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
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.