Morten Hjorth-Jensen
411 lines
15 KiB
HTML
411 lines
15 KiB
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'sections': [('Regression analysis, overarching aims', 2, None, '___sec0'),
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('Rewriting the fitting procedure as a linear algebra problem',
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('Generalizing the fitting procedure as a linear algebra problem',
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('Interpretations and optimizing our parameters',
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<!-- ------------------- main content ---------------------- -->
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<center><h1>Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis</h1></center> <!-- document title -->
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<p>
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<!-- author(s): Morten Hjorth-Jensen -->
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<center>
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<b>Morten Hjorth-Jensen</b> [1, 2]
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</center>
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<p>
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<!-- institution(s) -->
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<center>[1] <b>Department of Physics, University of Oslo</b></center>
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<center>[2] <b>Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University</b></center>
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<br>
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<p>
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<center><h4>Oct 17, 2017</h4></center> <!-- date -->
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<br>
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec0">Regression analysis, overarching aims </h2>
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<div class="alert alert-block alert-block alert-text-normal">
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<b></b>
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<p>
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<p>
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Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) varies as function of another variable or a set of such variables \( \hat{x} =[x_0, x_1,\dots, x_p]^T \).
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The first variable is called the <b>dependent</b>, the <b>outcome</b> or the <b>response</b> variable while the set of variables \( \hat{x} \) is called the independent variable, or the predictor variable or the explanatory variable.
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<p>
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A regression model aims at finding a likelihood function \( p(y\vert \hat{x}) \), that is the conditional distribution for \( y \) with a given \( \hat{x} \). The estimation of \( p(y\vert \hat{x}) \) is made using a data set with
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<ul>
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<li> \( n \) cases \( i = 0, 1, 2, \dots, n-1 \)</li>
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<li> Response (dependent or outcome) variable \( y_i \) with \( i = 0, 1, 2, \dots, n-1 \)</li>
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<li> \( p \) Explanatory (independent or predictor) variables \( \hat{x}_i=[x_{i0}, x_{i1}, \dots, x_{ip}] \) with \( i = 0, 1, 2, \dots, n-1 \)</li>
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</ul>
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The goal of the regression analysis is to extract/exploit relationship between \( y_i \) and \( \hat{x}_i \) in or to infer causal dependencies, approximations to the likelihood functions, functional relationships and to make predictions .
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</div>
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec1">General linear models </h2>
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<div class="alert alert-block alert-block alert-text-normal">
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<b></b>
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<p>
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Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \hat{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \hat{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function.
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<p>
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Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is
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$$
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y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_i x_i^j+\epsilon_i,
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$$
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where \( \epsilon_i \) is the error in our approximation.
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</div>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec2">Rewriting the fitting procedure as a linear algebra problem </h2>
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<div class="alert alert-block alert-block alert-text-normal">
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<b></b>
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<p>
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For every set of values \( y_i,x_i \) we have thus the corresponding set of equations
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$$
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\begin{align*}
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y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\
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y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\
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y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\
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\dots & \dots \\
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y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\epsilon_{n-1}.\\
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\end{align*}
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$$
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Defining the vectors
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$$
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\hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T,
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$$
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$$
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\hat{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T,
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$$
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$$
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\hat{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T,
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$$
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and the matrix
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$$
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\hat{X}=
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\begin{bmatrix}
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1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\
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1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\
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1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\
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\dots& \dots &\dots& \dots & \dots &\dots\\
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1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\
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\end{bmatrix}
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$$
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we can rewrite our equations as
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$$
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\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}.
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$$
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</div>
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec3">Generalizing the fitting procedure as a linear algebra problem </h2>
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<div class="alert alert-block alert-block alert-text-normal">
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<b></b>
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<p>
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We are obviously not limited to the above polynomial. We could replace the various powers of \( x \) with elements of Fourier series, that is, instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j x_i)} \), or time series or other orthogonal functions.
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For every set of values \( y_i,x_i \) we can then generalize the equations to
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$$
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\begin{align*}
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y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\
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y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\
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y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\
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\dots & \dots \\
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y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\
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\dots & \dots \\
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y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1}^{n-1,n-1}+\epsilon_{n-1}.\\
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\end{align*}
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$$
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We redefine in turn the matrix \( \hat{X} \) as
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$$
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\hat{X}=
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\begin{bmatrix}
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x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\
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x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\
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x_{20}& x_{21}^1 &x_{22}2& \dots & \dots &x_{2,n-1}\\
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\dots& \dots &\dots& \dots & \dots &\dots\\
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x_{n-1,00}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\
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\end{bmatrix}
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$$
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and without loss of generality we rewrite again our equations as
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$$
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\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}.
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$$
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The left-hand side of this equation forms know. Our error vector \( \hat{\epsilon} \) and the parameter vector \( \hat{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
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</div>
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec4">Optimizing our parameters </h2>
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<div class="alert alert-block alert-block alert-text-normal">
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<b></b>
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<p>
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We have defined the matrix \( \hat{X} \)
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$$
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\begin{align*}
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y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\
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y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\
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y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\
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\dots & \dots \\
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y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\
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\dots & \dots \\
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y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\
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\end{align*}
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$$
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We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as
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$$
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\hat{\tilde{y}}= \hat{X}\hat{\beta},
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$$
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and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parametrized values \( \tilde{y}_i \), namely
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$$
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Q(\hat{\beta})=\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\left(\hat{y}-\hat{\tilde{y}}\right)^T\left(\hat{y}-\hat{\tilde{y}}\right),
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$$
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or using the matrix \( \hat{X} \) as
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$$
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Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right).
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$$
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</div>
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec5">Interpretations and optimizing our parameters </h2>
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<div class="alert alert-block alert-block alert-text-normal">
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<b></b>
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<p>
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The function
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$$
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Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right),
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$$
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can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value of for example a numerical experiment. When linking below with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value
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$$
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y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i,
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$$
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where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till now we have treated \( y_i \) as the exact value. Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable.
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<p>
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In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring
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$$
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\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
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$$
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which results in
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$$
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\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)\right]=0,
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$$
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or in a matrix-vector form as
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$$
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\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right).
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$$
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</div>
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec6">Interpretations and optimizing our parameters </h2>
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<div class="alert alert-block alert-block alert-text-normal">
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<b></b>
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<p>
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We can rewrite
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$$
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\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right),
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$$
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as
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$$
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\hat{X}^T\hat{y} = \hat{X}^T\hat{X}\hat{\beta},
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$$
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and if the matrix \( \hat{X}^T\hat{X} \) is invertible we have the solution
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$$
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\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}.
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$$
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The residuals \( \hat{\epsilon} \) are in turn given by
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$$
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\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta},
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$$
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and with
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$$
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\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0,
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$$
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we have
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$$
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\hat{X}^T\hat{\epsilon}=\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0,
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$$
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meaning that the solution for \( \hat{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.
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</div>
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<p>
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<!-- !split --><br><br><br><br><br><br><br><br><br><br>
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<h2 id="___sec7">The singular value decompostion </h2>
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<div class="alert alert-block alert-block alert-text-normal">
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<b></b>
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<p>
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Here we derive the equations for the SVD.
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</div>
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<p>
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<!-- ------------------- end of main content --------------- -->
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<center style="font-size:80%">
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