3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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.)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.