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<title>PyXPlot Users' Guide: The Covariance Matrix</title>
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<div><h1 id="a0000000245">C.4 The Covariance Matrix</h1>
<p> <a name="a0000001676" id="a0000001676"></a> </p><p>The terms of the covariance matrix <img src="images/img-0822.png" alt="$V_{ij}$" style="vertical-align:-5px;
width:21px;
height:17px" class="math gen" /> are defined by: </p><table id="<plasTeX.TeXFragment object at 0xb25f6bc>" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="images/img-0823.png" alt="\begin{equation} V_{ij} = \left< \left(u_ i - u^0_ i\right) \left(u_ j - u^0_ j\right) \right> \label{eqn:def_ covar} \end{equation}" style="width:404px;
height:24px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"><span>(<span>C.8</span>)</span></td>
</tr>
</table><p>Its leading diagonal terms may be recognised as equalling the variances of each of our <img src="images/img-0794.png" alt="$n_\mathrm {u}$" style="vertical-align:-2px;
width:19px;
height:10px" class="math gen" /> variables; its cross terms measure the correlation between the variables. If a component <img src="images/img-0824.png" alt="$V_{ij} > 0$" style="vertical-align:-5px;
width:55px;
height:17px" class="math gen" />, it implies that higher estimates of the coefficient <img src="images/img-0821.png" alt="$u_ i$" style="vertical-align:-2px;
width:14px;
height:10px" class="math gen" /> make higher estimates of <img src="images/img-0825.png" alt="$u_ j$" style="vertical-align:-5px;
width:16px;
height:13px" class="math gen" /> more favourable also; if <img src="images/img-0826.png" alt="$V_{ij} < 0$" style="vertical-align:-5px;
width:55px;
height:17px" class="math gen" />, the converse is true. </p><p>It is a standard statistical result that <img src="images/img-0827.png" alt="$\mathbf{V} = (-\mathbf{A})^{-1}$" style="vertical-align:-4px;
width:101px;
height:20px" class="math gen" />. In the remainder of this section we prove this; readers who are willing to accept this may skip onto Section <a href="sec-correlation_matrix.html">C.5</a>. </p><p>Using <img src="images/img-0828.png" alt="$\Delta u_ i$" style="vertical-align:-2px;
width:28px;
height:14px" class="math gen" /> to denote <img src="images/img-0829.png" alt="$\left(u_ i - u^0_ i\right)$" style="vertical-align:-6px;
width:69px;
height:22px" class="math gen" />, we may proceed by rewriting Equation (<a></a>) as: </p><table id="a0000001677" cellpadding="7" width="100%" cellspacing="0" class="eqnarray">
<tr id="a0000001678">
<td style="width:40%"> </td>
<td style="vertical-align:middle; text-align:right"><img src="images/img-0830.png" alt="$\displaystyle V_{ij} $" style="vertical-align:-5px; width:21px; height:17px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:center"><img src="images/img-0058.png" alt="$\displaystyle = $" style="vertical-align:2px; width:12px; height:4px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:left"><img src="images/img-0831.png" alt="$\displaystyle \idotsint _{u_ i=-\infty }^{\infty } \Delta u_ i \Delta u_ j \mathrm{P}\left( \mathbf{u} | \left\{ \mathbf{x}_ i, f_ i, \sigma _ i \right\} \right) \, \mathrm{d}^{n_\mathrm {u}}\mathbf{u} $" style="vertical-align:-18px; width:348px; height:44px" class="math gen" /></td>
<td style="width:40%"> </td>
<td style="width:20%" class="eqnnum"><span>(<span>C.9</span>)</span></td>
</tr><tr id="a0000001679">
<td style="width:40%"> </td>
<td style="vertical-align:middle; text-align:right"><img src="images/img-0175.png" alt="$\displaystyle $" style="vertical-align:0px; width:1px; height:1px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:center"><img src="images/img-0058.png" alt="$\displaystyle = $" style="vertical-align:2px; width:12px; height:4px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:left"><img src="images/img-0832.png" alt="$\displaystyle \frac{ \idotsint _{u_ i=-\infty }^{\infty } \Delta u_ i \Delta u_ j \exp (-Q) \, \mathrm{d}^{n_\mathrm {u}}\mathbf{u} }{ \idotsint _{u_ i=-\infty }^{\infty } \exp (-Q) \, \mathrm{d}^{n_\mathrm {u}}\mathbf{u} } \nonumber $" style="vertical-align:-22px; width:278px; height:53px" class="math gen" /></td>
