CorrelationMatrix

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Revisjon per 25. mar. 2011 kl. 23:16 av Read@uio.no (diskusjon | bidrag) (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
[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 [math]\lt \delta_i\gt = 0[/math] and [math]\lt \delta_i*\delta_j\gt = \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]).


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].

Expanding the covariance matrix one finds [math]C_{ij}=\lt x_i*x_j\gt -\bar{x_i}\bar{x_j}\lt mathrm\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 [math]C_{12}=C_{21}=\beta*\lambda[/math]. Similar substitution for [math]C_{ij}[/math] gives 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].