# Does using a Cholesky decomposition to generate correlated samples preserve the distributions they are supposed to be generated from?

If I want to generate $1$ random number each from $n$ distributions whose values are correlated, I can follow this procedure:

1. Starting with the Covariance matrix $C$, where each element $ij$ shows the covariance between variables of the $i$'th and $j$'th distribution, determine a matrix $L$ such that $LL^T = C$ (i.e. Cholesky decomposition -- of the two possible solutions, $L$ would be the one that is in lower-triangular form and $L^T$ is its transpose).

2. Generate n independent samples from distributions that have a mean equal to zero and standard deviation equal to 1. Group these n samples into the vector $\overrightarrow z$.

3. The vector of correlated random samples, $\overrightarrow x$ can be determined by the equation $\overrightarrow x = L \times \overrightarrow z$.

It is not difficult to show that the covariance matrix of $\overrightarrow x$ is identical to $C$. I suppose these values can then be given their correct means by adding $\overrightarrow \mu$ to it, where $\overrightarrow \mu$ is the vector of means from each of the n distributions.

Regarding all this, I have a few questions:

1. As I've said above, the covariance matrix of $\overrightarrow x$ is identical to $C$. Furthermore, the expectation of $\overrightarrow x$, $E[\overrightarrow x]$ will be identical to $\overrightarrow \mu$. But does this hold for all other moments? That is, even though the mean and standard deviation is preserved, if we did this an infinite amount of time and collected all the samples from the $i$'th component of $\overrightarrow x$, can we be assured that we could rebuild the $i$'th distribution? (I want to know if using this method in any way distorts the values.)

2. Do the $n$ distributions have to be identical?

## 1 Answer

1. No. it doesn't. For example, if your original distributions are exponential, but you use standard normal distribution in step 2, then the generated samples will follow fundamentally different distributions than the original ones, although they may have the same first and second moments. For what I am doing, I generate the samples from each original distribution independently, translate and scale the samples so they have zero mean and unit std/var, using them as $\vec{z}$. I didn't proof they will have the moments, but intuitively, they will have the same shapes.

2. No. there is no such requirement for Cholesky decomposition.