FYS-STK4155/doc/pub/week37/ipynb/week37.ipynb

Week 37: Summary of Ridge and Lasso Regression and Resampling Methods

Morten Hjorth-Jensen, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

Date: Sep 15, 2022

Copyright 1999-2022, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license

Plans for week 37

• Thursday September 15: Summary of Ridge and Lasso with examples and statistical interpretation. Start resampling techniques and discussion of the bias-variance tradeoff.

  • Video of lecture

• Friday September 16: Resampling methods, bias-variance, overfitting, Cross-validation and Bootstrapping

Recommended Reading:

1. Lectures on Resampling methods (these lectures), see also lectures from week 36

2. Bishop 1.3 (cross-validation) and 3.2 (bias-variance tradeoff)

3. Hastie et al Chapter 7, here we recommend 7.1-7.5 and 7.10 (cross-validation) and 7.11 (bootstrap).

4. See also the excellent videos on the SVD at http://databookuw.com/page-2/page-4/. The texboook by Brunton and Kutz at http://databookuw.com is highly recommended

Thursday September 15, Summary of Ridge and Lasso Regression and start Resampling methods

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{\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]}.

Independent and Identically Distrubuted (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{\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}.

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 \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.

A new Cost Function

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}!

Bayes' Theorem

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.

Interpretations of Bayes' Theorem

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.

Test Function for what happens with OLS, Ridge and Lasso

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.

In [1]:

%matplotlib inline

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?

Rerunning the above code

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).

In [2]:

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)

Invoking Bayes' theorem

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}!

Ridge and Bayes

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?

Lasso and Bayes

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!

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: 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 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.

Resampling methods: Jackknife

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.

Jackknife code example

In [3]:

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)

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{\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.

Warning:

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