updating regression analysis slides by adding info on chi2

This commit is contained in:
Morten Hjorth-Jensen
2017-10-24 16:44:11 +02:00
parent 09cdc8de31
commit 2e5b3ce390
28 changed files with 2373 additions and 99 deletions
@@ -72,7 +72,13 @@ Automatically generated HTML file from DocOnce source
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@@ -121,7 +127,13 @@ MathJax.Hub.Config({
<!-- navigation toc: --> <li><a href="._Regression-bs009.html#___sec8" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs010.html#___sec9" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs011.html#___sec10" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The singular value decompostion</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs013.html#___sec12" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs014.html#___sec13" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs015.html#___sec14" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs016.html#___sec15" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs017.html#___sec16" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs018.html#___sec17" style="font-size: 80%;">The singular value decompostion</a></li>
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@@ -156,7 +168,7 @@ MathJax.Hub.Config({
<center>[2] <b>Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University</b></center>
<br>
<p>
<center><h4>Oct 18, 2017</h4></center> <!-- date -->
<center><h4>Oct 24, 2017</h4></center> <!-- date -->
<br>
<p>
@@ -180,7 +192,7 @@ MathJax.Hub.Config({
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<li><a href="._Regression-bs009.html">10</a></li>
<li><a href="">...</a></li>
<li><a href="._Regression-bs012.html">13</a></li>
<li><a href="._Regression-bs018.html">19</a></li>
<li><a href="._Regression-bs001.html">&raquo;</a></li>
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<!-- ------------------- end of main content --------------- -->
@@ -72,7 +72,13 @@ Automatically generated HTML file from DocOnce source
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@@ -121,7 +127,13 @@ MathJax.Hub.Config({
<!-- navigation toc: --> <li><a href="._Regression-bs009.html#___sec8" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs010.html#___sec9" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs011.html#___sec10" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The singular value decompostion</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs013.html#___sec12" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs014.html#___sec13" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs015.html#___sec14" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs016.html#___sec15" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs017.html#___sec16" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs018.html#___sec17" style="font-size: 80%;">The singular value decompostion</a></li>
</ul>
</li>
@@ -177,7 +189,7 @@ A regression model aims at finding a likelihood function \( p(y\vert \hat{x}) \)
<li><a href="._Regression-bs009.html">10</a></li>
<li><a href="._Regression-bs010.html">11</a></li>
<li><a href="">...</a></li>
<li><a href="._Regression-bs012.html">13</a></li>
<li><a href="._Regression-bs018.html">19</a></li>
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<!-- ------------------- end of main content --------------- -->
@@ -72,7 +72,13 @@ Automatically generated HTML file from DocOnce source
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<!-- navigation toc: --> <li><a href="._Regression-bs009.html#___sec8" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs010.html#___sec9" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs011.html#___sec10" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The singular value decompostion</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs013.html#___sec12" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs014.html#___sec13" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs015.html#___sec14" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs016.html#___sec15" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs017.html#___sec16" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs018.html#___sec17" style="font-size: 80%;">The singular value decompostion</a></li>
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</li>
@@ -174,7 +186,7 @@ where \( \epsilon_i \) is the error in our approximation.
