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Let $\mathbf{u}$ and $\mathbf{v}_i$ be an $M \times 1$ and $N \times 1$ vectors of unit norm, respectively. $\mathbf{u}$ is a column of a unitary matrix $\mathbf{U}$ and the $\mathbf{v}_i$ are columns of a unitary matrix $\mathbf{V}$. In addition, we define $\mathbf{F}$ as an $M \times N$ matrix with i.i.d. $\mathcal{CN}(0,1)$ elements, meaning that $\mathbf{F}$ is a $M \times N$ complex Gaussian matrix where each element has zero mean and unit variance.
I am looking for the distribution of the following sum $$\sum_{i=1}^d | \mathbf{u}^H \mathbf{F} \mathbf{v}_i |^2, $$ where $\mathbf{u}^H$ denotes the hermitian of $\mathbf{u}$.

I know that each $| \mathbf{u}^H \mathbf{F} \mathbf{v}_i |^2$ is exponentially distributed with parameter $1$. The problem is that I am not sure if these random variables are independent of each other.. Any idea ?

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Check whether $u^HFv_i$ are independent of $u^HFv_j$. This should be easy, since these two variables being linear combinations of normal variables are normal, so checking independence is the same as checking whether covariance is zero. If the variables are independent then their squares will be independent too.

Update: We have $Eu^HFv_i=0$ for each $i$, since $EF=0$. Thus

$$cov(u^HFv_i,u^HFv_j)=Eu^HFv_iv_j^HF^Hu,$$

Now since $v_i$ come from unitary matrix $V$, we have that $v_i^Hv_j=0$. After some algebra it is possible to see that this gives us

$$EFv_iv_j^HF^H=0.$$

The key is to try to write down the element of this matrix and see that either the products in that element are zero because of the iid elements, or the element is zero because $v_i^Hv_j=0$.

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  • $\begingroup$ Ex: $Cov(x_1,x_2)=\mathbb{E} \{ |u^H Fv_1 |^2 |u^H Fv_2 |^2 \} - \mathbb{E}\{ |u^H Fv_1 |^2 \} \mathbb{E}\{ |u^H Fv_2 |^2 \} =\mathbb{E} \{ |u^H Fv_1 |^2 |u^H Fv_2 |^2 \} -1$. How to continue ? $\endgroup$ – tam Nov 17 '15 at 8:24
  • $\begingroup$ Do not check the squares, check linear combinations, which I mentioned in my answers. $\endgroup$ – mpiktas Nov 17 '15 at 10:10
  • $\begingroup$ Do you mean $Cov(x_1,x_2)=\mathbb{E}\{ u^H F v_1 u^H F v_2 \}-\mathbb{E}\{ u^H F v_1 \} \mathbb{E}\{ u^H F v_2 \} $ ? Could you please detail your answer ? (because I am not familiar with this type of derivations) $\endgroup$ – tam Nov 17 '15 at 10:33

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