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TITLE: Week 36: Resampling techniques and Ordinary Least Square
AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo & Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
DATE: today
!split
===== Plans for week 36 =====
* Thursday: Statistics, probability theory and resampling methods
* Friday: Resampling methods and motivation for Ridge Regression
!split
===== Thursday September 3 =====
"Video of Lecture":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h20/forelesningsvideoer/LectureSept3.mp4?vrtx=view-as-webpage" and "handwritten notes":"https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/NotesSeptember3.pdf"
More material will be added here, see handwritten notes also.
!split
===== 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
o look at statistical properties, including a discussion of mean values, variance and the so-called bias-variance tradeoff
o introduce resampling techniques like cross-validation, bootstrapping and jackknife and more
This will allow us to link the standard linear algebra methods we have discussed above to a statistical interpretation of the methods.
!split
===== Resampling methods =====
!bblock
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,
o The _bootstrap method_
o 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.
!eblock
!split
===== Resampling approaches can be computationally expensive =====
!bblock
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 models 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.
!eblock
!split
===== Why resampling methods ? =====
!bblock Statistical analysis
* Our simulations can be treated as *computer experiments*. This is particularly the case for Monte Carlo methods
* The results can be analysed with the same statistical tools as we would use 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.
!eblock
!split
===== Statistical analysis =====
!bblock
* 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.
!eblock
!split
===== Linking the regression analysis with a statistical interpretation =====
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.:
!bt
\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*}
!et
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 $\mathbf{X}_{i,\ast} \, \bm{\beta}$ 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 $\bm{X}$ 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 $p$.
!split
===== 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 $f(\bm{x})$ and a normal distributed error $\bm{\varepsilon}\sim \mathcal{N}(0, \sigma^2)$
which describe our data
!bt
\[
\bm{y} = f(\bm{x})+\bm{\varepsilon}
\]
!et
We approximate this function with our model from the solution of the linear regression equations, that is our
function $f$ is approximated by $\bm{\tilde{y}}$ where we want to minimize $(\bm{y}-\bm{\tilde{y}})^2$, our MSE, with
!bt
\[
\bm{\tilde{y}} = \bm{X}\bm{\beta}.
\]
!et
!split
===== Expectation value and variance =====
We can calculate the expectation value of $\bm{y}$ for a given element $i$
!bt
\begin{align*}
\mathbb{E}(y_i) & =
\mathbb{E}(\mathbf{X}_{i, \ast} \, \bm{\beta}) + \mathbb{E}(\varepsilon_i)
\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta,
\end{align*}
!et
while
its variance is
!bt
\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} \, \bm{\beta})^2 \\ &
= \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \bm{\beta})^2 + 2 \varepsilon_i
\mathbf{X}_{i, \ast} \, \bm{\beta} + \varepsilon_i^2 ] - ( \mathbf{X}_{i,
\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \bm{\beta})^2 + 2
\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \bm{\beta} +
\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \bm{\beta})^2
\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \,
\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2.
\end{align*}
!et
Hence, $y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \bm{\beta}, \sigma^2)$, that is $\bm{y}$ follows a normal distribution with
mean value $\bm{X}\bm{\beta}$ and variance $\sigma^2$ (not be confused with the singular values of the SVD).
!split
===== Expectation value and variance for $\bm{\beta}$ =====
With the OLS expressions for the parameters $\bm{\beta}$ we can evaluate the expectation value
!bt
\[
\mathbb{E}(\bm{\beta}) = \mathbb{E}[ (\mathbf{X}^{\top} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbb{E}[ \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1} \mathbf{X}^{T}\mathbf{X}\bm{\beta}=\bm{\beta}.
\]
!et
This means that the estimator of the regression parameters is unbiased.
