# How to show that $X = LY$ where $Y\sim N(0,I)$?

Let $$X\sim MVN(0,\Sigma)$$ denote a random vector having the multivariate normal distribution with mean $$0$$ and covariance matrix $$\Sigma$$.

Suppose we want to sample from $$X\sim MVN(0,\Sigma)$$. Because $$\Sigma$$ is a positive semi-definite matrix (as it is a covariance matrix), there exists some matrix, $$L$$, such that $$\Sigma=LL',$$ where $$L'$$ is the transpose of $$L$$.

Observe now that we can write $$𝑋 = 𝐿𝑌$$, where $$Y\sim MVN(0,I)$$, where $$I$$ is the identity matrix.

Question: How to show the bolded equation? In particular, $$X= LY$$.

• Think about what $Cov[LY]$ would be. Also, are you aware that a linear combination of normally distributed random variables/vectors is, again, normally distributed? Sep 11 '19 at 5:37
• I have added the tag cholesky even though your question isn't specifically about that particular decomposition because it's (a) an example of such a decomposition and (b) many of the answers on site that already answer this question will have the tag, so it may make it easier to locate them. Sep 12 '19 at 6:10
• For some relevant posts see stats.stackexchange.com/a/238977/805 for example or stats.stackexchange.com/a/89830/805 Sep 12 '19 at 8:38

Opening on @baruuum's comment, let $$Y\sim MVN(0,I)$$; then $$E[LY]=LE[Y]=0$$, and \begin{align}\operatorname{var}(LY)&=\operatorname{cov}(LY,LY)=E[(LY)(LY)^T]-\underbrace{E[LY]E[LY]^T}_0\\&=E[LYY^TL]=L\underbrace{E[YY^T]}_{I}L^T=LL^T\end{align} So, by multiplying a MVN with zero-mean and idendity covariance with $$L$$ from left, you'll obtain a new jointly normal random vector with mean $$0$$ and covariance $$LL^T=\Sigma_X$$ as we wanted.