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

Answer 2

[John D'Errico]

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.

https://en.wikipedia.org/wiki/Inverse_distribution

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.

https://en.wikipedia.org/wiki/Complex_normal_distribution

https://en.wikipedia.org/wiki/Algebra_of_random_variables

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.

https://en.wikipedia.org/wiki/Cauchy_distribution

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.