We can also calculate the variance
The variance of $\bm{\beta}$ is
!bt
\begin{eqnarray*}
\mbox{Var}(\bm{\beta}) & = & \mathbb{E} \{ [\bm{\beta} - \mathbb{E}(\bm{\beta})] [\bm{\beta} - \mathbb{E}(\bm{\beta})]^{T} \}
\\
& = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \bm{\beta}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \bm{\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} \} - \bm{\beta} \, \bm{\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} \} - \bm{\beta} \, \bm{\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} - \bm{\beta} \, \bm{\beta}^{T}
\\
& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \{ \mathbf{X} \, \bm{\beta} \, \bm{\beta}^{T} \, \mathbf{X}^{T} + \sigma^2 \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \bm{\beta} \, \bm{\beta}^{T}
% \\
% & = & (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \bm{\beta} \, \bm{\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} - \bm{\beta} \bm{\beta}^T
\\
& = & \bm{\beta} \, \bm{\beta}^{T} + \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \bm{\beta} \, \bm{\beta}^{T}
\, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1},
\end{eqnarray*}
!et
where we have used that $\mathbb{E} (\mathbf{Y} \mathbf{Y}^{T}) =
\mathbf{X} \, \bm{\beta} \, \bm{\beta}^{T} \, \mathbf{X}^{T} +
\sigma^2 \, \mathbf{I}_{nn}$. From $\mbox{Var}(\bm{\beta}) = \sigma^2
\, (\mathbf{X}^{T} \mathbf{X})^{-1}$, one obtains an estimate of the
variance of the estimate of the $j$-th regression coefficient:
$\bm{\sigma}^2 (\bm{\beta}_j ) = \bm{\sigma}^2 \sqrt{
[(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} }$. 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 $\bm{\beta}$ and their variance
when we employ Ridge regression, allowing us again to define a confidence interval.
It is rather straightforward to show that
!bt
\[
\mathbb{E} \big[ \bm{\beta}^{\mathrm{Ridge}} \big]=(\mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X})\bm{\beta}^{\mathrm{OLS}}.
\]
!et
We see clearly that
$\mathbb{E} \big[ \bm{\beta}^{\mathrm{Ridge}} \big] \not= \bm{\beta}^{\mathrm{OLS}}$ for any $\lambda > 0$. We say then that the ridge estimator is biased.
We can also compute the variance as
!bt
\[
\mbox{Var}[\bm{\beta}^{\mathrm{Ridge}}]=\sigma^2[ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1} \mathbf{X}^{T} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T},
\]
!et
and it is easy to see that if the parameter $\lambda$ goes to infinity then the variance of Ridge parameters $\bm{\beta}$ goes to zero.
With this, we can compute the difference
!bt
\[
\mbox{Var}[\bm{\beta}^{\mathrm{OLS}}]-\mbox{Var}(\bm{\beta}^{\mathrm{Ridge}})=\sigma^2 [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}[ 2\lambda\mathbf{I} + \lambda^2 (\mathbf{X}^{T} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}.
\]
!et
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 $\bm{\beta}$ obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below.
!split
===== 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 us 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
o 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
o 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.
!split
===== Resampling methods: Jackknife and Bootstrap =====
Two famous
resampling methods are the _independent bootstrap_ and _the jackknife_.
The jackknife is a special case of the independent bootstrap. Still, the jackknife was made
popular prior to the independent bootstrap. And as the popularity of
the independent bootstrap soared, new variants, such as _the dependent bootstrap_.
The Jackknife and independent bootstrap work for
independent, identically distributed random variables.
If these conditions are not
satisfied, the methods will fail. Yet, it should be said that if the data are
independent, identically distributed, and we only want to estimate the
variance of $\overline{X}$ (which often is the case), then there is no
need for bootstrapping.
!split
===== Resampling methods: Jackknife =====
The Jackknife works by making many replicas of the estimator $\widehat{\theta}$.
The jackknife is a resampling method where we systematically leave out one observation from the vector of observed values $\bm{x} = (x_1,x_2,\cdots,X_n)$.
Let $\bm{x}_i$ denote the vector
!bt
\[
\bm{x}_i = (x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n),
\]
!et
which equals the vector $\bm{x}$ with the exception that observation
number $i$ is left out. Using this notation, define
$\widehat{\theta}_i$ to be the estimator
$\widehat{\theta}$ computed using $\vec{X}_i$.