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<li><a href="._Regression-bs011.html">12</a></li>
<li><a href="">...</a></li>
<li><a href="._Regression-bs012.html">13</a></li>
<li><a href="._Regression-bs018.html">19</a></li>
<li><a href="._Regression-bs003.html">&raquo;</a></li>
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<!-- ------------------- end of main content --------------- -->
@@ -72,7 +72,13 @@ Automatically generated HTML file from DocOnce source
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<!-- navigation toc: --> <li><a href="._Regression-bs009.html#___sec8" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs010.html#___sec9" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs011.html#___sec10" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The singular value decompostion</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs013.html#___sec12" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs014.html#___sec13" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs015.html#___sec14" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs016.html#___sec15" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs017.html#___sec16" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs018.html#___sec17" style="font-size: 80%;">The singular value decompostion</a></li>
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@@ -173,6 +185,8 @@ $$
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<li><a href="._Regression-bs012.html">13</a></li>
<li><a href="">...</a></li>
<li><a href="._Regression-bs018.html">19</a></li>
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@@ -72,7 +72,13 @@ Automatically generated HTML file from DocOnce source
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<!-- navigation toc: --> <li><a href="._Regression-bs009.html#___sec8" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs010.html#___sec9" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs011.html#___sec10" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The singular value decompostion</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs013.html#___sec12" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs014.html#___sec13" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs015.html#___sec14" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs016.html#___sec15" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs017.html#___sec16" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs018.html#___sec17" style="font-size: 80%;">The singular value decompostion</a></li>
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@@ -192,6 +204,9 @@ $$
<li><a href="._Regression-bs010.html">11</a></li>
<li><a href="._Regression-bs011.html">12</a></li>
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<li><a href="._Regression-bs018.html">19</a></li>
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@@ -72,7 +72,13 @@ Automatically generated HTML file from DocOnce source
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<!-- navigation toc: --> <li><a href="._Regression-bs010.html#___sec9" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs011.html#___sec10" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The singular value decompostion</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs013.html#___sec12" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs014.html#___sec13" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs015.html#___sec14" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs016.html#___sec15" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs017.html#___sec16" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs018.html#___sec17" style="font-size: 80%;">The singular value decompostion</a></li>
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@@ -176,6 +188,10 @@ $$
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<li><a href="._Regression-bs012.html">13</a></li>
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<li><a href="._Regression-bs018.html">19</a></li>
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<!-- navigation toc: --> <li><a href="._Regression-bs009.html#___sec8" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs010.html#___sec9" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs011.html#___sec10" style="font-size: 80%;">Interpretations and optimizing our parameters</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The singular value decompostion</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs013.html#___sec12" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs014.html#___sec13" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs015.html#___sec14" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs016.html#___sec15" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs017.html#___sec16" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs018.html#___sec17" style="font-size: 80%;">The singular value decompostion</a></li>
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@@ -181,6 +193,11 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo
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<li><a href="._Regression-bs011.html">12</a></li>
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<li><a href="._Regression-bs015.html">16</a></li>
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@@ -175,6 +187,12 @@ $$
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@@ -177,6 +189,13 @@ $$
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@@ -156,12 +168,12 @@ where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till n
<p>
In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring
$$
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
$$
which results in
$$
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0,
$$
or in a matrix-vector form as
@@ -192,6 +204,12 @@ $$
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@@ -179,6 +191,12 @@ $$
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@@ -180,6 +192,12 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r
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@@ -137,20 +149,28 @@ MathJax.Hub.Config({
<a name="part0012"></a>
<!-- !split -->
<h2 id="___sec11" class="anchor">The singular value decompostion </h2>
<h2 id="___sec11" class="anchor">The \( \chi^2 \) function </h2>
<div class="panel panel-default">
<div class="panel-body">
<p> <!-- subsequent paragraphs come in larger fonts, so start with a paragraph -->
How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then
<p>
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.
<p>
Introducing the standard deviation \( \sigma_i \) for each measurement \( y_i \), we define now the \( \chi^2 \) function as
$$
\hat{A} = \hat{U}\hat{\Sigma}\hat{V}
\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right),
$$
where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements.
<p>
</div>
</div>
<p>
<p>
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In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\hat{\beta}) \) by requiring
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,
$$
which results in
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,
$$
or in a matrix-vector form as
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right).
$$
where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \hat{b} \) with elements \( b_i = y_i/\sigma_i \).
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We can rewrite
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right),
$$
as
$$
\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta},
$$
and if the matrix \( \hat{A}^T\hat{A} \) is invertible we have the solution
$$
\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}.
$$
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<p>
If we then introduce the matrix
$$
\hat{H} = \hat{A}^T\hat{A},
$$
we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \hat{H} \) are \( h_{ij} \))
$$
\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}
$$
We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as
$$
\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2,
$$
resulting in
$$
\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!
$$
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The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write
$$
y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i.
$$
By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0,
$$
and
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0.
$$
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We define then
$$
\gamma = \sum_{i=0}^{1}\frac{1}{\sigma_i^2},
$$
$$
\gamma_x = \sum_{i=0}^{1}\frac{x_{i}}{\sigma_i^2},
$$
$$
\gamma_y = \sum_{i=0}^{1}\left(\frac{y_i}{\sigma_i^2}\right),
$$
$$
\gamma_{xx} = \sum_{i=0}^{1}\frac{x_ix_{i}}{\sigma_i^2},
$$
$$
\gamma_{xy} = \sum_{i=0}^{1}\frac{y_ix_{i}}{\sigma_i^2},
$$
and show that
$$
\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2},
$$
$$
\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.
$$
<p>
The LSM suffers often from both being underdetermined and overdetermined in the unknown coefficients \( \beta_i \). A better approach is to use the Singular Value Decomposition (SVD) method discussed below.
