I'm completely stuck on a problem involving a proof for the following: Suppose $X\sim$ MVNn$(\mu, \Sigma)$. If $A$ is a full rank $n\times n$ matrix, show that

$Y=AX\sim$ MVNn$(A\mu,A\Sigma A^T)$.

I don't know how to even start this problem. Is $X$ a matrix? How do you prove stuff about the mean and variance? Just looking for how to get started on this because I am lost.

  • 2
    $\begingroup$ $A$ is a matrix and $X$ is a vector. How you go about proving this depends a great deal on what you already know. It could be as trivial as computing the mean vector and the covariance matrix of $AX$, then saying that $AX$ follows a multivariate Normal distribution. If you don't have that, you may need to find the distribution of each component of $AX$, finding its mean and giving the entry of the covariance matrix at any row/column combination, then use the fact that the sum of jointly Normal random variables is also Normal. $\endgroup$
    – cgmil
    Commented Oct 21, 2021 at 4:47
  • $\begingroup$ Or if you know about characteristic functions that might be a good way to go $\endgroup$
    – Ali
    Commented Oct 21, 2021 at 8:09

1 Answer 1


We have that $X=(X_1,...,X_n)$ and each $X_i \sim N(\mu_i,\sigma^2_i)$

Given $A$ is a deterministic matrix, we have that $$E(Y)=E(AX)=AE(X)=A\mu$$

Since $E$ is a linear operator, i.e., $E(aX+bY)=aE(X)+bE(Y)$ given $X,Y$ are random.

I think that it's easier to think about it in a one dimensional case. Let $a\in \mathbb{R}$ and $X\sim N(\mu,\sigma^2)$, where $X,\mu,\sigma^2\in\mathbb{R}$. What would the expectation of $aX$ be? Well, since $a$ is a scalar, it would just scale the values of $X$ by $a$, so this means $\mu$ is also scaled by $a$. If you draw out a simple example for $X$ and scale it by some constant, you will see the mean is also shifted by $a$

For the variance, it's not that trivial as with the expectation. We have that $$\text{Var}(Y)=\text{Var}(AX)=E\left[(AX-E(AX))(AX-E(AX))^T\right]\\= E\left[(AX-AE(X))(AX-AE(X))^T\right]\\=E\left[\left((AX-AE(X)\right)\left((AX)^T-(AE(X))^T\right)\right]\\=E\left[((AX-E(X))(X^TA^T-E(X)^TA^T)\right]\\= E[A(X-E(X))(X-E(X))^TA^T]\\=AE[(X-E(X))(X-E(X))^T]A^T\\=A\text{Var}(X)A^T=A\Sigma A^T$$


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