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I have two probability distributions $p(x) = N_1 x^{-\tau} \exp(-\frac{x}{x_0})$ and $p(y) = N_2 y^{-\kappa} \exp(-\frac{y}{y_0})$. $N_1$ and $N_2$ are just normalization constants and $x>0$, $y>0$, $x_0>0$, $y_0>0$, $\tau>0$, and $\kappa>0$. I want to know what is the distribution of x*y. I am not interested in the normalization constant.

I tried many things via characteristic functions or convolution, but could not find any analytic solution. I suspect there is no analytic solution, because of the nasty integral involved. Thus, I want to know if there is a clever approximation. For example, prove that the new distribution is again a power law for small x*y, and the exponential cut-off on the new distribution is $>x_0$ and $>y_0$. Empirically, I find if I take the product of multiple $x$, let's say $m$-times, with distribution $p(x)$ as above, the resulting distribution approaches a power law. Can this be proven for the limit $m\rightarrow \infty$?

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Shouldn't that be $x^{-\tau}e^{-x_0/x}$? The functions you've written down aren't actually integrable at small $x$. – userN Oct 18 '12 at 3:35

So your random variables $X$ and $Y$ have Gamma distributions with scale parameters $x_0$ and $y_0$ and shape parameters $1-\tau$ and $1-\kappa$ respectively (of course, you must assume $\tau < 1$ and $\kappa < 1$ for these to be well-defined). I presume you are also assuming $X$ and $Y$ are independent. Maple says the probability density function of $Z=XY$ is $$ f_Z(z) = \frac{2}{\Gamma(\tau) \Gamma(\kappa)} {z}^{{\tau/2}+{\kappa/2}-1}{y_{{0}}}^{-{\tau/2}-{\kappa/2}}{x_{{0}}}^{-{\tau/2}-{\kappa/2}} {{\rm K}_{\kappa-\tau}\left({\frac {2 \sqrt {z}}{\sqrt {x_{{0}}y_{{0}}}}}\right)} $$ for $z > 0$, where $K_{\kappa-\tau}$ is a modified Bessel function of the second kind.

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Thanks. I also came across the modified Bessel function of the second kind. Is there a good approximation of K? – Heiko Hoffmann Oct 18 '12 at 0:56
For example, series expansion around $0$: for non-integer $\nu$, $$K_\nu(x) = \sum _{k=0}^{\infty }{\frac {{2}^{\nu-2\,k-1}\pi \,{x}^{-\nu+2\,k}} { \left( k-\nu \right) !\,k!\,\sin \left( \pi \,\nu \right) }}-{ \frac {{2}^{-\nu-2\,k-1}\pi \,{x}^{\nu+2\,k}}{ \left( \nu+k \right) !\,k !\,\sin \left( \pi \,\nu \right) }} $$ See also[nu%2Cx] – Robert Israel Oct 18 '12 at 2:40

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