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('The $\\chi^2$ function', 2, None, '___sec16'),
('The singular value decompostion', 2, None, '___sec17')]}
end of tocinfo -->
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<!-- navigation toc: --> <li><a href="._Regression-bs003.html#___sec2" style="font-size: 80%;">Rewriting the fitting procedure as a linear algebra problem</a></li>
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<h2 id="___sec17" class="anchor">The singular value decompostion </h2>
<div class="panel panel-default">
<div class="panel-body">
<p> <!-- subsequent paragraphs come in larger fonts, so start with a paragraph -->
How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then
$$
\hat{A} = \hat{U}\hat{\Sigma}\hat{V}
$$
</div>
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<p>
<p>
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+18 -6
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@@ -72,7 +72,13 @@ Automatically generated HTML file from DocOnce source
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'___sec10'),
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('The $\\chi^2$ function', 2, None, '___sec11'),
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('The $\\chi^2$ function', 2, None, '___sec13'),
('The $\\chi^2$ function', 2, None, '___sec14'),
('The $\\chi^2$ function', 2, None, '___sec15'),
('The $\\chi^2$ function', 2, None, '___sec16'),
('The singular value decompostion', 2, None, '___sec17')]}
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@@ -121,7 +127,13 @@ MathJax.Hub.Config({
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<!-- navigation toc: --> <li><a href="._Regression-bs012.html#___sec11" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
<!-- navigation toc: --> <li><a href="._Regression-bs013.html#___sec12" style="font-size: 80%;">The \( \chi^2 \) function</a></li>
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@@ -156,7 +168,7 @@ MathJax.Hub.Config({
<center>[2] <b>Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University</b></center>
<br>
<p>
<center><h4>Oct 18, 2017</h4></center> <!-- date -->
<center><h4>Oct 24, 2017</h4></center> <!-- date -->
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@@ -180,7 +192,7 @@ MathJax.Hub.Config({
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+209 -7
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@@ -1,4 +1,3 @@
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@@ -117,8 +116,8 @@ MathJax.Hub.Config({
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@@ -148,7 +147,7 @@ MathJax.Hub.Config({
<center>[2] <b>Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University</b></center>
<br>
<p>&nbsp;<br>
<center><h4>Oct 18, 2017</h4></center> <!-- date -->
<center><h4>Oct 24, 2017</h4></center> <!-- date -->
<br>
<p>
@@ -404,14 +403,14 @@ where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till n
In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring
<p>&nbsp;<br>
$$
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
$$
<p>&nbsp;<br>
which results in
<p>&nbsp;<br>
$$
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0,
$$
<p>&nbsp;<br>
@@ -492,7 +491,210 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r
<section>
<h2 id="___sec11">The singular value decompostion </h2>
<h2 id="___sec11">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
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.
<p>
Introducing the standard deviation \( \sigma_i \) for each measurement \( y_i \), we define now the \( \chi^2 \) function as
<p>&nbsp;<br>
$$
\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right),
$$
<p>&nbsp;<br>
where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements.
</div>
</section>
<section>
<h2 id="___sec12">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\hat{\beta}) \) by requiring
<p>&nbsp;<br>
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,
$$
<p>&nbsp;<br>
which results in
<p>&nbsp;<br>
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,
$$
<p>&nbsp;<br>
or in a matrix-vector form as
<p>&nbsp;<br>
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right).
$$
<p>&nbsp;<br>
where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \hat{b} \) with elements \( b_i = y_i/\sigma_i \).
</div>
</section>
<section>
<h2 id="___sec13">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
We can rewrite
<p>&nbsp;<br>
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right),
$$
<p>&nbsp;<br>
as
<p>&nbsp;<br>
$$
\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta},
$$
<p>&nbsp;<br>
and if the matrix \( \hat{A}^T\hat{A} \) is invertible we have the solution
<p>&nbsp;<br>
$$
\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}.
$$
<p>&nbsp;<br>
</div>
</section>
<section>
<h2 id="___sec14">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
If we then introduce the matrix
<p>&nbsp;<br>
$$
\hat{H} = \hat{A}^T\hat{A},
$$
<p>&nbsp;<br>
we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \hat{H} \) are \( h_{ij} \))
<p>&nbsp;<br>
$$
\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}
$$
<p>&nbsp;<br>
We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as
<p>&nbsp;<br>
$$
\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2,
$$
<p>&nbsp;<br>
resulting in
<p>&nbsp;<br>
$$
\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!
