Expected value of complex gaussian

1. Mar 22, 2013

jashua

What is the expected value of the following expression

$exp(|z+\mu|)$,

where $\mu$ is a real constant and $z=x+jy$ such that $x$ and $y$ are independent Gaussian random variables each with zero mean and $\sigma^2$ variance.

When I try to take the expectation, I couldn't obtain a gaussian integral, so I couldn't take the expectation. So, can we obtain the expected value of the above exponential in a closed form?

2. Mar 22, 2013

Staff: Mentor

After a substitution $y=r\cos(\phi)$, $x=r \sin(\phi)-\mu$, the one of the two integrals looks possible. I did not check the other integral.

3. Mar 22, 2013

jashua

Thank you for your reply. Then, I have carried out the following integral:

$\frac{1}{2\pi\sigma^2} \int_{0}^{2\pi}\int_{0}^{∞} exp({-\frac{r^2-2r\mu sin(\phi) + 2\sigma^2r + \mu^2}{2\sigma^2}})rdrd\phi$

The result is

$exp({-\frac{\mu^2}{2\sigma^2}})\frac{1}{\sqrt{2\pi\sigma^2}}\int_0^{2\pi} (\mu sin(\phi) - \sigma^2) exp(\frac{(\mu sin(\phi) - \sigma^2)^2}{2\sigma^2})Q(-\frac{\mu sin(\phi) - \sigma^2}{\sigma})d\phi$

If I've correctly found this result, even the first integral does not have a closed form due to the Q-function. So, any idea to find an approximate closed form for this integral?

Last edited: Mar 23, 2013
4. Mar 23, 2013

Staff: Mentor

How do you get that Q function? A linear substitution simplifies the first integral to $(r+r_0)e^{-r^2}$ (neglecting prefactors), which gives a Gaussian integral and a component with an antiderivative.

5. Mar 23, 2013

jashua

First, let me correct my result that I have given above. The correct version will be as follows:

$exp({-\frac{\mu^2}{2\sigma^2}})(1+\frac{1}{\sqrt{2\pi\sigma^2}}\int_0^{2\pi} (\mu sin(\phi) - \sigma^2) exp(\frac{(\mu sin(\phi) - \sigma^2)^2}{2\sigma^2})Q(-\frac{\mu sin(\phi) - \sigma^2}{\sigma})d\phi)$

The Q-function has the following form:

$Q(x)=\frac{1}{\sqrt{2\pi}}\int_{x}^{∞}exp(-r^2/2)dr$

So, the term corresponding to $r_0e^{-r^2}$ can be expressed by the Q-function. If we go other way around, I mean if we take integral w.r.t. $\phi$ first, then we get a Bessel function of the first kind. In either case, we don't have a closed form. So, can we get an approximate closed form?

Last edited: Mar 23, 2013
6. Mar 23, 2013

Staff: Mentor

I don't see how you get the Q function.
$$\int_0^\infty r_0 e^{-r^2} dr = r_0 \int_0^\infty e^{-r^2} dr = r_0 \frac{\sqrt{\pi}}{2}$$

Edit: Oh, that is a result of the linear substitution.
Okay, I see... and I don't know how to avoid that.

7. Mar 23, 2013

jashua

The lower limit does not become 0 if you expand the expression. I mean the linear substitution changes the lower limit.