifying
+assumptions built into the method. The second term represents the
+variance of the chosen model and finally the last terms is variance of
+the error \( \boldsymbol{\epsilon} \).
+ To derive this equation, we need to recall that the variance of \( \boldsymbol{y} \) and \( \boldsymbol{\epsilon} \) are both equal to \( \sigma^2 \). The mean value of \( \boldsymbol{\epsilon} \) is by definition equal to zero. Furthermore, the function \( f \) is not a stochastics variable, idem for \( \boldsymbol{\tilde{y}} \).
+We use a more compact notation in terms of the expectation value
+ and adding and subtracting \( \mathbb{E}\left[\boldsymbol{\tilde{y}}\right] \) we get which, using the abovementioned expectation values can be rewritten as that is the rewriting in terms of the so-called bias, the variance of the model \( \boldsymbol{\tilde{y}} \) and the variance of \( \boldsymbol{\epsilon} \).
@@ -317,6 +373,8 @@ MathJax.Hub.Config({
@@ -386,6 +321,7 @@ plt.show()
The bias-variance tradeoff summarizes the fundamental tension in
-machine learning, particularly supervised learning, between the
-complexity of a model and the amount of training data needed to train
-it. Since data is often limited, in practice it is often useful to
-use a less-complex model with higher bias, that is a model whose asymptotic
-performance is worse than another model because it is easier to
-train and less sensitive to sampling noise arising from having a
-finite-sized training dataset (smaller variance).
- The above equations tell us that in
-order to minimize the expected test error, we need to select a
-statistical learning method that simultaneously achieves low variance
-and low bias. Note that variance is inherently a nonnegative quantity,
-and squared bias is also nonnegative. Hence, we see that the expected
-test MSE can never lie below \( Var(\epsilon) \), the irreducible error.
- What do we mean by the variance and bias of a statistical learning
-method? The variance refers to the amount by which our model would change if we
-estimated it using a different training data set. Since the training
-data are used to fit the statistical learning method, different
-training data sets will result in a different estimate. But ideally the
-estimate for our model should not vary too much between training
-sets. However, if a method has high variance then small changes in
-the training data can result in large changes in the model. In general, more
-flexible statistical methods have higher variance.
- You may also find this recent article of interest.
@@ -339,6 +381,7 @@ flexible statistical methods have higher variance.
The bias-variance tradeoff summarizes the fundamental tension in
+machine learning, particularly supervised learning, between the
+complexity of a model and the amount of training data needed to train
+it. Since data is often limited, in practice it is often useful to
+use a less-complex model with higher bias, that is a model whose asymptotic
+performance is worse than another model because it is easier to
+train and less sensitive to sampling noise arising from having a
+finite-sized training dataset (smaller variance).
+ The above equations tell us that in
+order to minimize the expected test error, we need to select a
+statistical learning method that simultaneously achieves low variance
+and low bias. Note that variance is inherently a nonnegative quantity,
+and squared bias is also nonnegative. Hence, we see that the expected
+test MSE can never lie below \( Var(\epsilon) \), the irreducible error.
+ What do we mean by the variance and bias of a statistical learning
+method? The variance refers to the amount by which our model would change if we
+estimated it using a different training data set. Since the training
+data are used to fit the statistical learning method, different
+training data sets will result in a different estimate. But ideally the
+estimate for our model should not vary too much between training
+sets. However, if a method has high variance then small changes in
+the training data can result in large changes in the model. In general, more
+flexible statistical methods have higher variance.
+ You may also find this recent article of interest.
@@ -400,6 +343,7 @@ plt.show()
When the repetitive splitting of the data set is done randomly,
-samples may accidently end up in a fast majority of the splits in
-either training or test set. Such samples may have an unbalanced
-influence on either model building or prediction evaluation. To avoid
-this \( k \)-fold cross-validation structures the data splitting. The
-samples are divided into \( k \) more or less equally sized exhaustive and
-mutually exclusive subsets. In turn (at each split) one of these
-subsets plays the role of the test set while the union of the
-remaining subsets constitutes the training set. Such a splitting
-warrants a balanced representation of each sample in both training and
-test set over the splits. Still the division into the \( k \) subsets
-involves a degree of randomness. This may be fully excluded when
-choosing \( k=n \). This particular case is referred to as leave-one-out
-cross-validation (LOOCV).