$$
<p>&nbsp;<br>
</div>
</section>
<section>
<h2 id="___sec15">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write
<p>&nbsp;<br>
$$
y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i.
$$
<p>&nbsp;<br>
By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by
<p>&nbsp;<br>
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0,
$$
<p>&nbsp;<br>
and
<p>&nbsp;<br>
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0.
$$
<p>&nbsp;<br>
</div>
</section>
<section>
<h2 id="___sec16">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
We define then
<p>&nbsp;<br>
$$
\gamma = \sum_{i=0}^{1}\frac{1}{\sigma_i^2},
$$
<p>&nbsp;<br>
<p>&nbsp;<br>
$$
\gamma_x = \sum_{i=0}^{1}\frac{x_{i}}{\sigma_i^2},
$$
<p>&nbsp;<br>
<p>&nbsp;<br>
$$
\gamma_y = \sum_{i=0}^{1}\left(\frac{y_i}{\sigma_i^2}\right),
$$
<p>&nbsp;<br>
<p>&nbsp;<br>
$$
\gamma_{xx} = \sum_{i=0}^{1}\frac{x_ix_{i}}{\sigma_i^2},
$$
<p>&nbsp;<br>
<p>&nbsp;<br>
$$
\gamma_{xy} = \sum_{i=0}^{1}\frac{y_ix_{i}}{\sigma_i^2},
$$
<p>&nbsp;<br>
and show that
<p>&nbsp;<br>
$$
\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2},
$$
<p>&nbsp;<br>
<p>&nbsp;<br>
$$
\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.
$$
<p>&nbsp;<br>
<p>
The LSM suffers often from both being underdetermined and overdetermined in the unknown coefficients \( \beta_i \). A better approach is to use the Singular Value Decomposition (SVD) method discussed below.
</div>
</section>
<section>
<h2 id="___sec17">The singular value decompostion </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
@@ -92,7 +92,13 @@ div { text-align: justify; text-justify: inter-word; }
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<body>
@@ -107,8 +113,8 @@ MathJax.Hub.Config({
}
});
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@@ -134,7 +140,7 @@ MathJax.Hub.Config({
<center>[2] <b>Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University</b></center>
<br>
<p>
<center><h4>Oct 18, 2017</h4></center> <!-- date -->
<center><h4>Oct 24, 2017</h4></center> <!-- date -->
<br>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
@@ -360,12 +366,12 @@ where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till n
<p>
In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring
$$
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
$$
which results in
$$
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0,
$$
or in a matrix-vector form as
@@ -434,7 +440,184 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec11">The singular value decompostion </h2>
<h2 id="___sec11">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
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.
<p>
Introducing the standard deviation \( \sigma_i \) for each measurement \( y_i \), we define now the \( \chi^2 \) function as
$$
\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right),
$$
where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements.
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec12">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\hat{\beta}) \) by requiring
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,
$$
which results in
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,
$$
or in a matrix-vector form as
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right).
$$
where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \hat{b} \) with elements \( b_i = y_i/\sigma_i \).
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec13">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
We can rewrite
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right),
$$
as
$$
\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta},
$$
and if the matrix \( \hat{A}^T\hat{A} \) is invertible we have the solution
$$
\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}.
$$
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec14">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
If we then introduce the matrix
$$
\hat{H} = \hat{A}^T\hat{A},
$$
we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \hat{H} \) are \( h_{ij} \))
$$
\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}
$$
We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as
$$
\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2,
$$
resulting in
$$
\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!
$$
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec15">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write
$$
y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i.
$$
By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0,
$$
and
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0.
$$
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec16">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
We define then
$$
\gamma = \sum_{i=0}^{1}\frac{1}{\sigma_i^2},
$$
$$
\gamma_x = \sum_{i=0}^{1}\frac{x_{i}}{\sigma_i^2},
$$
$$
\gamma_y = \sum_{i=0}^{1}\left(\frac{y_i}{\sigma_i^2}\right),
$$
$$
\gamma_{xx} = \sum_{i=0}^{1}\frac{x_ix_{i}}{\sigma_i^2},
$$
$$
\gamma_{xy} = \sum_{i=0}^{1}\frac{y_ix_{i}}{\sigma_i^2},
$$
and show that
$$
\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2},
$$
$$
\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.
$$
<p>
The LSM suffers often from both being underdetermined and overdetermined in the unknown coefficients \( \beta_i \). A better approach is to use the Singular Value Decomposition (SVD) method discussed below.
