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This question is related to my previous question in here

So I was trying to simulate correlated $\chi^2(1)$ random variables given the desired co-variance matrix. However, it seems like the only possible route was the following

Given the co-variance matrix $R$, then we perform the cholesky decomposition such that $$ R=LL^t $$ Then we simulate a number of independent vector $A$ with $N\sim(0,1)$, we can then simulate $B$ using $$ LA=B $$ where B should be correlated $N\sim(0,1)$, then we can obtain the correlated $\chi^2(1)$ variables.

When playing around with the variables, I thought that given $$ LA=B $$ and that the correlated $\chi^2$ are simply $B^2$, then by squaring the equation, I should be able to get the corresponding independent $\chi^2$ variables e.g. $$ LAA^tL^t=BB^t $$ However, I note that $AA^t$ and $BB^t$ are both square matrix, so I am not sure whether if they are both $\chi^2$ distributed. What interested me most is that: given a set of correlated random variables that are normally distributed, we can simply use the cholesky distribution to orthogonalizing the variables or un-correlate them. Will there be similar simple equations that we can use to perform the same trick in orthogonalizing $\chi^2$ or even non-centric $\chi^2$ variables?

kjetil b halvorsen
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Sam
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    For simulating correlated $\chi^2(1)$ variates, why not just square a bivariate Normal distribution? – whuber Feb 23 '15 at 18:01
  • I can, however, I am interested in whether if there is any method that can simulate correlated $\chi^2(1)$ from independent $\chi^2(1)$ variables. – Sam Feb 24 '15 at 04:09
  • Sure: multiply the square roots of the $\chi^2(1)$ variates by independent Rademacher variables. The result is a collection of iid standard Normal variates. Use a linear combination of them to generate a multivariate Normal variable and square its components. The relationship between the two covariance matrices is simple, but requires all correlations among the $\chi^2$ variates to be non-negative. – whuber Dec 14 '22 at 21:19
  • You might find this related answer using the spectral decomposition to be of assistance. – Ben Dec 14 '22 at 22:42

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