Average of Dirac Delta-Function over Double Gaussian Variables

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SUMMARY

The discussion focuses on calculating the average of a Dirac delta-function, specifically \(\delta(y - y_n)\), over two normally distributed variables \(z_m^{(n)}\) and \(v_m^{(n)}\). The user employs the Fourier integral representation of the delta function and expresses \(y_n\) in terms of \(z\) and \(v\) using the formula \(y_n = \frac{\sqrt{\alpha_n}}{\pi} \sum_{k \neq 0} \frac{Re[z_m^{(n)*} v_m^{(n)}]}{k}\). The challenge arises in simplifying the integral involving Gaussian distributions of \(z\) and \(v\), particularly in handling the real part of the product between \(v\) and the conjugate of \(z\). The discussion highlights potential notational issues and the divergence of the series involved.

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fast_eddie
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I need to work out an expression for the average of a Dirac delta-function
\delta(y-y_n)
over two normally distributed variables: z_m^{(n)}, v_m^{(n)}

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

\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}

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

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

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:

\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

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.
 
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fast_eddie said:
I need to work out an expression for the average of a Dirac delta-function
\delta(y-y_n)
over two normally distributed variables: z_m^{(n)}, v_m^{(n)}

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

\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}

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

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

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:

\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

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.

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
f(x) = g(x) = \frac{1}{\sqrt{2 \pi x}} e^{-x^2/2}\; (x = u \text{ or }v). These are not Gaussians; true Gaussians would be
f(x) = g(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}, 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
\int \delta(y - z(u,v)) f(u) g(v) \; du \, dv. Your function z(u,v) happens to be given in terms of some type of series, but it looks like the series is divergent:
z(u,v) = \frac{\sqrt{\alpha}}{\pi} \sum_{k \neq 0} \frac{\text{Re}(u^* v)}{k}<br /> = \frac{\sqrt{\alpha} \:\text{Re}(u^* v)}{\pi} \sum_{k \neq 0} \frac{1}{k}. 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:
Thanks for the reply, I realize 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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