!split
===== Jackknife code example =====
!bc pycod
from numpy import *
from numpy.random import randint, randn
from time import time
def jackknife(data, stat):
n = len(data);t = zeros(n); inds = arange(n); t0 = time()
## 'jackknifing' by leaving out an observation for each i
for i in range(n):
t[i] = stat(delete(data,i) )
# analysis
print("Runtime: %g sec" % (time()-t0)); print("Jackknife Statistics :")
print("original bias std. error")
print("%8g %14g %15g" % (stat(data),(n-1)*mean(t)/n, (n*var(t))**.5))
return t
# Returns mean of data samples
def stat(data):
return mean(data)
mu, sigma = 100, 15
datapoints = 10000
x = mu + sigma*random.randn(datapoints)
# jackknife returns the data sample
t = jackknife(x, stat)
!ec
!split
===== Resampling methods: Bootstrap =====
!bblock
Bootstrapping is a nonparametric approach to statistical inference
that substitutes computation for more traditional distributional
assumptions and asymptotic results. Bootstrapping offers a number of
advantages:
o The bootstrap is quite general, although there are some cases in which it fails.
o 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.
o It is possible to apply the bootstrap to statistics with sampling distributions that are difficult to derive, even asymptotically.
o It is relatively simple to apply the bootstrap to complex data-collection plans (such as stratified and clustered samples).
!eblock
!split
===== Resampling methods: Bootstrap background =====
Since $\widehat{\theta} = \widehat{\theta}(\bm{X})$ is a function of random variables,
$\widehat{\theta}$ itself must be a random variable. Thus it has
a pdf, call this function $p(\bm{t})$. The aim of the bootstrap is to
estimate $p(\bm{t})$ by the relative frequency of
$\widehat{\theta}$. You can think of this as using a histogram
in the place of $p(\bm{t})$. If the relative frequency closely
resembles $p(\vec{t})$, then using numerics, it is straight forward to
estimate all the interesting parameters of $p(\bm{t})$ using point
estimators.
!split
===== 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 straight forward to do this by:
o Drawing lots of numbers from $p(x)$, suppose we call one such set of numbers $(X_1^*, X_2^*, \cdots, X_n^*)$.
o Then using these numbers, we could compute a replica of $\widehat{\theta}$ called $\widehat{\theta}^*$.
By repeated use of (1) and (2), many
estimates of $\widehat{\theta}$ could have been obtained. The
idea is to use the relative frequency of $\widehat{\theta}^*$
(think of a histogram) as an estimate of $p(\bm{t})$.
!split
===== 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":"https://projecteuclid.org/euclid.aos/1176344552" 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.
Instead of generating the histogram for the relative
frequency of the observation $X_i$, just draw the values
$(X_1^*,X_2^*,\cdots,X_n^*)$ with replacement from the vector
$\bm{X}$.
!split
===== Resampling methods: Bootstrap steps =====
The independent bootstrap works like this:
o Draw with replacement $n$ numbers for the observed variables $\bm{x} = (x_1,x_2,\cdots,x_n)$.
o Define a vector $\bm{x}^*$ containing the values which were drawn from $\bm{x}$.
o Using the vector $\bm{x}^*$ compute $\widehat{\theta}^*$ by evaluating $\widehat \theta$ under the observations $\bm{x}^*$.
o 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
$\widehat \theta ^*$.
!split
===== 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.
