It is well known that if we have n i.i.d. observations of a normal random variable, then Cochran's theorem tells us that:

$\frac{(n-1)s^2}{\sigma^2} \widetilde{} χ^2_{n-1}$

But what if the samples are not i.i.d., but they are correlated? Is there any expression or theorem that gives us an insight on the distribution of the sample variance, assuming that we know a-priori the correlation matrix between the samples?

For example assume that the samples are extracted from a realization of a stationary gaussian random field and that the correlation function $\rho(\tau)$ is completely known.


1 Answer 1


Of course if you know the correlation matrix you can derive an expression for sample variance distribution - but I think it will not be a closed form distribution.

Let $X$ be your sample. You can express it as $X = AY+m$, where Y is iid standard normal, $A$ is a matrix and $m$ is deterministic mean.

In such a setting a variable of interest is $$\sum{(X_i-\bar{X})^2}=||X-\bar{X}||^2=||(AY+m-\bar{A}Y-m)||^2=||(A-\bar{A})Y||^2$$

Mean $m$ cancels out. You finally get a weighted sum of squares of standard normal variable (note that weights are not generally equal, so it is not a chi-squared distribution) and possibly some cross-products. If you know your correlation matrix you may derive matrix A (eg. Cholesky decomposition) and find the parameters, but I am not sure if this is useful for practical purpose.

  • 2
    $\begingroup$ +1 Because the square of a standard Normal has a $\chi^2$ distribution, which is a Gamma distribution, the distribution of a weighted sum of such independent variables can be found using techniques described at stats.stackexchange.com/questions/72479. The "cross products" will disappear upon using a suitable change of variable (use an eigenbasis). $\endgroup$
    – whuber
    Commented Jan 16, 2016 at 15:06
  • $\begingroup$ This is very interesting, could I ask you or @whuber to give me some more details on this approach and on the derivation of the A matrix? Basically is the A matrix just the transformation that allows me to move from correlated variables to independent ones? $\endgroup$
    – xanz
    Commented Jan 17, 2016 at 22:27
  • $\begingroup$ @whuber. Ok now I'm at the point where $\parallel(A-\bar{A})Y\parallel^2$. I get the A matrix by computing the eigenvalues and eigenvectors of the covariance matrix and multiplying the eigenvectors by the square root of the corresponding ordered eigenvalues. Now how do I cancel out the cross products, i.e. how do I change basis? $\endgroup$
    – xanz
    Commented Jan 18, 2016 at 19:29
  • $\begingroup$ Ok, nevermind, I think I got the solution to the problem. Thank you all $\endgroup$
    – xanz
    Commented Jan 19, 2016 at 17:13
  • $\begingroup$ Can anyone please comment on $m$ - is it the sample mean or expected value of $X$? they might not be the same, and if $m$ is the sample mean, I don't understand how we can ensure $Y$ will have zero mean. $\endgroup$
    – yoki
    Commented Mar 27, 2016 at 14:19

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