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I am having trouble in finding out the resulting PDF of the product of normal and Cauchy distributions. It turns out that we have a general formula for calculating the PDF of product of two random distribution but the integral is not converging . Also, I tried using Mellin Transform method but it is getting too complicated. If anybody has any idea about how to approach the problem, please share it with me.

Thanks

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  • $\begingroup$ Are these centered? $\endgroup$
    – Douglas Zare
    Commented Jan 4, 2015 at 22:43
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    $\begingroup$ Product means product of random variables with those distributions? $\endgroup$
    – Igor Rivin
    Commented Jan 5, 2015 at 4:50
  • $\begingroup$ normal distribution is not centered. It has mean not equal to zero. Also, cauchy distribution is also not standard. $\endgroup$
    – user64634
    Commented Jan 5, 2015 at 21:59

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Since the normal distribution is a $ t $ distribution with $\infty $ degrees of freedom, perhaps you could find the pdf of the product of two $ t $ distributed random variables and then take a limit.

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Assuming the normal is centered with variance $s$ and the Cauchy distribution has parameters $a, b$, combining this Wikipedia page and Mathematica gives

$$ \text{ConditionalExpression}\left[\frac{i \left(\frac{e^{-\frac{z^2}{2 s^2 (a-i b)^2}} \left(2 \pi \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s-i b s)}\right)+2 \text{Ei}\left(\frac{z^2}{2 (a-i b)^2 s^2}\right)+\log \left(\frac{s^2 (a-i b)^2}{z^2}\right)-\log \left(\frac{z^2}{s^2 (a-i b)^2}\right)-4 \log (-a+i b)-2 \log \left(\frac{s^2}{z^2}\right)\right)}{a-i b}-\frac{e^{-\frac{z^2}{2 s^2 (a+i b)^2}} \left(2 \pi \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s+i b s)}\right)+2 \text{Ei}\left(\frac{z^2}{2 (a+i b)^2 s^2}\right)-\log \left(\frac{1}{(a+i b)^2}\right)+4 \log \left(-\frac{1}{a+i b}\right)+\log \left((a+i b)^2\right)\right)}{a+i b}\right)}{8 \sqrt{2} \pi ^{3/2} s}-\frac{i \left(\frac{e^{-\frac{z^2}{2 s^2 (a-i b)^2}} \left(2 \pi \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s-i b s)}\right)-2 \text{Ei}\left(\frac{z^2}{2 (a-i b)^2 s^2}\right)+\log \left(\frac{1}{(a-i b)^2}\right)+4 \log (a-i b)-\log \left((a-i b)^2\right)\right)}{a-i b}+\frac{e^{-\frac{z^2}{2 s^2 (a+i b)^2}} \left(-2 \pi \text{erfi}\left(\frac{\left| z\right| }{\sqrt{2} (a s+i b s)}\right)+2 \text{Ei}\left(\frac{z^2}{2 (a+i b)^2 s^2}\right)-\log \left(\frac{1}{(a+i b)^2}\right)-4 \log (a+i b)+\log \left((a+i b)^2\right)\right)}{a+i b}\right)}{8 \sqrt{2} \pi ^{3/2} s},b\neq 0\land a=0\right] $$

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    $\begingroup$ I tried using MAPLE, the result is not zero. and We do have a valide PDF curve as per MAPLE $\endgroup$
    – user64634
    Commented Jan 5, 2015 at 22:02

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