Expectation of inverse of complex Gaussian variables

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Hi there, I just have a mathematic problem. If we consider a Gaussian complex random variable vector , where each element in follows zeros mean and variance γ. Is there any close form with γ for ? where is the norm-2 operation. I have asked the same question in MathOverflow at https://mathoverflow.net/questions/436733/expectation-of-inverse-of-complex-gaussian-variables?noredirect=1#comment1125524_436733.
the people in mathoverflow showed that this expectation is infinity mathematically. But in matlab, we can find out that the above expectation can converge to a certain value. So there must be some error that I couldn't find out and misunderstanding. Really appreciate for any comments!
clc;close all;clear all;
num_loop=5000;
N=5;Eh=0;
for i=1:num_loop
h=sqrt(1/2)*(randn(N,1)+1i*randn(N,1));
Eh=Eh+1/norm(h)^2;
end
Eh=Eh/num_loop
  8 个评论
Paul
Paul 2023-1-3
编辑:Paul 2023-1-3
Then I guess you disagree with the commen on the mathoverflow page?
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Matt J
Matt J 2023-1-3
编辑:Matt J 2023-1-3
So there must be some error that I couldn't find out and misunderstanding.
The misunderstanding is that the expectation is infinite for when n=1, but for higher dimensions, it is finite. The general formula can be derived by adapting the material from here, leading to,
The integral can be evaluated for n>1 by integration by parts.
  22 个评论
mingcheng nie
mingcheng nie 2024-10-15
@Matt J@Paul Sorry guys I have an add-on question. What if in this question, n > 1, but each entry in has different variance, i.e., , is different for i= 1....n. Here, each entry of is still complex Gaussian with zero mean. How would the answer presented above changed in this siutation? I still assume there exist a closed-form solution as from the simulation I got fixed results.
Thank you guys so much for your patience!
Paul
Paul 2024-10-27,13:54
编辑:Paul 2024-10-27,16:08
If you can find the probability density function for the sum of independent, normal random variables that are not identically distributed, then you can try to proceed as shown above. Maybe there is a closed form expression. As to finding that density function, maybe this link will be of use. I think, but am not positive, that the Generalized Chi-Squared Distribution is what you're looking for.

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