!bc pycod
from numpy import *
from numpy.random import randint, randn
from time import time
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
# Returns mean of bootstrap samples
def stat(data):
return mean(data)
# Bootstrap algorithm
def bootstrap(data, statistic, R):
t = zeros(R); n = len(data); inds = arange(n); t0 = time()
# non-parametric bootstrap
for i in range(R):
t[i] = statistic(data[randint(0,n,n)])
# analysis
print("Runtime: %g sec" % (time()-t0)); print("Bootstrap Statistics :")
print("original bias std. error")
print("%8g %8g %14g %15g" % (statistic(data), std(data),mean(t),std(t)))
return t
mu, sigma = 100, 15
datapoints = 10000
x = mu + sigma*random.randn(datapoints)
# bootstrap returns the data sample
t = bootstrap(x, stat, datapoints)
# the histogram of the bootstrapped data
n, binsboot, patches = plt.hist(t, 50, normed=1, facecolor='red', alpha=0.75)
# add a 'best fit' line
y = mlab.normpdf( binsboot, mean(t), std(t))
lt = plt.plot(binsboot, y, 'r--', linewidth=1)
plt.xlabel('Smarts')
plt.ylabel('Probability')
plt.axis([99.5, 100.6, 0, 3.0])
plt.grid(True)
plt.show()
!ec
!split
===== Various steps in cross-validation =====
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).
!split
===== How to set up the cross-validation for Ridge and/or Lasso =====
* Define a range of interest for the penalty parameter.
* Divide the data set into training and test set comprising samples $\{1, \ldots, n\} \setminus i$ and $\{ i \}$, respectively.
* Fit the linear regression model by means of ridge estimation for each $\lambda$ in the grid using the training set, and the corresponding estimate of the error variance $\bm{\sigma}_{-i}^2(\lambda)$, as
!bt
\begin{align*}
\bm{\beta}_{-i}(\lambda) & = ( \bm{X}_{-i, \ast}^{T}
\bm{X}_{-i, \ast} + \lambda \bm{I}_{pp})^{-1}
\bm{X}_{-i, \ast}^{T} \bm{y}_{-i}
\end{align*}
!et
* Evaluate the prediction performance of these models on the test set by $\log\{L[y_i, \bm{X}_{i, \ast}; \bm{\beta}_{-i}(\lambda), \bm{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|y_i - \bm{X}_{i, \ast} \bm{\beta}_{-i}(\lambda)|$, the relative error, the error squared or the R2 score function.
* Repeat the first three steps such that each sample plays the role of the test set once.
* Average the prediction performances of the test sets at each grid point of the penalty bias/parameter. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as
!bt
\begin{align*}
\frac{1}{n} \sum_{i = 1}^n \log\{L[y_i, \mathbf{X}_{i, \ast}; \bm{\beta}_{-i}(\lambda), \bm{\sigma}_{-i}^2(\lambda)]\}.
\end{align*}
!et
!split
===== Cross-validation in brief =====
For the various values of $k$
o shuffle the dataset randomly.
o Split the dataset into $k$ groups.
o For each unique group:
o Decide which group to use as set for test data
o Take the remaining groups as a training data set
o Fit a model on the training set and evaluate it on the test set
o Retain the evaluation score and discard the model
o Summarize the model using the sample of model evaluation scores
!split
===== Code Example for Cross-validation and $k$-fold Cross-validation =====
The code here uses Ridge regression with cross-validation (CV) resampling and $k$-fold CV in order to fit a specific polynomial.
!bc pycod
import numpy as np
import matplotlib.pyplot as plt
from sklearn.model_selection import KFold
from sklearn.linear_model import Ridge
from sklearn.model_selection import cross_val_score
from sklearn.preprocessing import PolynomialFeatures
# A seed just to ensure that the random numbers are the same for every run.
# Useful for eventual debugging.
np.random.seed(3155)