-
@@ -322,6 +404,7 @@ cross-validation (LOOCV).
For more discussions of Ridge regression and calculation of averages, Wessel van Wieringen's article is highly recommended.
Why resampling methods
+Rerunning the above code
-
-
-import numpy as np
+import pandas as pd
+from IPython.display import display
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+from sklearn import linear_model
+
+# Make data set.
+n = 1000
+x = np.random.rand(n)
+y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.randn(n)
+
+Maxpolydegree = 5
+X = np.zeros((len(x),Maxpolydegree))
+X[:,0] = 1.0
+
+for polydegree in range(1, Maxpolydegree+1):
+ for degree in range(polydegree):
+ X[:,degree] = x**(degree)
+
+
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
+
+# Decide which values of lambda to use
+nlambdas = 5
+lambdas = np.logspace(-3, 2, nlambdas)
+for i in range(nlambdas):
+ lmb = lambdas[i]
+ # Make the fit using Ridge only
+ RegRidge = linear_model.Ridge(lmb,fit_intercept=False)
+ RegRidge.fit(X_train,y_train)
+ # and then make the prediction
+ ypredictRidge = RegRidge.predict(X_test)
+ Coeffs = np.array(RegRidge.coef_)
+ BetaValues = pd.DataFrame(Coeffs)
+ BetaValues.columns = ['beta']
+ display(BetaValues)
+
+Resampling methods
-Why resampling methods
-
-
-Resampling approaches can be computationally expensive
+Resampling methods
+
+Why resampling methods ?
+Resampling approaches can be computationally expensive
-
+Statistical analysis
+Why resampling methods ?
-
-
-Resampling methods
+Statistical analysis
+
-
-
+
+
+
+Resampling methods: Bootstrap
-Resampling methods
+
+
-
-The Central Limit Theorem
-
-Resampling methods: Bootstrap
+
+
+Finding the Limit
+The Central Limit Theorem
-Rewriting the \( \delta \)-function
+Finding the Limit
-Identifying Terms
+Rewriting the \( \delta \)-function
-Wrapping it up
+Identifying Terms
-Confidence Intervals
+Wrapping it up
-Standard Approach based on the Normal Distribution
+Confidence Intervals
-Resampling methods: Bootstrap background
+Standard Approach based on the Normal Distribution
-Resampling methods: More Bootstrap background
+Resampling methods: Bootstrap background
-
-
-Resampling methods: Bootstrap approach
+Resampling methods: More Bootstrap background
-
+
+Resampling methods: Bootstrap steps
+Resampling methods: Bootstrap approach
-
-
-Code example for the Bootstrap method
+Resampling methods: Bootstrap steps
-
+
+import numpy as np
-from time import time
-from scipy.stats import norm
-import matplotlib.pyplot as plt
-
-# Returns mean of bootstrap samples
-# Bootstrap algorithm
-def bootstrap(data, datapoints):
- t = np.zeros(datapoints)
- n = len(data)
- # non-parametric bootstrap
- for i in range(datapoints):
- t[i] = np.mean(data[np.random.randint(0,n,n)])
- # analysis
- print("Bootstrap Statistics :")
- print("original bias std. error")
- print("%8g %8g %14g %15g" % (np.mean(data), np.std(data),np.mean(t),np.std(t)))
- return t
-
-# We set the mean value to 100 and the standard deviation to 15
-mu, sigma = 100, 15
-datapoints = 10000
-# We generate random numbers according to the normal distribution
-x = mu + sigma*np.random.randn(datapoints)
-# bootstrap returns the data sample
-t = bootstrap(x, datapoints)
-
-Plotting the Histogram
+Code example for the Bootstrap method
+
+# the histogram of the bootstrapped data (normalized data if density = True)
-n, binsboot, patches = plt.hist(t, 50, density=True, facecolor='red', alpha=0.75)
-# add a 'best fit' line
-y = norm.pdf(binsboot, np.mean(t), np.std(t))
-lt = plt.plot(binsboot, y, 'b', linewidth=1)
-plt.xlabel('x')
-plt.ylabel('Probability')
-plt.grid(True)
-plt.show()
+
import numpy as np
+from time import time
+from scipy.stats import norm
+import matplotlib.pyplot as plt
+
+# Returns mean of bootstrap samples
+# Bootstrap algorithm
+def bootstrap(data, datapoints):
+ t = np.zeros(datapoints)
+ n = len(data)
+ # non-parametric bootstrap
+ for i in range(datapoints):
+ t[i] = np.mean(data[np.random.randint(0,n,n)])
+ # analysis
+ print("Bootstrap Statistics :")
+ print("original bias std. error")
+ print("%8g %8g %14g %15g" % (np.mean(data), np.std(data),np.mean(t),np.std(t)))
+ return t
+
+# We set the mean value to 100 and the standard deviation to 15
+mu, sigma = 100, 15
+datapoints = 10000
+# We generate random numbers according to the normal distribution
+x = mu + sigma*np.random.randn(datapoints)
