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Let $\mathbf{u}$ and $\mathbf{v}$ 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 $\mathbf{v}$ is a column of a unitray matrix $\mathbf{V}$. In addition, we define $\mathbf{G}$ as an $M \times N$ matrix with i.i.d $\mathcal{CN}(0,1)$ elements. I am looking for the distribution of the scalar given by $$| \mathbf{u}^H \mathbf{G} \mathbf{v} |^2, $$ where $\mathbf{u}^H$ denotes the hermitian of $\mathbf{u}$.

Thank you!

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If $G$ has centered, Gaussian i.i.d entries, then each row is multivariate Gaussian distributed. If $u$ is any vector, then each component in $Gu$ is a linear combination of the components of a multivariate Gaussian, and so is a Gaussian distributed vector of mean zero. Each component of this vector is still independent of the others, so this resulting vector is also multivariate Gaussian distributed. Consequently, for any vector $v$, $v^t(Gu)$ is also a linear combination of components of a multivariate gaussian, and so is a centered univariate Gaussian.

If we let $\sigma$ be the standard deviation of this resulting univaraite normal, then by $\frac{1}{\sigma} v^t G u $ is a unit normal, and by definition, $\frac{1}{\sigma} \left| v^t G u \right|^2$ is $\chi^2_1$ distributed.

To compute $\sigma$ we can apply the following result

If $X_1 \sim \mathcal{N}(0, \sigma_1)$ and $X_2 \sim \mathcal{N}(0, \sigma_2)$ are two independent Gaussian distributions, then $aX_1 + bX_2$ is also Gaussian, with mean zero, and standard deviation $\sqrt{a^2 \sigma_1^2 + b^2 \sigma_2^2}$.

The expression for the standard deviation can be derived by expanding $\mathrm{cov}(aX_1 + bX_2, aX_1 + bX_2)$.

Using this result, each of the components of the vector $Gu$ is an independent, centered normal with standard deviation

$$ \sqrt{\sum_i u_i^2}. $$

Using it once again, the final univariate normal $v^t G u$ is a centered Gaussian with standard deviation

$$ \sqrt{\sum_j \left( v_i^2 \sum_i u_i^2 \right)} .$$

In your case, with $u$ and $v$ unit vectors, this of course reduces to

$$ \sqrt{\sum_j \left( v_j^2 \sum_i u_i^2 \right)} = \sqrt{\sum_j \left( v_j^2 \right)} = 1 $$

so $v^t G u$ is a standard normal. Hence $\left| v^t G u \right|^2 \sim \chi^2_1$.

Here's a short demonstration in R you can play with

set.seed(154)

normalize <- function(v) {v / sqrt(sum(v*v))}
u <- normalize(c(.25, .75))
v <- normalize(c(.1, .4))

samples <- c()
for(i in 1:10000) {
  G <- rnorm(4)
  dim(G) <- c(2,2)
  x <- t(u) %*% G %*% c(v)
  samples <- c(samples, x)
}

hist(samples, breaks = 50)

enter image description here

And a validation of the standard deviation calculation

sd(samples)
[1] 0.9955483
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