# Proving properties of the Multivariate Normal Distribution?

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.

• $A$ 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. Commented Oct 21, 2021 at 4:47
• Or if you know about characteristic functions that might be a good way to go
– Ali
Commented Oct 21, 2021 at 8:09

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 $$\text{Var}(Y)=\text{Var}(AX)=E\left[(AX-E(AX))(AX-E(AX))^T\right]\\= E\left[(AX-AE(X))(AX-AE(X))^T\right]\\=E\left[\left((AX-AE(X)\right)\left((AX)^T-(AE(X))^T\right)\right]\\=E\left[((AX-E(X))(X^TA^T-E(X)^TA^T)\right]\\= E[A(X-E(X))(X-E(X))^TA^T]\\=AE[(X-E(X))(X-E(X))^T]A^T\\=A\text{Var}(X)A^T=A\Sigma A^T$$