+# bootstrap returns the data sample
+t = bootstrap(x, datapoints)
The bias-variance tradeoff
+Plotting the Histogram
-# the histogram of the bootstrapped data (normalized data if density = True)
+n, binsboot, patches = plt.hist(t, 50, density=True, facecolor='red', alpha=0.75)
+# add a 'best fit' line
+y = norm.pdf(binsboot, np.mean(t), np.std(t))
+lt = plt.plot(binsboot, y, 'b', linewidth=1)
+plt.xlabel('x')
+plt.ylabel('Probability')
+plt.grid(True)
+plt.show()
+
+A way to Read the Bias-Variance Tradeoff
+The bias-variance tradeoff
-
-
+Example code for Bias-Variance tradeoff
-
-
-import matplotlib.pyplot as plt
-import numpy as np
-from sklearn.linear_model import LinearRegression, Ridge, Lasso
-from sklearn.preprocessing import PolynomialFeatures
-from sklearn.model_selection import train_test_split
-from sklearn.pipeline import make_pipeline
-from sklearn.utils import resample
-
-np.random.seed(2018)
-
-n = 500
-n_boostraps = 100
-degree = 18 # A quite high value, just to show.
-noise = 0.1
-
-# Make data set.
-x = np.linspace(-1, 3, n).reshape(-1, 1)
-y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2) + np.random.normal(0, 0.1, x.shape)
-
-# Hold out some test data that is never used in training.
-x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)
-
-# Combine x transformation and model into one operation.
-# Not neccesary, but convenient.
-model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
-
-# The following (m x n_bootstraps) matrix holds the column vectors y_pred
-# for each bootstrap iteration.
-y_pred = np.empty((y_test.shape[0], n_boostraps))
-for i in range(n_boostraps):
- x_, y_ = resample(x_train, y_train)
-
- # Evaluate the new model on the same test data each time.
- y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel()
-
-# Note: Expectations and variances taken w.r.t. different training
-# data sets, hence the axis=1. Subsequent means are taken across the test data
-# set in order to obtain a total value, but before this we have error/bias/variance
-# calculated per data point in the test set.
-# Note 2: The use of keepdims=True is important in the calculation of bias as this
-# maintains the column vector form. Dropping this yields very unexpected results.
-error = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
-bias = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
-variance = np.mean( np.var(y_pred, axis=1, keepdims=True) )
-print('Error:', error)
-print('Bias^2:', bias)
-print('Var:', variance)
-print('{} >= {} + {} = {}'.format(error, bias, variance, bias+variance))
-
-plt.plot(x[::5, :], y[::5, :], label='f(x)')
-plt.scatter(x_test, y_test, label='Data points')
-plt.scatter(x_test, np.mean(y_pred, axis=1), label='Pred')
-plt.legend()
-plt.show()
-
-A way to Read the Bias-Variance Tradeoff
+
+
Understanding what happens
+Example code for Bias-Variance tradeoff
Summing up
+
+Understanding what happens
-import matplotlib.pyplot as plt
+import numpy as np
+from sklearn.linear_model import LinearRegression, Ridge, Lasso
+from sklearn.preprocessing import PolynomialFeatures
+from sklearn.model_selection import train_test_split
+from sklearn.pipeline import make_pipeline
+from sklearn.utils import resample
-
+Another Example from Scikit-Learn's Repository
+
+Summing up
-
-"""
-============================
-Underfitting vs. Overfitting
-============================
+
-Various steps in cross-validation
+
+Another Example from Scikit-Learn's Repository
+
+
+"""
+============================
+Underfitting vs. Overfitting
+============================
+
+This example demonstrates the problems of underfitting and overfitting and
+how we can use linear regression with polynomial features to approximate
+nonlinear functions. The plot shows the function that we want to approximate,
+which is a part of the cosine function. In addition, the samples from the
+real function and the approximations of different models are displayed. The
+models have polynomial features of different degrees. We can see that a
+linear function (polynomial with degree 1) is not sufficient to fit the
+training samples. This is called **underfitting**. A polynomial of degree 4
+approximates the true function almost perfectly. However, for higher degrees
+the model will **overfit** the training data, i.e. it learns the noise of the
+training data.