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec17">The singular value decompostion </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
+190 -7
View File
@@ -97,7 +97,13 @@ div { text-align: justify; text-justify: inter-word; }
2,
None,
'___sec10'),
('The singular value decompostion', 2, None, '___sec11')]}
('The $\\chi^2$ function', 2, None, '___sec11'),
('The $\\chi^2$ function', 2, None, '___sec12'),
('The $\\chi^2$ function', 2, None, '___sec13'),
('The $\\chi^2$ function', 2, None, '___sec14'),
('The $\\chi^2$ function', 2, None, '___sec15'),
('The $\\chi^2$ function', 2, None, '___sec16'),
('The singular value decompostion', 2, None, '___sec17')]}
end of tocinfo -->
<body>
@@ -112,8 +118,8 @@ MathJax.Hub.Config({
}
});
</script>
<script type="text/javascript" async
src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.1/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
<script type="text/javascript"
src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
</script>
@@ -139,7 +145,7 @@ MathJax.Hub.Config({
<center>[2] <b>Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University</b></center>
<br>
<p>
<center><h4>Oct 18, 2017</h4></center> <!-- date -->
<center><h4>Oct 24, 2017</h4></center> <!-- date -->
<br>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
@@ -365,12 +371,12 @@ where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till n
<p>
In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring
$$
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
$$
which results in
$$
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0,
$$
or in a matrix-vector form as
@@ -439,7 +445,184 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec11">The singular value decompostion </h2>
<h2 id="___sec11">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
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.
<p>
Introducing the standard deviation \( \sigma_i \) for each measurement \( y_i \), we define now the \( \chi^2 \) function as
$$
\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right),
$$
where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements.
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec12">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\hat{\beta}) \) by requiring
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,
$$
which results in
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,
$$
or in a matrix-vector form as
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right).
$$
where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \hat{b} \) with elements \( b_i = y_i/\sigma_i \).
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec13">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
We can rewrite
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right),
$$
as
$$
\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta},
$$
and if the matrix \( \hat{A}^T\hat{A} \) is invertible we have the solution
$$
\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}.
$$
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec14">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
If we then introduce the matrix
$$
\hat{H} = \hat{A}^T\hat{A},
$$
we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \hat{H} \) are \( h_{ij} \))
$$
\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}
$$
We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as
$$
\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2,
$$
resulting in
$$
\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!
$$
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec15">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write
$$
y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i.
$$
By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0,
$$
and
$$
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0.
$$
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec16">The \( \chi^2 \) function </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>
We define then
$$
\gamma = \sum_{i=0}^{1}\frac{1}{\sigma_i^2},
$$
$$
\gamma_x = \sum_{i=0}^{1}\frac{x_{i}}{\sigma_i^2},
$$
$$
\gamma_y = \sum_{i=0}^{1}\left(\frac{y_i}{\sigma_i^2}\right),
$$
$$
\gamma_{xx} = \sum_{i=0}^{1}\frac{x_ix_{i}}{\sigma_i^2},
$$
$$
\gamma_{xy} = \sum_{i=0}^{1}\frac{y_ix_{i}}{\sigma_i^2},
$$
and show that
$$
\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2},
$$
$$
\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.
$$
<p>
The LSM suffers often from both being underdetermined and overdetermined in the unknown coefficients \( \beta_i \). A better approach is to use the Singular Value Decomposition (SVD) method discussed below.
</div>
<p>
<!-- !split --><br><br><br><br><br><br><br><br><br><br>
<h2 id="___sec17">The singular value decompostion </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
Binary file not shown.
Binary file not shown.
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+165 -10
View File
@@ -198,13 +198,13 @@ where $\langle y_i \rangle$ is the mean value. Keep in mind also that till now
In order to find the parameters $\beta_i$ we will then minimize the spread of $Q(\hat{\beta})$ by requiring
!bt
\[
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
\]
!et
which results in
!bt
\[
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)\right]=0,
\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0,
\]
!et
or in a matrix-vector form as
@@ -267,8 +267,169 @@ meaning that the solution for $\hat{\beta}$ is the one which minimizes the resid
!eblock
!split
===== The $\chi^2$ function =====
!bblock
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.
