Expectation of inverse of complex Gaussian variables

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

[Bruno Luong]

People in mathoverflow are correct. Your code does not compute the expextation. It estimates the expectation for random variable that has an expectation, which is NOT the case.

[Matt J]

@Bruno Luong Still, though, the empirical means do seem to converge to something as we take more and more samples. It would be good to know if there is a mathematical reason for that, and if that limit can be predicted, even if it is not the statistical mean.

M=1e7;
N=5;


 h=sqrt(1/2)*(randn(N,M)+1i*randn(N,M));

 x=1./vecnorm(h).^2;

 Eh=cumsum(x,2)./(1:M);

 loglog(Eh)

[Bruno Luong]

@Matt J I think randn cannot capture the math correctness (true), sine it is diverge theoreticlly because normal random variable can get arrbitrally small create singular 1/r close to 0. However for whatever reason, randn rarely create number smaller than this

r=randn(1,1e7);
min(abs(r))

ans = 1.5919e-07

which is very far from arbitrary small to my book. The imperfect of randn() does not matter in most case, matter here.

Note that rand has the same imperfect filling toward 0

r=rand(1,1e7);
min(abs(r))

ans = 5.5373e-08

[Paul]

As I understand the linked mathoverflow page, if h is a complex scalar then the expectation is infinite (or does not exist?). But if h is a complex vector, then the expectation is finite.

[Bruno Luong]

@Paul, the singularity is 1/r when r goes to 0 for all cases. So the expectation is Inf for all cases as long as P(0) > 0.

[John D'Errico]

I see that @Bruno Luong agrees with me.

The problem is, if you compute a sample mean from a Monte Carlo simulation, you will get some finite (and random) result. But that does not tell you anything about the population mean. And you are asking how to compute an EXPECTATION, so the population mean. And that has no finite value.

[Paul]

@Bruno Luong

Then I guess you disagree with the commen on the mathoverflow page?

link to comment (make sure to click on "Show 6 more comments"

[Bruno Luong]

@Paul I see I miss to read the comment, thanks. I stand corrected.

Answer 1

[Matt J]

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.

[Bruno Luong]

Thinking more I believe you are correct with r^(n-3) = r^-2 * r^(n-1). The first term is from 1/h^2 the second from volumic integration in r. Well done Matt.

[Paul]

Why do you think there is an error or misunderstanding?

That r^(2n-3) is exactly what's stated in the comments section in the mathoverflow page for the case of vector h. Are you sure that it shouldn't be 2/gamma(N) out front?

N=5;
Gamma = 0.5

Gamma = 0.5000

2/gamma(N)*integral(@(r) exp(-r.^2/2/Gamma).*r.^(2*N-3),0,inf)

ans = 0.2500

Matches result above.

[Matt J]

Why do you think there is an error or misunderstanding?

Well, a misunderstanding on our part, because only you seem to have seen the additional comments!

Are you sure that it shouldn't be 2/gamma(N) out front?

Fixed it.

[Paul]

Looking through those comments I learned more about LatEx formatting than math!

Looks like there is a closed form solution:

syms r sigma real positive
syms n integer positive
E = int(exp(-r^2/2/sigma^2)*r^(2*n-3),r,0,inf)/sigma^(2*n)/2^(n-1)/gamma(n)

E = 

E = simplify(E)

E = 

simplify(subs(E,[n sigma],[5 sqrt(1/2)]))

ans = 

[Bruno Luong]

btw, OT question : does Answer supports TeX/LaTeX ?

[Matt J]

does Answer supports TeX/LaTeX ?

Some subset of it, yes. You insert it using the sigma toolbar button,

[Bruno Luong]

Thanks I miss it.

[Paul]

For anyone like me that doesn't know LaTex, just want point out the useful latex command that I sometimes use if I need a LaTex expression.

[mingcheng nie]

Thank you all guys for your wonderful arguments, answers and explanantions!!

Now I know that when I have vector , then the solution of the formula in the question is finite and has closed-form.

[Paul]

@Matt J

In your development is there an assumption that all of the random variables are independent? Rerunning the numerical experiment appears to show that assumption makes a difference.

Assume h is 3 x 1, all RVs are independent, standard normal.

M=1e7;
N=3;
sigma = 1;
h = sigma*(randn(N,M)+1i*randn(N,M));
x = 1./vecnorm(h).^2;
Eh = cumsum(x,2)./(1:M);
figure
loglog(Eh)
hold on

Closed form expression, matches blue curve

1/2/sigma^2/(N-1)

ans = 0.2500

Make the real RVs not independent, same for the imaginary RVs.

Sigma = sigma^2*[1 .8 .9; .8 1 .7;.9 .7 1];
Xr = mvnrnd(zeros(1,3),Sigma,M);
Xi = mvnrnd(zeros(1,3),Sigma,M);
h = (Xr + 1i*Xi).';
x = 1./vecnorm(h).^2;
Eh = cumsum(x,2)./(1:M);
loglog(Eh)

[Matt J]

Yes, that was the assumption.

[Paul]

Alternative derivation, which at heart is probably the same as @Matt J's, but just makes use of known results rather than deriving from scratch.

Let c(x,k) be the pdf of the Chi-square distribution with k degrees of freedom

syms sigma real positive
syms k n integer positive
syms x real positive
c(x,k) = x^(k/2-1)*exp(-x/2)/2^(k/2)/gamma(k/2)

c(x, k) = 

Let Xi, i = 1:k be i.i.d. normal random variables with mean = 0 and variance sigma^2.

Let Z = sum(Xi^2). The pdf of Z is then

z(x,k) = c(x/sigma^2,k)/sigma^2;

Let H = 1/Z. The expected value of H is

Eh = simplify(int(simplify(z(x,k)/x),x,0,inf))

Eh = 

If the vector h is n x 1, then k = 2*n

Eh = simplify(subs(Eh,k,2*n))

Eh = 

[Matt J]

Pretty amazing that it simplifies that much!

[Bruno Luong]

Wait a minute, I though the expectation is Inf for n=2 from the integral with r^(n-3) inside the integral, so it diverges like a log. Why I can't see this in the simplified formula?

[Paul]

n is the dimension of the vector, not the number of variables, which is 2*n (real + imaginary). So if h is a scalar, then n = 1 and Eh is undefined.

[mingcheng nie]

Dear Pual@Paul, I just a bit confuse that may I know if is a complex Guassian variables vector, which means 5 elements in this vector, then should it be n=2 or n=5? Here n=2 is due to h is real+imaginary; n=5 is that you memtioned 'if h is n×1....'.

[mingcheng nie]

Hi guys@Matt J, @Paul,I have one more question that if the variance of Guassian complex is random, not fixed, for example, if we consider the exponential delay profile ,where we have a total number of P of τ and they are independently uniformly distributed from . Then can can we compute the close form ?

[Matt J]

If h is no longer Gaussian, it's a whole new problem.

[Paul]

If h is a 5 x 1 vectory, then n = 5.

[mingcheng nie]

@Matt J 'If h is no longer Gaussian, it's a whole new problem.' h is still Guassian, but its variance is no longer a constant for n variables, it will follow the exponential delay profile.

@Paul 'If h is a 5 x 1 vectory, then n = 5.' Ok got it, thank you

[mingcheng nie]

@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]

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.