FYS-STK4155/doc/LectureNotes/_build/jupyter_execute/week38.ipynb

Week 38: Statistical analysis, bias-variance tradeoff and resampling methods

Morten Hjorth-Jensen, Department of Physics and Center for Computing in Science Education, University of Oslo, Norway

Date: September 15-19, 2025

Plans for week 38, lecture Monday September 15

Material for the lecture on Monday September 15.

1. Statistical interpretation of OLS and various expectation values

2. Resampling techniques, Bootstrap and cross validation and bias-variance tradeoff

3. The material we did not cover last week, that is on more advanced methods for updating the learning rate, are covered by its own video. We will briefly discuss these topics at the beginning of the lecture and during the lab sessions. See video on ADAgrad, RMSprop and ADAM (material from last week not covered during lecture) at https://youtu.be/J_41Hld6tTU

4. Video of Lecture

5. Whiteboard notes

Readings and Videos

1. Raschka et al, pages 175-192

2. Hastie et al Chapter 7, here we recommend 7.1-7.5 and 7.10 (cross-validation) and 7.11 (bootstrap). See https://link.springer.com/book/10.1007/978-0-387-84858-7.

3. Video on bias-variance tradeoff

4. Video on Bootstrapping

5. Video on cross validation

For the lab session, the following video on cross validation (from 2024), could be helpful, see https://www.youtube.com/watch?v=T9jjWsmsd1o

Linking the regression analysis with a statistical interpretation

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 \theta.

On of the advantages 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{\theta} 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.

Assumptions made

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

\boldsymbol{y} = f(\boldsymbol{x})+\boldsymbol{\varepsilon}

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{\theta}.

Expectation value and variance

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{\theta}) + \mathbb{E}(\varepsilon_i)
\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \theta, 
\end{align*}

while its variance is

\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} \,
\theta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\theta})^2 \\ &
= \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \boldsymbol{\theta})^2 + 2 \varepsilon_i
\mathbf{X}_{i, \ast} \, \boldsymbol{\theta} + \varepsilon_i^2 ] - ( \mathbf{X}_{i,
\ast} \, \theta)^2 \\  & = ( \mathbf{X}_{i, \ast} \, \boldsymbol{\theta})^2 + 2
\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \boldsymbol{\theta} +
\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \boldsymbol{\theta})^2 
\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \,
\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2.  
\end{align*}

Hence, y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \boldsymbol{\theta}, \sigma^2), that is \boldsymbol{y} follows a normal distribution with mean value \boldsymbol{X}\boldsymbol{\theta} and variance \sigma^2 (not be confused with the singular values of the SVD).

Expectation value and variance for \boldsymbol{\theta}

With the OLS expressions for the optimal parameters \boldsymbol{\hat{\theta}} we can evaluate the expectation value

\mathbb{E}(\boldsymbol{\hat{\theta}}) = \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{\theta}=\boldsymbol{\theta}.

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{\theta}} is

\begin{eqnarray*}
\mbox{Var}(\boldsymbol{\hat{\theta}}) & = & \mathbb{E} \{ [\boldsymbol{\theta} - \mathbb{E}(\boldsymbol{\theta})] [\boldsymbol{\theta} - \mathbb{E}(\boldsymbol{\theta})]^{T} \}
\\
& = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\theta}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\theta}]^{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{\theta} \, \boldsymbol{\theta}^{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{\theta} \, \boldsymbol{\theta}^{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{\theta} \, \boldsymbol{\theta}^{T}
\\
& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \{ \mathbf{X} \, \boldsymbol{\theta} \, \boldsymbol{\theta}^{T} \,  \mathbf{X}^{T} + \sigma^2 \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\theta} \, \boldsymbol{\theta}^{T}
% \\
% & = & (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \boldsymbol{\theta} \, \boldsymbol{\theta}^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{\theta} \boldsymbol{\theta}^T
\\
& = & \boldsymbol{\theta} \, \boldsymbol{\theta}^{T}  + \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\theta} \, \boldsymbol{\theta}^{T}
\, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1},
\end{eqnarray*}

where we have used that $\mathbb{E} (\mathbf{Y} \mathbf{Y}^{T}) = \mathbf{X} , \boldsymbol{\theta} , \boldsymbol{\theta}^{T} , \mathbf{X}^{T} + \sigma^2 , \mathbf{I}_{nn}$. From $\mbox{Var}(\boldsymbol{\theta}) = \sigma^2 , (\mathbf{X}^{T} \mathbf{X})^{-1}$, one obtains an estimate of the variance of the estimate of the $j$-th regression coefficient: \boldsymbol{\sigma}^2 (\boldsymbol{\theta}_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{\theta} 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{\theta}^{\mathrm{Ridge}} \big]=(\mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X})\boldsymbol{\theta}^{\mathrm{OLS}}.

