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I understand that when one have multiple independent variable that follows $N\sim(0,1)$, denoted as $A$ if we have a correlation matrix $R$, we can generate correlated variables $B$ that are normally distributed using the Cholesky decomposition as stated here: Given the Cholesky decomposition of $R$ be $$ R=LL^t $$ Then we have $$ LA=B $$

However, is there any way for one to simulate correlated $\chi^2$ given multiple independent $\chi^2(1)$ variables?

I have tried to apply the cholesky directly to the $R^2$ and solve the following: $$ R^2=LL^t $$ $$ LA^2=B^2 $$ which doesn't give the expected result because $B^2$ isn't $\chi^2$ distributed although they do have the expected correlation $R^2$. I have also try to squaring the whole equation such that I have $$ LAA^tL^t =BB^t $$ Yet the result is difficult to interpret as $BB^t$ is a square matrix.

So is there any way to generate correlated $\chi^2$ variables given multiple independent $\chi^2(1)$ variables? Or will $A$ need to be following some other distribution for $B$ to be correlated $\chi^2$ distribution?

Thank you

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Sam
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  • could Wishart distribution be of some help? – Math-fun Feb 18 '15 at 07:01
  • Thank you @Mehdi, I will definitely check on Wishart distribution. However, this is the first time I heard about this distribution, is it possible to show the use of Wishart distribution in simulating correlated chi square in matrix form??? – Sam Feb 18 '15 at 07:08
  • I just checked closely, sorry for possibly a bad comment, it seems Wishart does not offer an straight forward way to your simulation task. But what if you generate two, say, correlated normals and form two chi out of them, then you have correlated chi squared random variables with some correlation coefficient which is a "calculable" function of the correlation structure of the original normals. this should work. – Math-fun Feb 18 '15 at 07:29

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Here is a little sketch that might be helpful.

Let $\displaystyle (X_1,X_2)$ be bivariate mean zero normal rvs with unit variance and correlation $\displaystyle \rho$ and with density $\displaystyle f(x_1,x_2)$. Now define $\displaystyle Y_1=X_1^2$ and $\displaystyle Y_2=X_2^2$, each has, clearly, $\displaystyle \chi^2_{(1)}$. \begin{align} corr(Y_1,Y_2)&=\frac{EX_1^2 X_2^2-EX_1^2EX_2^2}{\sqrt{var(X_1^2)var(X_2^2)}}\\ &=\frac{\int\int x_1^2x_2^2f(x_1,x_2)dx_1dx_2-1}{2}\\ &=\rho^2 \end{align}

Therefore if you want to generate two chi distributed rvs with correlation $\displaystyle \rho^2$, you need to generate a mean zero bivariate normal with correlation $\rho$ and unit variance then square them. This could be generalized to multivariate normals, but things get a bit messy.

Math-fun
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  • But then is it possible for me to obtain the $A$? I am currently interested in the matrix equation $RA=B$. With the process mentioned, I can simulate correlated $\chi^2$ variable but can't seems to get the $A$... – Sam Feb 18 '15 at 10:27