# Generate the data.
nsamples = 100
x = np.random.randn(nsamples)
y = 3*x**2 + np.random.randn(nsamples)
## Cross-validation on Ridge regression using KFold only
# Decide degree on polynomial to fit
poly = PolynomialFeatures(degree = 6)
# Decide which values of lambda to use
nlambdas = 500
lambdas = np.logspace(-3, 5, nlambdas)
# Initialize a KFold instance
k = 5
kfold = KFold(n_splits = k)
# Perform the cross-validation to estimate MSE
scores_KFold = np.zeros((nlambdas, k))
i = 0
for lmb in lambdas:
ridge = Ridge(alpha = lmb)
j = 0
for train_inds, test_inds in kfold.split(x):
xtrain = x[train_inds]
ytrain = y[train_inds]
xtest = x[test_inds]
ytest = y[test_inds]
Xtrain = poly.fit_transform(xtrain[:, np.newaxis])
ridge.fit(Xtrain, ytrain[:, np.newaxis])
Xtest = poly.fit_transform(xtest[:, np.newaxis])
ypred = ridge.predict(Xtest)
scores_KFold[i,j] = np.sum((ypred - ytest[:, np.newaxis])**2)/np.size(ypred)
j += 1
i += 1
estimated_mse_KFold = np.mean(scores_KFold, axis = 1)
## Cross-validation using cross_val_score from sklearn along with KFold
# kfold is an instance initialized above as:
# kfold = KFold(n_splits = k)
estimated_mse_sklearn = np.zeros(nlambdas)
i = 0
for lmb in lambdas:
ridge = Ridge(alpha = lmb)
X = poly.fit_transform(x[:, np.newaxis])
estimated_mse_folds = cross_val_score(ridge, X, y[:, np.newaxis], scoring='neg_mean_squared_error', cv=kfold)
# cross_val_score return an array containing the estimated negative mse for every fold.
# we have to the the mean of every array in order to get an estimate of the mse of the model
estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)
i += 1
## Plot and compare the slightly different ways to perform cross-validation
plt.figure()
plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')
plt.plot(np.log10(lambdas), estimated_mse_KFold, 'r--', label = 'KFold')
plt.xlabel('log10(lambda)')
plt.ylabel('mse')
plt.legend()
plt.show()
!ec
!split
===== 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 $\mathcal{L}$ consisting of the data
$\mathbf{X}_\mathcal{L}=\{(y_j, \boldsymbol{x}_j), j=0\ldots n-1\}$.
Let us assume that the true data is generated from a noisy model
!bt
\[
\bm{y}=f(\boldsymbol{x}) + \bm{\epsilon}
\]
!et
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
$\bm{\beta}$ and the design matrix $\bm{X}$ which embody our model,
that is $\bm{\tilde{y}}=\bm{X}\bm{\beta}$.
Thereafter we found the parameters $\bm{\beta}$ by optimizing the means squared error via the so-called cost function
!bt
\[
C(\bm{X},\bm{\beta}) =\frac{1}{n}\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2=\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right].
\]
!et
We can rewrite this as
!bt
\[
\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right]=\frac{1}{n}\sum_i(f_i-\mathbb{E}\left[\bm{\tilde{y}}\right])^2+\frac{1}{n}\sum_i(\tilde{y}_i-\mathbb{E}\left[\bm{\tilde{y}}\right])^2+\sigma^2.
\]
!et
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 $\bm{\epsilon}$.
To derive this equation, we need to recall that the variance of $\bm{y}$ and $\bm{\epsilon}$ are both equal to $\sigma^2$. The mean value of $\bm{\epsilon}$ is by definition equal to zero. Furthermore, the function $f$ is not a stochastics variable, idem for $\bm{\tilde{y}}$.
We use a more compact notation in terms of the expectation value
!bt
\[
\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right]=\mathbb{E}\left[(\bm{f}+\bm{\epsilon}-\bm{\tilde{y}})^2\right],
\]
!et
and adding and subtracting $\mathbb{E}\left[\bm{\tilde{y}}\right]$ we get
!bt
\[
\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right]=\mathbb{E}\left[(\bm{f}+\bm{\epsilon}-\bm{\tilde{y}}+\mathbb{E}\left[\bm{\tilde{y}}\right]-\mathbb{E}\left[\bm{\tilde{y}}\right])^2\right],
\]
!et
which, using the abovementioned expectation values can be rewritten as
!bt
\[
\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right]=\mathbb{E}\left[(\bm{y}-\mathbb{E}\left[\bm{\tilde{y}}\right])^2\right]+\mathrm{Var}\left[\bm{\tilde{y}}\right]+\sigma^2,
\]
!et
that is the rewriting in terms of the so-called bias, the variance of the model $\bm{\tilde{y}}$ and the variance of $\bm{\epsilon}$.
