2
votes
Variance of product of Gaussian random variables
The first thing to say is that if we define a new random variable $X_i$=$h_ir_i$, then each possible $X_i$,$X_j$ where $i\neq j$, will be independent.
Therefore, we are able to say
$$Var(\sum_i^nX_i)=\...
2
votes
Accepted
Variance of product of Gaussian random variables
First just consider the individual components, which are gaussian r.v., call them $r,h$, $$r\sim N(\mu,\sigma^2),h\sim N(0,\sigma_h^2)$$
$$
Var(rh)=\mathbb E(r^2h^2)-\mathbb E(rh)^2=\mathbb E(r^2)\...
2
votes
Accepted
Why $[(\mathbf{I}_N-\mathbf{A}^\top \mathbf{A})\mathbf{x}]$ is Gaussian with i.i.d. Gaussian $\mathbf{A}$?
Partial answer. I hope I do not make mistakes.
The vectors $A^\top Ax$ and $(I-A^\top Ax)$ are NOT gaussian, and my impression is that the gaussian character is just an asymptotic law as $1 << M ...
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