Dear all, I might just be blind, so forgive me if it is a trivial question. Given two normally distributed variables $x_1$ and $x_2$ (with zero mean), their correlation $c$ can be estimated from the samples $x_1 x_2$, $c = E[x_1 x_2]$ (where E denotes the expectation value). Now assume I want to estimate the square of the correlation, $c^2$. Unfortunately, the expectation value of $E[(x_1x_2)^2]$ for two zero-mean Gaussians is simply the product of their variances, so $E[x_1^2x_2^2] = \sigma_1^2\sigma_2^2$ where $\sigma_i^2$ is the variance of $x_i$. So how do I compute $c^2$ directly as an expectation value of independent samples?

Thanks so much! Wieland

  • $\begingroup$ $c$ is called the covariance, not correlation, between $x_1$ and $x_2$. $\endgroup$ – Lutz Mattner Mar 28 '13 at 14:48

$E[x_1^2 x_2^2]=\sigma_1^2 \sigma_2^2 + 2 c^2$

(an application of Isserlis theorem)

  • $\begingroup$ Thanks Carlo for pointing me to Isserlis theorem! Indeed I missed a 2$c^2$ here. This expression is still not completely what I wanted, but I should be able to use it for optimization purposes. Ideally I searched for an expectation value that is zero if the correlations are zero and positive otherwise. $\endgroup$ – Wieland Mar 28 '13 at 11:41

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