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 $$ y_i\sim \mathcal{N}(\boldsymbol{X}_{i,*}\boldsymbol{\beta}, \sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}. $$
We assume now that the various \( y_i \) values are stochastically distributed according to the above Gaussian distribution. We define this distribution as $$ p(y_i, \boldsymbol{X}\vert\boldsymbol{\beta})=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}, $$ 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 $$ \boldsymbol{D}=[(x_0,y_0), (x_1,y_1),\dots, (x_{n-1},y_{n-1})]. $$ 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 \( \boldsymbol{\beta} \).
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 \( \beta \) 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.
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.
We could now define a new cost function to minimize, namely the negative logarithm of the above PDF $$ C(\boldsymbol{\beta}=-\log{\prod_{i=0}^{n-1}p(y_i,\boldsymbol{X}\vert\boldsymbol{\beta})}=-\sum_{i=0}^{n-1}\log{p(y_i,\boldsymbol{X}\vert\boldsymbol{\beta})}, $$ 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 $$ \boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right) =0, $$ which leads to the well-known OLS equation for the optimal paramters \( \beta \) $$ \hat{\boldsymbol{\beta}}^{\mathrm{OLS}}=\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}! $$
If we combine the conditional probability with the marginal probability and the standard product rule, we have $$ p(X\vert Y)= \frac{p(X,Y)}{p(Y)}, $$ 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 \( Y \).
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.
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.
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.
For Ridge and Lasso regression, the higher order parameters will typically be reduced, providing thereby less fluctuations from one order to another one.
import numpy as np
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 = 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):
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?
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 \( \lambda \).
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).
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):
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)
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 $$ p(\boldsymbol{\beta}\vert\boldsymbol{D}). $$
Bayes' theorem comes to our rescue here since (omitting the normalization constant) $$ p(\boldsymbol{\beta}\vert\boldsymbol{D})\propto p(\boldsymbol{D}\vert\boldsymbol{\beta})p(\boldsymbol{\beta}). $$
We have a model for \( p(\boldsymbol{D}\vert\boldsymbol{\beta}) \) but need one for the prior \( p(\boldsymbol{\beta} \)!
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 $$ p(\boldsymbol{\beta})=\prod_{j=0}^{p-1}\exp{\left(-\frac{\beta_j^2}{2\tau^2}\right)}. $$
Our posterior probability becomes then (omitting the normalization factor which is just a constant) $$ p(\boldsymbol{\beta\vert\boldsymbol{D})}=\prod_{i=0}^{n-1}\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left[-\frac{(y_i-\boldsymbol{X}_{i,*}\boldsymbol{\beta})^2}{2\sigma^2}\right]}\prod_{j=0}^{p-1}\exp{\left(-\frac{\beta_j^2}{2\tau^2}\right)}. $$
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 $$ C(\boldsymbol{\beta})=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\frac{1}{2\tau^2}\vert\vert\boldsymbol{\beta}\vert\vert_2^2, $$ and replacing \( 1/2\tau^2 \) with \( \lambda \) we have $$ C(\boldsymbol{\beta})=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_2^2, $$ which is our Ridge cost function! Nice, isn't it?
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 $$ C(\boldsymbol{\beta}=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\frac{1}{\tau}\vert\vert\boldsymbol{\beta}\vert\vert_1, $$ and replacing \( 1/\tau \) with \( \lambda \) we have $$ C(\boldsymbol{\beta}=\frac{\vert\vert (\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\vert\vert_2^2}{2\sigma^2}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_1, $$ which is our Lasso cost function!
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
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,
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.
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
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 have also been developed..
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.
The Jackknife works by making many replicas of the estimator \( \widehat{\beta} \). The jackknife is a resampling method where we systematically leave out one observation from the vector of observed values \( \boldsymbol{x} = (x_1,x_2,\cdots,X_n) \). Let \( \boldsymbol{x}_i \) denote the vector $$ \boldsymbol{x}_i = (x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n), $$
which equals the vector \( \boldsymbol{x} \) with the exception that observation number \( i \) is left out. Using this notation, define \( \widehat{\beta}_i \) to be the estimator \( \widehat{\beta} \) computed using \( \vec{X}_i \).
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)
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:
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.
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 \( z \).
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.
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 $$ \int_{-\infty}^{\infty}dxp(x)\exp{\left(iq(\mu-x)/m\right)}= \int_{-\infty}^{\infty}dxp(x) \left[1+\frac{iq(\mu-x)}{m}-\frac{q^2(\mu-x)^2}{2m^2}+\dots\right]. $$
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 \( p(x) \).
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 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.
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.
We found also that the variance of the estimate of the \( j \)-th regression coefficient is \( \boldsymbol{\sigma}^2 (\boldsymbol{\beta}_j ) = \boldsymbol{\sigma}^2 [(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} \).
This quantity will be used to 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 $$ \left(\mu_{\beta}\pm \frac{z\sigma_{\beta}}{\sqrt{n}}\right), $$
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.
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.
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:
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.
The independent bootstrap works like this:
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.
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.
# 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()
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 \( \boldsymbol{\beta} \) 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 $$ C(\boldsymbol{X},\boldsymbol{\beta}) =\frac{1}{n}\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2=\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]. $$
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 $$ \mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{f}+\boldsymbol{\epsilon}-\boldsymbol{\tilde{y}})^2\right], $$ and adding and subtracting \( \mathbb{E}\left[\boldsymbol{\tilde{y}}\right] \) we get $$ \mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{f}+\boldsymbol{\epsilon}-\boldsymbol{\tilde{y}}+\mathbb{E}\left[\boldsymbol{\tilde{y}}\right]-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2\right], $$ which, using the abovementioned expectation values can be rewritten as $$ \mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{y}-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2\right]+\mathrm{Var}\left[\boldsymbol{\tilde{y}}\right]+\sigma^2, $$ that is the rewriting in terms of the so-called bias, the variance of the model \( \boldsymbol{\tilde{y}} \) and the variance of \( \boldsymbol{\epsilon} \).
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()
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()
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.
"""
============================
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()
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).
For the various values of \( k \)
The code here uses Ridge regression with cross-validation (CV) resampling and \( k \)-fold CV in order to fit a specific polynomial.
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()
# 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=False).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()
Note that we kept the intercept column in the fitting here. This means that we need to set the intercept in the call to the Scikit-Learn function as False. Alternatively, we could have set up the design matrix \( X \) without the first column of ones.
In this example we keep the intercept column again but add cross-validation in order to estimate the best possible value of the means squared error.
# 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(fit_intercept=False)
# 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()