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Let $v\in \mathbb C^n$ be an $n \times 1$ complex unit norm vector (i.e. $\sum|v_i|^2=1$). Also, let $G$ be an $m \times n$ matrix with i.i.d. $\mathcal{CN}(0,1)$ elements, meaning that each element $G_{ij}$ follows a standard complex normal distribution i.e. is a complex number with real and imaginary parts independently distributed as $\mathcal N(0,1/2)$.

What is the expectation $ \mathbb{E}\{G v v^H G^H\} $, where $v^H$ denotes the conjugate transpose of $v$? Note that this is a $m\times m$ matrix, and I need to compute the expected value of this matrix-valued random variable.

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Let $w = Gv$. Its elements are $w_j = \sum G_{ij}v_i$ and the expected value you want to compute is $\mathbb E[ww^*]$. So let's write it down:

\begin{align} \mathbb E[w_j w^*_k] &= \mathbb E\left[\sum_{i=1}^n G_{ij}v_i \sum_{l=1}^n G_{lk}^*v^*_l\right]=\sum_{i=1}^n\sum_{l=1}^nv_iv_l^*\mathbb E[G_{ij}G_{lk}^*]\\ &=\sum_{i=1}^n\sum_{l=1}^nv_iv_l^*\delta_{il}\delta_{jk} = \delta_{jk}\cdot\sum_{i=1}^n|v_i|^2 = \delta_{jk}, \end{align} which is zero when $j\ne k$ and one when $j=k$, i.e. $\mathbb E[ww^*]=\mathbf I$.

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  • $\begingroup$ Assume that the mean is $\mu$ ($\ne 0$) and the variance $=v$. Could you plz tell me if it is possible to calculate the expectation in this case? thank you! $\endgroup$
    – tam
    Commented Jan 22, 2016 at 17:15
  • $\begingroup$ Mean of what and variance and of what? $\endgroup$
    – amoeba
    Commented Jan 22, 2016 at 17:34
  • $\begingroup$ of the elements of $G$. $\mu$ and $v$ instead of $0$ and $1$. $\endgroup$
    – tam
    Commented Jan 22, 2016 at 17:37
  • $\begingroup$ Well, if the elements are independent then $\mathbb E[G_{ij}G_{lk}^*] = \mathbb E[G_{ij}]\mathbb E[G_{lk}^*]=\mu^2$ for two different elements and $\mu^2+v$ for the same element. So it seems that the expected matrix will have $\mu^2+v$ on the diagonal and $\mu^2$ elsewhere. $\endgroup$
    – amoeba
    Commented Jan 22, 2016 at 22:05
  • $\begingroup$ Let's take $n=2$. Then, $\sum_{i=1}^2 \sum_{l=1}^2 v_iv^*_l \mathbb{E}[G_{ij}G^*_{lk}]$. For $j \ne k$ we have: $v_1(v_1^*+v_2^*) \mu^2 + v_2 (v_1^*+v_2^*)\mu^2 = (1+v_1v_2^*+v_2v_1^*)\mu^2 \ne \mu^2$ or I am wrong ? $\endgroup$
    – tam
    Commented Jan 22, 2016 at 23:58

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