+We evaluate quantitatively **overfitting** / **underfitting** by using
+cross-validation. We calculate the mean squared error (MSE) on the validation
+set, the higher, the less likely the model generalizes correctly from the
+training data.
+"""
+
+print(__doc__)
+
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.pipeline import Pipeline
+from sklearn.preprocessing import PolynomialFeatures
+from sklearn.linear_model import LinearRegression
+from sklearn.model_selection import cross_val_score
+
+
+def true_fun(X):
+ return np.cos(1.5 * np.pi * X)
+
+np.random.seed(0)
+
+n_samples = 30
+degrees = [1, 4, 15]
+
+X = np.sort(np.random.rand(n_samples))
+y = true_fun(X) + np.random.randn(n_samples) * 0.1
+
+plt.figure(figsize=(14, 5))
+for i in range(len(degrees)):
+ ax = plt.subplot(1, len(degrees), i + 1)
+ plt.setp(ax, xticks=(), yticks=())
+
+ polynomial_features = PolynomialFeatures(degree=degrees[i],
+ include_bias=False)
+ linear_regression = LinearRegression()
+ pipeline = Pipeline([("polynomial_features", polynomial_features),
+ ("linear_regression", linear_regression)])
+ pipeline.fit(X[:, np.newaxis], y)
+
+ # Evaluate the models using crossvalidation
+ scores = cross_val_score(pipeline, X[:, np.newaxis], y,
+ scoring="neg_mean_squared_error", cv=10)
+
+ X_test = np.linspace(0, 1, 100)
+ plt.plot(X_test, pipeline.predict(X_test[:, np.newaxis]), label="Model")
+ plt.plot(X_test, true_fun(X_test), label="True function")
+ plt.scatter(X, y, edgecolor='b', s=20, label="Samples")
+ plt.xlabel("x")
+ plt.ylabel("y")
+ plt.xlim((0, 1))
+ plt.ylim((-2, 2))
+ plt.legend(loc="best")
+ plt.title("Degree {}\nMSE = {:.2e}(+/- {:.2e})".format(
+ degrees[i], -scores.mean(), scores.std()))
+plt.show()
+
+
Sep 10, 2023
+Sep 11, 2023
+
+
@@ -328,7 +335,7 @@ MathJax.Hub.Config({
Sep 10, 2023
+Sep 11, 2023
+
+
-
+Material from last week and relevant for the weekly exercises
+Linking the regression analysis with a statistical interpretation
@@ -432,6 +443,8 @@ $$
matrix product is non-negative definite.
This means the variance we obtain with the standard OLS will always for \( \lambda > 0 \) be larger than the variance of \( \boldsymbol{\beta} \) obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below.
+
+
+
+
Plans for week 37
@@ -266,6 +272,7 @@ MathJax.Hub.Config({
+Material from last week and relevant for the weekly exercises
+
Linking the regression analysis with a statistical interpretation
@@ -461,6 +473,8 @@ matrix product is non-negative definite.
This means the variance we obtain with the standard OLS will always for \( \lambda > 0 \) be larger than the variance of \( \boldsymbol{\beta} \) obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below.
For more discussions of Ridge regression and calculation of averages, Wessel van Wieringen's article is highly recommended.
+
For more discussions of Ridge regression and calculation of averages, Wessel van Wieringen's article is highly recommended.
+