# 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? – baruuum 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. – Glen_b -Reinstate Monica 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 – Glen_b -Reinstate Monica 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.