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.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1250137/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1250137/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1250142/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1250147/image.png)
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 Comments
Paul
on 3 Jan 2023
Edited: Paul
on 3 Jan 2023
Then I guess you disagree with the commen on the mathoverflow page?
link to comment (make sure to click on "Show 6 more comments"
Accepted Answer
Matt J
on 3 Jan 2023
Edited: Matt J
on 3 Jan 2023
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,
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1250362/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1250437/image.png)
The integral can be evaluated for n>1 by integration by parts.
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More Answers (1)
John D'Errico
on 3 Jan 2023
Edited: John D'Errico
on 3 Jan 2023
Sorry. I think you have a problem. Essentially, the moments of that variable are undefined. They have no computable value. I was pretty sure I recalled that fact, so I had to check.
It is in fact not an uncommon result. You should understand this is the case, because the normal distribtion has a significant probability of your value being near or very near to zero. Anytime that happens, the mean ends up being undefined as a parameter.
Do some reading in the link I showed. For example, we see there that the reciprocal normal distribution has all of its moments undefined. Now you are hoping to compute something slightly different, in that you want to compute the mean (i.e., the FIRST central moment) of the inverse of a complex Gaussian random variable.
Now, when I say undefined, I should be specific. Here, undefined means greater than any fixed value, or what we commonly call inf. That is different from a reslut like 0/0, which is returned as NaN, because you can make a convincing argument that 0/0 is ANY number you wish. It could be 0, 1, 2 inf, pi, ANYTHING. So we return a value of NaN when we try to compute 0.0, but inf when we compute 1/0.
Do you need more arguments for this case? Consider what is the mean (expectation) of a Cauchy random variable.
Again, if you read carefuly, you will see the mean is undefined. Yes, you can compute the sample mean, of samples of such a random variable. But that result is meaningless. Just as you could use a Monte Carlo simulation to compute the mean of your inverse of a complex Gaussian, you will get a valid result. But it will be a meaningless result, since the population mean of that transformation is undefined.
Does it matter here? Probably not. I'd need to spend a minute or so to decide which case applies to the expectation you want to compute. But it will have no finite value, of that I am totally confidant.
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