!split
===== Example code for Bias-Variance tradeoff =====
!bc pycod
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()
!ec
!split
===== Understanding what happens =====
!bc pycod
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 = 40
n_boostraps = 100
maxdegree = 14
# Make data set.
x = np.linspace(-3, 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)
error = np.zeros(maxdegree)
bias = np.zeros(maxdegree)
variance = np.zeros(maxdegree)
polydegree = np.zeros(maxdegree)
x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)
for degree in range(maxdegree):
model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
y_pred = np.empty((y_test.shape[0], n_boostraps))
for i in range(n_boostraps):
x_, y_ = resample(x_train, y_train)
y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel()
polydegree[degree] = degree
error[degree] = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) )
bias[degree] = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 )
variance[degree] = np.mean( np.var(y_pred, axis=1, keepdims=True) )
print('Polynomial degree:', degree)
print('Error:', error[degree])
print('Bias^2:', bias[degree])
print('Var:', variance[degree])
print('{} >= {} + {} = {}'.format(error[degree], bias[degree], variance[degree], bias[degree]+variance[degree]))
plt.plot(polydegree, error, label='Error')
plt.plot(polydegree, bias, label='bias')
plt.plot(polydegree, variance, label='Variance')
plt.legend()
plt.show()
!ec
!split
===== Summing up =====
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":"https://www.pnas.org/content/116/32/15849" of interest.
!split
===== Another Example from Scikit-Learn's Repository =====
!bc pycod
"""
============================
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()
!ec
!split
===== More examples on bootstrap and cross-validation and errors =====
!bc pycod
# Common imports
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression, Ridge, Lasso
from sklearn.model_selection import train_test_split
from sklearn.utils import resample
from sklearn.metrics import mean_squared_error
# Where to save the figures and data files
PROJECT_ROOT_DIR = "Results"
FIGURE_ID = "Results/FigureFiles"
DATA_ID = "DataFiles/"
if not os.path.exists(PROJECT_ROOT_DIR):
os.mkdir(PROJECT_ROOT_DIR)
if not os.path.exists(FIGURE_ID):
os.makedirs(FIGURE_ID)
if not os.path.exists(DATA_ID):
os.makedirs(DATA_ID)
def image_path(fig_id):
return os.path.join(FIGURE_ID, fig_id)
def data_path(dat_id):
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("EoS.csv"),'r')
# Read the EoS data as csv file and organize the data into two arrays with density and energies
EoS = pd.read_csv(infile, names=('Density', 'Energy'))
EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
EoS = EoS.dropna()
Energies = EoS['Energy']
Density = EoS['Density']
# The design matrix now as function of various polytrops
Maxpolydegree = 30
X = np.zeros((len(Density),Maxpolydegree))
X[:,0] = 1.0
testerror = np.zeros(Maxpolydegree)
trainingerror = np.zeros(Maxpolydegree)
polynomial = np.zeros(Maxpolydegree)
trials = 100
for polydegree in range(1, Maxpolydegree):
polynomial[polydegree] = polydegree
for degree in range(polydegree):
X[:,degree] = Density**(degree/3.0)
# loop over trials in order to estimate the expectation value of the MSE
testerror[polydegree] = 0.0
trainingerror[polydegree] = 0.0
for samples in range(trials):
x_train, x_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)
model = LinearRegression(fit_intercept=True).fit(x_train, y_train)
ypred = model.predict(x_train)
ytilde = model.predict(x_test)
testerror[polydegree] += mean_squared_error(y_test, ytilde)
trainingerror[polydegree] += mean_squared_error(y_train, ypred)
testerror[polydegree] /= trials