Introducing the standard deviation $\sigma_i$ for each measurement $y_i$, we define now the $\chi^2$ function as
!bt
\[
\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right),
\]
!et
where the matrix $\hat{\Sigma}$ is a diagonal matrix with $\sigma_i$ as matrix elements.
!eblock
!split
===== The $\chi^2$ function =====
!bblock
In order to find the parameters $\beta_i$ we will then minimize the spread of $\chi^2(\hat{\beta})$ by requiring
!bt
\[
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,
\]
!et
which results in
!bt
\[
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,
\]
!et
or in a matrix-vector form as
!bt
\[
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right).
\]
!et
where we have defined the matrix $\hat{A} =\hat{X}/\hat{\Sigma}$ with matrix elements $a_{ij} = x_{ij}/\sigma_i$ and the vector $\hat{b}$ with elements $b_i = y_i/\sigma_i$.
!eblock
!split
===== The $\chi^2$ function =====
!bblock
We can rewrite
!bt
\[
\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right),
\]
!et
as
!bt
\[
\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta},
\]
!et
and if the matrix $\hat{A}^T\hat{A}$ is invertible we have the solution
!bt
\[
\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}.
\]
!et
!eblock
!split
===== The $\chi^2$ function =====
!bblock
If we then introduce the matrix
!bt
\[
\hat{H} = \hat{A}^T\hat{A},
\]
!et
we have then the following expression for the parameters $\beta_j$ (the matrix elements of $\hat{H}$ are $h_{ij}$)
!bt
\[
\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}
\]
!et
We state without proof the expression for the uncertainty in the parameters $\beta_j$ as
!bt
\[
\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2,
\]
!et
resulting in
!bt
\[
\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!
\]
!et
!eblock
!split
===== The $\chi^2$ function =====
!bblock
The first step here is to approximate the function $y$ with a first-order polynomial, that is we write
!bt
\[
y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i.
\]
!et
By computing the derivatives of $\chi^2$ with respect to $\beta_0$ and $\beta_1$ show that these are given by
!bt
\[
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0,
\]
!et
and
!bt
\[
\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0.
\]
!et
!eblock
!split
===== The $\chi^2$ function =====
!bblock
We define then
!bt
\[
\gamma = \sum_{i=0}^{1}\frac{1}{\sigma_i^2},
\]
!et
!bt
\[
\gamma_x = \sum_{i=0}^{1}\frac{x_{i}}{\sigma_i^2},
\]
!et
!bt
\[
\gamma_y = \sum_{i=0}^{1}\left(\frac{y_i}{\sigma_i^2}\right),
\]
!et
!bt
\[
\gamma_{xx} = \sum_{i=0}^{1}\frac{x_ix_{i}}{\sigma_i^2},
\]
!et
!bt
\[
\gamma_{xy} = \sum_{i=0}^{1}\frac{y_ix_{i}}{\sigma_i^2},
\]
!et
and show that
!bt
\[
\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2},
\]
!et
!bt
\[
\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.
\]
!et
The LSM suffers often from both being underdetermined and overdetermined in the unknown coefficients $\beta_i$. A better approach is to use the Singular Value Decomposition (SVD) method discussed below.
!eblock
@@ -287,13 +448,7 @@ How can we use the singular value decomposition to find the parameters $\beta_j$
Suppose M is a m × n matrix whose entries come from the field K, which is either the field of real numbers or the field of complex numbers. Then there exists a factorization, called a singular value decomposition of M, of the form
{\displaystyle \mathbf {M} =\mathbf {U} {\boldsymbol {\Sigma }}\mathbf {V} ^{*}} \mathbf {M} =\mathbf {U} {\boldsymbol {\Sigma }}\mathbf {V} ^{*}
where
U is an m × m unitary matrix (if K = {\displaystyle \mathbb {R} } \mathbb {R} , unitary matrices are orthogonal matrices),
Σ is a diagonal m × n matrix with non-negative real numbers on the diagonal,
V is an n × n unitary matrix over K, and
V is the conjugate transpose of V.
The diagonal entries σi of Σ are known as the singular values of M. A common convention is to list the singular values in descending order. In this case, the diagonal matrix, Σ, is uniquely determined by M (though not the matrices U and V, see below).