Let $w \in \mathbb C^M$ be a unit norm complex vector. Also, let $s \in \mathbb C^M$ be a unit norm complex vector independent of $w$. We assume that $s$ and $w$ are i.i.d. isotropic vectors.

I am looking for the distribution of $s^* w$, where $s^*$ denotes the conjugate transpose of $s$. I also want to deduce the distribution of $|s^* w|$ and $|s^* w|^2$.

I saw the claim that $|s^* w|^2\sim\mathsf{Beta}(1,M-1)$ but without a proof.

There is a similar thread Distribution of a scalar product of two random unit vectors in $\mathbb{R}^D$ that shows that for real vectors scalar product $t =s^\top w$ is distributed such that $$(t+1)/2 \sim \mathsf{Beta}\big((D−1)/2,(D−1)/2\big),$$ but I am not sure how to generalize that to complex vectors.

  • $\begingroup$ @amoeba In the accepted answer, the distribution of $(t+1)/2$ is calculated and not of $t$ (which is the dot product). Aslo, here I have complex vectors. Maybe the this result can be used and I can deduce the distribution in the complex domain ? $\endgroup$ – tam Nov 19 '15 at 7:36
  • $\begingroup$ @amoeba I would appreciate if you could vote to reopen the question. Most of these questions came up in my research (sometimes I just need to confirm my expectation for the result). I hope this is fine and I am not disturbing the Cross Validated community :) $\endgroup$ – tam Nov 20 '15 at 12:07
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    $\begingroup$ Actually, I need the distribution of $s^* w$. And I want to see if $|s^* w|^2$ is Beta$(1,M-2)$ distributed; I found an author of a paper using this formula without providing the proof. $\endgroup$ – tam Nov 20 '15 at 12:17
  • $\begingroup$ After writing complex vectors in terms of their real and imaginary parts, this question becomes identical to the other one you reference. $\endgroup$ – whuber Nov 20 '15 at 15:22
  • $\begingroup$ @whuber Let us define $a=b=(D-1)/2$. From your answer, we can deduce that $t=2u-1$, so $t$ follows the general Beta distribution: $f(t)=\frac{(x+1)^{a-1} (1-x)^{a-1} }{ B(a,a)2^{2a-1} }= \frac{ (1-x^2)^{a-1} }{ B(a,a) 2^{2a-1} }$ (using this). It seems that this result is in contradiction with this result (where we get a beta distribution). Any idea ? $\endgroup$ – tam Nov 27 '15 at 17:07

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