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

Question

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 $ = $, 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 05:37
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, Sep 12 '19 at 06:10
For some relevant posts see https://stats.stackexchange.com/a/238977/805 for example or https://stats.stackexchange.com/a/89830/805 — Glen_b, Sep 12 '19 at 08:38

score 4 · Accepted Answer · answered Sep 11 '19 at 05:46

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.

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

1 Answers1