What does it mean to say that the random $d$-vector $X$ has a multivariate normal distribution with mean $\mu$ and covariance matrix $\Sigma$ ?

Suppose that $X \sim N_{d}\left(0, \sigma^{2} I_{d}\right)$, and that for each $j=1, \ldots, J, A_{j}$ is a $d_{j} \times d$ matrix. Suppose further that

$$A_{j} A_{i}^{T}=0$$

for $j \neq i$. Prove that the random vectors $Y_{j} \equiv A_{j} X$ are independent, and that $Y \equiv\left(Y_{1}^{T}, \ldots, Y_{J}^{T}\right)^{T}$ has a multivariate normal distribution.

[Hint: Random vectors are independent if their joint $M G F$ is the product of their individual MGFs.]

If $Z_{1}, \ldots, Z_{n}$ is an independent sample from a univariate $N\left(\mu, \sigma^{2}\right)$ distribution, prove that the sample variance $S_{Z Z} \equiv(n-1)^{-1} \sum_{i=1}^{n}\left(Z_{i}-\bar{Z}\right)^{2}$ and the sample mean $\bar{Z} \equiv$ $n^{-1} \sum_{i=1}^{n} Z_{i}$ are independent.