How to determine the mean and variance of a transformed normal distribution

Question

Suppose that we have a vector $\vec{m}\sim\mathcal{N}(0,\sigma^2 I)$ and a constant $\vec{r}$. What is the distribution of $2\vec{r}^T\vec{m}$? I wasn't too sure how to approach this because we're dealing with vectors.

Answer 1

Multivariate normal distribution has the following property:

If $X:p\times1 \sim N_{p}(\mu,\Sigma)$ and let $a^{T}$ be vector of real numbers, then $a^{T}X\sim N_{1}(a^{T}\mu, a^{T}\Sigma a)$

Mean = $E(a^{T}X)=a^{T}E(X)=a^{T}\mu$

Variance: \begin{eqnarray*} E\left\{(a^{T}X-E(a^{T}X)\right\}^{2}&=&E\left\{(a^{T}X-E(a^{T}X)(a^{T}X-E(a^{T}X)\right\}\\ &=&E\left\{[a^{T}X-E(a^{T}X)][X^{T}a-E(X^{T}a)]\right\}\\ &=&E\left\{a^{T}[X-E(X)][X^{T}-E(X^{T})]a\right\}\\ &=&a^{T}E\left([X-E(X)][X^{T}-E(X^{T})]\right)a\\ &=&a^{T}E\left([X-E(X)][X-E(X]^{T}\right)a\\ &=&a^{T}\Sigma a \end{eqnarray*}

Answer 2

The answer is $2r^Tm\sim N(0,4\sigma^2\sum_{i=1}^kr^2_i)$ assuming $r$ and $m$ are $k$ dimensional vector.

It is very trial task. From the covariance structure of $m$, it can be said that the components of $m$ are independent. Then rest are followed.