<td style="width:40%"> </td>
<td style="width:20%" class="eqnnum"> </td>
</tr>
</table><p>The normalisation factor in the denominator of this expression, which we denote as <img src="images/img-0833.png" alt="$Z$" style="vertical-align:0px;
width:12px;
height:12px" class="math gen" />, the <i class="itshape">partition function</i>, may be evaluated by <img src="images/img-0794.png" alt="$n_\mathrm {u}$" style="vertical-align:-2px;
width:19px;
height:10px" class="math gen" />-dimensional Gaussian integration, and is a standard result: </p><table id="a0000001680" cellpadding="7" width="100%" cellspacing="0" class="eqnarray">
<tr id="a0000001681">
<td style="width:40%"> </td>
<td style="vertical-align:middle; text-align:right"><img src="images/img-0834.png" alt="$\displaystyle Z $" style="vertical-align:0px; width:12px; height:12px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:center"><img src="images/img-0058.png" alt="$\displaystyle = $" style="vertical-align:2px; width:12px; height:4px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:left"><img src="images/img-0835.png" alt="$\displaystyle \idotsint _{u_ i=-\infty }^{\infty } \exp \left(\frac{1}{2} \Delta \mathbf{u}^\mathbf {T} \mathbf{A} \Delta \mathbf{u} \right) \, \mathrm{d}^{n_\mathrm {u}}\mathbf{u} $" style="vertical-align:-18px; width:305px; height:45px" class="math gen" /></td>
<td style="width:40%"> </td>
<td style="width:20%" class="eqnnum"><span>(<span>C.10</span>)</span></td>
</tr><tr id="a0000001682">
<td style="width:40%"> </td>
<td style="vertical-align:middle; text-align:right"><img src="images/img-0175.png" alt="$\displaystyle $" style="vertical-align:0px; width:1px; height:1px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:center"><img src="images/img-0058.png" alt="$\displaystyle = $" style="vertical-align:2px; width:12px; height:4px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:left"><img src="images/img-0836.png" alt="$\displaystyle \frac{(2\pi )^{n_\mathrm {u}/2}}{\mathrm{Det}(\mathbf{-A})} \nonumber $" style="vertical-align:-17px; width:74px; height:47px" class="math gen" /></td>
<td style="width:40%"> </td>
<td style="width:20%" class="eqnnum"> </td>
</tr>
</table><p>Differentiating <img src="images/img-0837.png" alt="$\log _ e(Z)$" style="vertical-align:-4px;
width:57px;
height:18px" class="math gen" /> with respect of any given component of the Hessian matrix <img src="images/img-0838.png" alt="$A_{ij}$" style="vertical-align:-5px;
width:25px;
height:17px" class="math gen" /> yields: </p><table id="a0000001683" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="images/img-0839.png" alt="\begin{equation} -2 \frac{\partial }{\partial A_{ij}} \left[ \log _ e(Z) \right] = \frac{1}{Z} \idotsint _{u_ i=-\infty }^{\infty } \Delta u_ i \Delta u_ j \exp (-Q) \, \mathrm{d}^{n_\mathrm {u}}\mathbf{u} \end{equation}" style="width:547px;
height:44px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"><span>(<span>C.11</span>)</span></td>
</tr>
</table><p>which we may identify as equalling <img src="images/img-0822.png" alt="$V_{ij}$" style="vertical-align:-5px;
width:21px;
height:17px" class="math gen" />: </p><table id="<plasTeX.TeXFragment object at 0xb279b3c>" cellpadding="7" width="100%" cellspacing="0" class="eqnarray">
<tr id="a0000001684">
<td style="width:40%"> </td>
<td style="vertical-align:middle; text-align:right"><img src="images/img-0840.png" alt="$\displaystyle \label{eqa:v_ zrelate} V_{ij} $" style="vertical-align:-5px; width:21px; height:17px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:center"><img src="images/img-0058.png" alt="$\displaystyle = $" style="vertical-align:2px; width:12px; height:4px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:left"><img src="images/img-0841.png" alt="$\displaystyle -2 \frac{\partial }{\partial A_{ij}} \left[ \log _ e(Z) \right] $" style="vertical-align:-19px; width:132px; height:44px" class="math gen" /></td>
