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(Ny side: The covariance matrix is <br> <math>C_{ij} = <(x_i-\bar{x_i})(x_j-\bar{x_j})></math>.<br> Suppose <math>\delta_1,\ \delta_2,\ \delta_3</math> are three independent sources of normally-distr...)
 
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The covariance matrix is <br>
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The covariance matrix is <br> <math>C_{ij} = <(x_i-\bar{x_i})(x_j-\bar{x_j})></math>.<br> Suppose <math>\delta_1,\ \delta_2,\ \delta_3</math> are three independent sources of normally-distributed unit fluctuations (with <span class="texhtml"> &lt; δ<sub>''i''</sub> &gt;  = 0</span> and <span class="texhtml"> &lt; δ<sub>''i''</sub> * δ<sub>''j''</sub> &gt;  = δ<sub>''i''''j'''</sub></span>'''and where <span class="texhtml">δ<sub>''i'''</sub></span>'''''j'' is the kronikker delta function (1 for <span class="texhtml">''i'' = ''j''</span> and zero for <math>i\neq j</math>).  
<math>C_{ij} = <(x_i-\bar{x_i})(x_j-\bar{x_j})></math>.<br>
 
Suppose <math>\delta_1,\ \delta_2,\ \delta_3</math> are three independent sources of normally-distributed unit fluctuations (with <math><\delta_i> = 0</math>
 
and <math><\delta_i*\delta_j> = \delta_{ij}</math> and where <math>\delta_{ij}</math> is the kronikker delta function (1 for <math>i=j</math> and zero for <math>i\neq j</math>).
 
  
 
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<br> Pseudo-measurements of <span class="texhtml">(''x''<sub>1</sub>,''x''<sub>2</sub>)</span> can be generated from the expressions <span class="texhtml">(''x''<sub>1</sub>,''x''<sub>2</sub>) = (''x''<sub>10</sub> + α * δ<sub>1</sub> + β * δ<sub>2</sub>,''x''<sub>20</sub> + γ * δ<sub>3</sub> + λ * δ<sub>2</sub>)</span>.  
Pseudo-measurements of <math>(x_1,x_2)</math> can be generated from the expressions <math>(x_1,x_2)=(x_{10}+\alpha*\delta_1+\beta*\delta_2,x_{20}+\gamma*\delta_3+\lambda*\delta_2)</math>.  
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<blockquote>Expanding the covariance matrix one finds <math>C_{ij}=<x_i*x_j>-\bar{x_i}\bar{x_j}<math>.<br>
 
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Expanding the covariance matrix one finds  
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Using the properties of <math><\delta_i*\delta_j></math> above and <math><x_i>=\bar{x_i}</math>, one finds &lt;x_1*x_2&gt; = x_{10}*x_{20}+\beta*\lambda&lt;/math&gt; so that <span class="texhtml">''C''<sub>12</sub> = ''C''<sub>21</sub> = β * λ</span>. Similar substitution for <span class="texhtml">''C''<sub>''i''''j'''</sub></span>'''gives for the diagonal matrix elements <math>C_{11}=x_{10}^2+\alpha^2+\beta^2-x_{10}^2</math><br> and <br> <math>C_{22}=x_{20}^2+\gamma^2+\lambda^2-x_{10}^2</math>. so that C= '''
<math>C_{ij}=<x_i*x_j>-\bar{x_i}\bar{x_j}<mathrm>.  
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{| width="200" cellspacing="1" cellpadding="1" border="1"
Using the properties of <math><\delta_i*\delta_j></math> above and <math><x_i>=\bar{x_i}</math>, one finds <x_1*x_2> = x_{10}*x_{20}+\beta*\lambda</math> so that <math>C_{12}=C_{21}=\beta*\lambda</math>. Similar substitution for <math>C_{ij}</math> gives for the diagonal matrix elements
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<math>C_{11}=x_{10}^2+\alpha^2+\beta^2-x_{10}^2</math> and
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| <span class="texhtml">α<sup>2</sup> + β<sup>2</sup></span>
<math>C_{22}=x_{20}^2+\gamma^2+\lambda^2-x_{10}^2</math>.
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| <span class="texhtml">β * λ</span>
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| <math>\beta*\lambda</math>
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| <math>\gamma^2+\lambda^2</math>
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|}
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</blockquote>

Revisjonen fra 25. mar. 2011 kl. 23:21

The covariance matrix is
[math]C_{ij} = \lt (x_i-\bar{x_i})(x_j-\bar{x_j})\gt [/math].
Suppose [math]\delta_1,\ \delta_2,\ \delta_3[/math] are three independent sources of normally-distributed unit fluctuations (with < δi > = 0 and < δi * δj > = δi'jand where δij is the kronikker delta function (1 for i = j and zero for [math]i\neq j[/math]).


Pseudo-measurements of (x1,x2) can be generated from the expressions (x1,x2) = (x10 + α * δ1 + β * δ2,x20 + γ * δ3 + λ * δ2).

Expanding the covariance matrix one finds [math]C_{ij}=\lt x_i*x_j\gt -\bar{x_i}\bar{x_j}\lt math\gt .\lt br\gt Using the properties of \lt math\gt \lt \delta_i*\delta_j\gt [/math] above and [math]\lt x_i\gt =\bar{x_i}[/math], one finds <x_1*x_2> = x_{10}*x_{20}+\beta*\lambda</math> so that C12 = C21 = β * λ. Similar substitution for Ci'jgives for the diagonal matrix elements [math]C_{11}=x_{10}^2+\alpha^2+\beta^2-x_{10}^2[/math]
and
[math]C_{22}=x_{20}^2+\gamma^2+\lambda^2-x_{10}^2[/math]. so that C=
α2 + β2 β * λ
[math]\beta*\lambda[/math] [math]\gamma^2+\lambda^2[/math]