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, $X= LY$.

  • $\begingroup$ 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? $\endgroup$ – baruuum Sep 11 '19 at 5:37
  • $\begingroup$ 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. $\endgroup$ – Glen_b Sep 12 '19 at 6:10
  • $\begingroup$ For some relevant posts see stats.stackexchange.com/a/238977/805 for example or stats.stackexchange.com/a/89830/805 $\endgroup$ – Glen_b 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.


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