trainingerror[polydegree] /= trials
print("Degree of polynomial: %3d"% polynomial[polydegree])
print("Mean squared error on training data: %.8f" % trainingerror[polydegree])
print("Mean squared error on test data: %.8f" % testerror[polydegree])
plt.plot(polynomial, np.log10(trainingerror), label='Training Error')
plt.plot(polynomial, np.log10(testerror), label='Test Error')
plt.xlabel('Polynomial degree')
plt.ylabel('log10[MSE]')
plt.legend()
plt.show()
!ec
!split
===== The same example but now with cross-validation =====
!bc pycod
# Common imports
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression, Ridge, Lasso
from sklearn.metrics import mean_squared_error
from sklearn.model_selection import KFold
from sklearn.model_selection import cross_val_score
# Where to save the figures and data files
PROJECT_ROOT_DIR = "Results"
FIGURE_ID = "Results/FigureFiles"
DATA_ID = "DataFiles/"
if not os.path.exists(PROJECT_ROOT_DIR):
os.mkdir(PROJECT_ROOT_DIR)
if not os.path.exists(FIGURE_ID):
os.makedirs(FIGURE_ID)
if not os.path.exists(DATA_ID):
os.makedirs(DATA_ID)
def image_path(fig_id):
return os.path.join(FIGURE_ID, fig_id)
def data_path(dat_id):
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("EoS.csv"),'r')
# Read the EoS data as csv file and organize the data into two arrays with density and energies
EoS = pd.read_csv(infile, names=('Density', 'Energy'))
EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
EoS = EoS.dropna()
Energies = EoS['Energy']
Density = EoS['Density']
# The design matrix now as function of various polytrops
Maxpolydegree = 30
X = np.zeros((len(Density),Maxpolydegree))
X[:,0] = 1.0
estimated_mse_sklearn = np.zeros(Maxpolydegree)
polynomial = np.zeros(Maxpolydegree)
k =5
kfold = KFold(n_splits = k)
for polydegree in range(1, Maxpolydegree):
polynomial[polydegree] = polydegree
for degree in range(polydegree):
X[:,degree] = Density**(degree/3.0)
OLS = LinearRegression()
# loop over trials in order to estimate the expectation value of the MSE
estimated_mse_folds = cross_val_score(OLS, X, Energies, scoring='neg_mean_squared_error', cv=kfold)
#[:, np.newaxis]
estimated_mse_sklearn[polydegree] = np.mean(-estimated_mse_folds)
plt.plot(polynomial, np.log10(estimated_mse_sklearn), label='Test Error')
plt.xlabel('Polynomial degree')
plt.ylabel('log10[MSE]')
plt.legend()
plt.show()
!ec
!split
===== Cross-validation with Ridge =====
!bc pycod
import numpy as np
import matplotlib.pyplot as plt
from sklearn.model_selection import KFold
from sklearn.linear_model import Ridge
from sklearn.model_selection import cross_val_score
from sklearn.preprocessing import PolynomialFeatures
# A seed just to ensure that the random numbers are the same for every run.
np.random.seed(3155)
# Generate the data.
n = 100
x = np.linspace(-3, 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)
# Decide degree on polynomial to fit
poly = PolynomialFeatures(degree = 10)
# Decide which values of lambda to use
nlambdas = 500
lambdas = np.logspace(-3, 5, nlambdas)
# Initialize a KFold instance
k = 5
kfold = KFold(n_splits = k)
estimated_mse_sklearn = np.zeros(nlambdas)
i = 0
for lmb in lambdas:
ridge = Ridge(alpha = lmb)
estimated_mse_folds = cross_val_score(ridge, x, y, scoring='neg_mean_squared_error', cv=kfold)
estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)
i += 1
plt.figure()
plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
!ec
!split
===== Friday September 4 =====
"Video of Lecture":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h20/forelesningsvideoer/LectureSeptember4.mp4?vrtx=view-as-webpage" and "handwritten notes":"https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/NotesSeptember4.pdf"
More material will be added here, see handwritten notes also.