We see clearly that \mathbb{E} \big[ \boldsymbol{\theta}^{\mathrm{Ridge}} \big] \not= \boldsymbol{\theta}^{\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{\theta}^{\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{\theta} goes to zero.

With this, we can compute the difference

\mbox{Var}[\boldsymbol{\theta}^{\mathrm{OLS}}]-\mbox{Var}(\boldsymbol{\theta}^{\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{\theta} obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below.

Deriving OLS from a probability distribution

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{\theta}} 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{\theta}}. This means that a single output y_i is given by the Gaussian distribution

y_i\sim \mathcal{N}(\boldsymbol{X}_{i,*}\boldsymbol{\theta}, \sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\theta})^2}{2\sigma^2}\right]}.

Independent and Identically Distributed (iid)

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{\theta})=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\theta})^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{\theta}.

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{\theta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\theta})^2}{2\sigma^2}\right]}=\prod_{i=0}^{n-1}p(y_i,\boldsymbol{X}\vert\boldsymbol{\theta}).

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{\theta})=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\theta})^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{\theta}.

Maximum Likelihood Estimation (MLE)

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 will assume here that our events are given by the above Gaussian distribution and we will determine the optimal parameters \theta 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.

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.

Note also that maximization/minimization of the logarithm of the PDF is equivalent to the maximization/minimization of the function itself.

A new Cost Function

We could now define a new cost function to minimize, namely the negative logarithm of the above PDF

C(\boldsymbol{\theta})=-\log{\prod_{i=0}^{n-1}p(y_i,\boldsymbol{X}\vert\boldsymbol{\theta})}=-\sum_{i=0}^{n-1}\log{p(y_i,\boldsymbol{X}\vert\boldsymbol{\theta})},

which becomes

C(\boldsymbol{\theta})=\frac{n}{2}\log{2\pi\sigma^2}+\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\theta})\vert\vert_2^2}{2\sigma^2}.

Taking the derivative of the new cost function with respect to the parameters \theta we recognize our familiar OLS equation, namely

\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\theta}\right) =0,

which leads to the well-known OLS equation for the optimal paramters \theta

\hat{\boldsymbol{\theta}}^{\mathrm{OLS}}=\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}!

Next week we will make a similar analysis for Ridge and Lasso regression

Why resampling methods

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

1. look at statistical properties, including a discussion of mean values, variance and the so-called bias-variance tradeoff

2. introduce resampling techniques like cross-validation, bootstrapping and jackknife and more

and discuss how to select a given model (one of the difficult parts in machine learning).

Resampling methods

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,

1. The bootstrap method

2. and Cross-Validation

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

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.

Why resampling methods ?

Statistical analysis.

• Our simulations can be treated as computer experiments. This is particularly the case for Monte Carlo methods which are widely used in statistical analyses.

• The results can be analysed with the same statistical tools as we would use when analysing experimental data.

• As in all experiments, we are looking for expectation values and an estimate of how accurate they are, i.e., possible sources for errors.

Statistical analysis

• As in other experiments, many numerical experiments have two classes of errors:

  • Statistical errors

  • Systematical errors

• Statistical errors can be estimated using standard tools from statistics

• Systematical errors are method specific and must be treated differently from case to case.

Resampling methods

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

1. prediction error or simply the test error \mathrm{Err_{Test}}, where we have a fixed training set and the test error is the MSE arising from the data reserved for testing. We discuss also the

2. training error \mathrm{Err_{Train}}, which is the average loss over the training data.

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.

Resampling methods: Bootstrap

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:

1. The bootstrap is quite general, although there are some cases in which it fails.

2. Because it does not require distributional assumptions (such as normally distributed errors), the bootstrap can provide more accurate inferences when the data are not well behaved or when the sample size is small.

3. It is possible to apply the bootstrap to statistics with sampling distributions that are difficult to derive, even asymptotically.

4. It is relatively simple to apply the bootstrap to complex data-collection plans (such as stratified and clustered samples).

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.

The Central Limit Theorem

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

z=\frac{x_1+x_2+\dots+x_m}{m},

the question we pose is which is the PDF of the new variable z.

Finding the Limit

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

\tilde{p}(z)=\int dx_1p(x_1)\int dx_2p(x_2)\dots\int dx_mp(x_m)
    \delta(z-\frac{x_1+x_2+\dots+x_m}{m}),

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.

