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.
\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.