# Expected value of square of Euclidean norm of a gaussian random vector

If $X$ is a $p \times 1$ gaussian random vector with such that $X \sim \mathcal{N}(0,\Sigma)$. What is the expected value of the square of the euclidean norm i.e $E[\|AX\|_2^2]$? Here $A$ is a $n \times p$ matrix.

Setting $Y=AX$ we have $Y\sim\mathcal N(0,A\Sigma A^T)$ and:

$$\mathsf E\|AX\|_2^2=\operatorname{\mathsf E} Y^TY = \sum_{i=1}^n \operatorname{\mathsf E} Y_i^2 = \sum_{i=1}^n \operatorname{\mathsf{Var}}Y_i=\operatorname{\mathsf{tr}}(A\Sigma A^T)$$

Let $X^T=(X_1,X_2\dots,X_p)$ and let the entries of $A$ be denoted by $a_{i,j}$,$i=1,2,\dots n$ and $j=1,2,\dots,p$.

Then the $i^{th}$ elment of the $n$ vector $AX$ is

$$\sum_{j=1}^p X_ja_{ij}.$$

The expectation of the square of the same is

$$\sum_{k=1}^p\sum_{l=1}^pE[X_kX_l]a_{ik}a_{il}.$$

The expectations above can be calculated if $\Sigma$ is known. (The expectation vector is a zero vector.)

• It apparently follows a generalized chi square distribution. So I guess it might have a more straight forward formula ? –  redenzione11 Nov 21 '17 at 10:26
• Chi squared belongs to the sum of the squares of independent standard normal distributions. – zoli Nov 21 '17 at 10:28