Rewriting the $\delta$-function

If we use the integral expression for the $\delta$-function

\delta(z-\frac{x_1+x_2+\dots+x_m}{m})=\frac{1}{2\pi}\int_{-\infty}^{\infty}
   dq\exp{\left(iq(z-\frac{x_1+x_2+\dots+x_m}{m})\right)},

and inserting e^{i\mu q-i\mu q} where \mu is the mean value we arrive at

\tilde{p}(z)=\frac{1}{2\pi}\int_{-\infty}^{\infty}
   dq\exp{\left(iq(z-\mu)\right)}\left[\int_{-\infty}^{\infty}
   dxp(x)\exp{\left(iq(\mu-x)/m\right)}\right]^m,

with the integral over x resulting in

\int_{-\infty}^{\infty}dxp(x)\exp{\left(iq(\mu-x)/m\right)}=
  \int_{-\infty}^{\infty}dxp(x)
   \left[1+\frac{iq(\mu-x)}{m}-\frac{q^2(\mu-x)^2}{2m^2}+\dots\right].

Identifying Terms

The second term on the rhs disappears since this is just the mean and employing the definition of \sigma^2 we have

\int_{-\infty}^{\infty}dxp(x)e^{\left(iq(\mu-x)/m\right)}=
  1-\frac{q^2\sigma^2}{2m^2}+\dots,

resulting in

\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

\tilde{p}(z)=\frac{1}{\sqrt{2\pi}(\sigma/\sqrt{m})}
    \exp{\left(-\frac{(z-\mu)^2}{2(\sigma/\sqrt{m})^2}\right)},

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).

Wrapping it up

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

\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)

\sigma_m\approx 
\frac{\sigma}{\sqrt{m-1}}.

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.

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).

Confidence Intervals

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{\theta} from linear regression.

With the OLS expressions for the parameters \boldsymbol{\theta} we found \mathbb{E}(\boldsymbol{\theta}) = \boldsymbol{\theta}, which means that the estimator of the regression parameters is unbiased.

In the exercises this week we show that the variance of the estimate of the $j$-th regression coefficient is \boldsymbol{\sigma}^2 (\boldsymbol{\theta}_j ) = \boldsymbol{\sigma}^2 [(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} .

This quantity can be used to construct a confidence interval for the estimates.

Standard Approach based on the Normal Distribution

We will assume that the parameters \theta follow a normal distribution. We can then define the confidence interval. Here we will be using as shorthands \mu_{\theta} for the above mean value and \sigma_{\theta} for the standard deviation. We have then a confidence interval

\left(\mu_{\theta}\pm \frac{z\sigma_{\theta}}{\sqrt{n}}\right),

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.

Resampling methods: Bootstrap background

Since \widehat{\theta} = \widehat{\theta}(\boldsymbol{X}) is a function of random variables, \widehat{\theta} 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{\theta}. 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.

Resampling methods: More Bootstrap background

In the case that \widehat{\theta} 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:

1. Drawing lots of numbers from p(x), suppose we call one such set of numbers (X_1^*, X_2^*, \cdots, X_n^*).

2. Then using these numbers, we could compute a replica of \widehat{\theta} called \widehat{\theta}^*.

By repeated use of the above two points, many estimates of \widehat{\theta} can be obtained. The idea is to use the relative frequency of \widehat{\theta}^* (think of a histogram) as an estimate of p(\boldsymbol{t}).

Resampling methods: Bootstrap approach

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?

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.

Resampling methods: Bootstrap steps

The independent bootstrap works like this:

1. Draw with replacement n numbers for the observed variables \boldsymbol{x} = (x_1,x_2,\cdots,x_n).

2. Define a vector \boldsymbol{x}^* containing the values which were drawn from \boldsymbol{x}.

3. Using the vector \boldsymbol{x}^* compute \widehat{\theta}^* by evaluating \widehat \theta under the observations \boldsymbol{x}^*.

4. Repeat this process k times.

When you are done, you can draw a histogram of the relative frequency of \widehat \theta^*. 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{\theta}^*. Instead you use the estimators corresponding to the statistic of interest. For example, if you are interested in estimating the variance of $\widehat \theta$, apply the etsimator \widehat \sigma^2 to the values \widehat \theta^*.

Code example for the Bootstrap method

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.

In [1]:

%matplotlib inline

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)

We see that our new variance and from that the standard deviation, agrees with the central limit theorem.

Plotting the Histogram

In [2]:

# 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()

The bias-variance tradeoff

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

\boldsymbol{y}=f(\boldsymbol{x}) + \boldsymbol{\epsilon}

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{\theta} and the design matrix \boldsymbol{X} which embody our model, that is \boldsymbol{\tilde{y}}=\boldsymbol{X}\boldsymbol{\theta}.

Thereafter we found the parameters \boldsymbol{\theta} by optimizing the means squared error via the so-called cost function

C(\boldsymbol{X},\boldsymbol{\theta}) =\frac{1}{n}\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2=\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right].

We can rewrite this as

\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\frac{1}{n}\sum_i(f_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2+\frac{1}{n}\sum_i(\tilde{y}_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2+\sigma^2.

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

\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{f}+\boldsymbol{\epsilon}-\boldsymbol{\tilde{y}})^2\right],

Warning:

Output truncated. This notebook contains too many cells to display efficiently.