<td style="width:40%"> </td>
<td style="width:20%" class="eqnnum"><span>(<span>C.12</span>)</span></td>
</tr><tr id="a0000001685">
<td style="width:40%"> </td>
<td style="vertical-align:middle; text-align:right"><img src="images/img-0175.png" alt="$\displaystyle $" style="vertical-align:0px; width:1px; height:1px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:center"><img src="images/img-0058.png" alt="$\displaystyle = $" style="vertical-align:2px; width:12px; height:4px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:left"><img src="images/img-0842.png" alt="$\displaystyle -2 \frac{\partial }{\partial A_{ij}} \left[ \log _ e((2\pi )^{n_\mathrm {u}/2}) - \log _ e(\mathrm{Det}(\mathbf{-A})) \right] \nonumber $" style="vertical-align:-19px; width:329px; height:44px" class="math gen" /></td>
<td style="width:40%"> </td>
<td style="width:20%" class="eqnnum"> </td>
</tr><tr id="a0000001686">
<td style="width:40%"> </td>
<td style="vertical-align:middle; text-align:right"><img src="images/img-0175.png" alt="$\displaystyle $" style="vertical-align:0px; width:1px; height:1px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:center"><img src="images/img-0058.png" alt="$\displaystyle = $" style="vertical-align:2px; width:12px; height:4px" class="math gen" /></td>
<td style="vertical-align:middle; text-align:left"><img src="images/img-0843.png" alt="$\displaystyle 2 \frac{\partial }{\partial A_{ij}} \left[ \log _ e(\mathrm{Det}(\mathbf{-A})) \right] \nonumber $" style="vertical-align:-19px; width:177px; height:44px" class="math gen" /></td>
<td style="width:40%"> </td>
<td style="width:20%" class="eqnnum"> </td>
</tr>
</table><p>This expression may be simplified by recalling that the determinant of a matrix is equal to the scalar product of any of its rows with its cofactors, yielding the result: </p><table id="a0000001687" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="images/img-0844.png" alt="\begin{equation} \frac{\partial }{\partial A_{ij}} \left[\mathrm{Det}(\mathbf{-A})\right] = -a_{ij} \end{equation}" style="width:389px;
height:44px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"><span>(<span>C.13</span>)</span></td>
</tr>
</table><p>where <img src="images/img-0845.png" alt="$a_{ij}$" style="vertical-align:-5px;
width:21px;
height:13px" class="math gen" /> is the cofactor of <img src="images/img-0838.png" alt="$A_{ij}$" style="vertical-align:-5px;
width:25px;
height:17px" class="math gen" />. Substituting this into Equation (<a></a>) yields: </p><table id="a0000001688" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="images/img-0846.png" alt="\begin{equation} V_{ij} = \frac{-a_{ij}}{\mathrm{Det}(\mathbf{-A})} \end{equation}" style="width:358px;
height:37px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"><span>(<span>C.14</span>)</span></td>
</tr>
</table><p>Recalling that the adjoint <img src="images/img-0847.png" alt="$\mathbf{A}^\dagger $" style="vertical-align:0px;
width:22px;
height:16px" class="math gen" /> of the Hessian matrix is the matrix of cofactors of its transpose, and that <img src="images/img-0812.png" alt="$\mathbf{A}$" style="vertical-align:0px;
width:15px;
height:12px" class="math gen" /> is symmetric, we may write: </p><table id="a0000001689" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="images/img-0848.png" alt="\begin{equation} V_{ij} = \frac{-\mathbf{A}^\dagger }{\mathrm{Det}(\mathbf{-A})} \equiv (-\mathbf{A})^{-1} \end{equation}" style="width:402px;
height:44px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"><span>(<span>C.15</span>)</span></td>
</tr>
</table><p>which proves the result stated earlier. </p></div>
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