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Expected value of complex gaussian

  1. Mar 22, 2013 #1
    What is the expected value of the following expression


    where [itex]\mu[/itex] is a real constant and [itex]z=x+jy[/itex] such that [itex]x[/itex] and [itex]y[/itex] are independent Gaussian random variables each with zero mean and [itex]\sigma^2[/itex] 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. jcsd
  3. Mar 22, 2013 #2


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    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.
  4. Mar 22, 2013 #3
    Thank you for your reply. Then, I have carried out the following integral:

    [itex]\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[/itex]

    The result is

    [itex]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 [/itex]

    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
  5. Mar 23, 2013 #4


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    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.
  6. Mar 23, 2013 #5
    First, let me correct my result that I have given above. The correct version will be as follows:

    [itex]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) [/itex]

    The Q-function has the following form:


    So, the term corresponding to [itex]r_0e^{-r^2}[/itex] can be expressed by the Q-function. If we go other way around, I mean if we take integral w.r.t. [itex]\phi[/itex] 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
  7. Mar 23, 2013 #6


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    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.
  8. Mar 23, 2013 #7
    The lower limit does not become 0 if you expand the expression. I mean the linear substitution changes the lower limit.
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