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Double Gaussian integral

  1. Jan 31, 2013 #1
    I need to work out an expression for the average of a Dirac delta-function
    over two normally distributed variables: [tex]z_m^{(n)}, v_m^{(n)}[/tex]

    So I take the Fourier integral representation of the delta function:

    [tex]\delta(y-y_n)=\int \frac{d\omega}{2\pi} e^{i\omega(y-y_n)} =\int \frac{d\omega}{2\pi} e^{i\omega y}e^{-i\omega y_n} [/tex]

    And I already know from a previous calculation that I can express the y_n's in terms of z's and v's:

    [tex]y_n = \frac{\sqrt{\alpha_n}}{\pi} \sum_{k ≠ 0} \frac{Re[z_m^{(n)*} v_m^{(n)}]}{k}[/tex]

    Where the alpha can essentially be regarded as a coefficient for our purposes. So I substitute this into my integral above, ignoring the exp(iωy) part for the moment, and write out the expression for the average over the variables z and v:

    [tex]\int e^{-i\omega \frac{\sqrt{\alpha_n}}{\pi} \sum_{k ≠ 0} \frac{Re[z_m^{(n)*} v_m^{(n)}]}{k}} \frac{e^{-\frac{(z^{(n)}_m)^2}{2}}}{\sqrt{2\pi z^{(n)}_m}} \frac{e^{-\frac{(v^{(n)}_m)^2}{2}}}{\sqrt{2\pi v^{(n)}_m}} d z^{(n)}_m dv^{(n)}_m[/tex]

    where I've introduced the Gaussian distributions of z and v to take the average over these variables. And here is where I am stuck. I am pretty sure that I must do something like make a change of variables in order to simplify this integral, with terms that will go to 1 as they are just the integral over a probability distribution, and some infinite product term will be left over. The exact step to take next in order to achieve this is where I am stuck, as I don't really know what to do with the Real part of the product between v and the conjugate of z to simplify the exponential in the integrand. Any help or tips would be greatly appreciated, thanks.
  2. jcsd
  3. Jan 31, 2013 #2

    Ray Vickson

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    Your question is hard to follow because of notational issues, etc., so let me re-phrase it. You have two random variables U and V with probability densities f(u) and g(v) (although, in this case, it looks like f and g are the same function). I have dropped the 'n' superscript amd 'm' subscripts, as they serve no useful purpose here (but may be needed somewhere else---if so, put them back after completing the calculation.) You say that f(u) and g(v) are Gaussian functions, but that is not what you wrote later: you have
    [tex] f(x) = g(x) = \frac{1}{\sqrt{2 \pi x}} e^{-x^2/2}\; (x = u \text{ or }v).[/tex] These are not Gaussians; true Gaussians would be
    [tex] f(x) = g(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2},[/tex] with no ##\sqrt{x}## in the normalizing factors. So, I don't know whether you just made a typo, or whether your f and g are not truly Gaussians.

    Anyway, you also have a variable you call ##y_n##, which is a function of u and v; let's call it ##z(u,v)##. Finally, you have a parameter, y, and you want to evaluate
    [tex] \int \delta(y - z(u,v)) f(u) g(v) \; du \, dv.[/tex] Your function z(u,v) happens to be given in terms of some type of series, but it looks like the series is divergent:
    [tex] z(u,v) = \frac{\sqrt{\alpha}}{\pi} \sum_{k \neq 0} \frac{\text{Re}(u^* v)}{k}
    = \frac{\sqrt{\alpha} \:\text{Re}(u^* v)}{\pi} \sum_{k \neq 0} \frac{1}{k}.[/tex] That series is divergent, or possibly undefined, so I have no idea how you are supposed to get your function z(u,v).

    Finally, the last expression you wrote has no 'y' in it, and I do not see where it went.
    Last edited: Jan 31, 2013
  4. Jan 31, 2013 #3
    Thanks for the reply, I realise it is hard for me to explain the question without writing up lots of the context that came before it. I will see if I can try to make more sense of it for myself and then if I can phrase it more clearly here, thank you anyway.
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