pdf of power of rayleigh variables I'm trying to compute the pdf of the power of a rayleigh distributed random variable. So, let $X$ be distributed as Rayleigh 
$$ X \sim \frac{x}{\sigma^2} e^{-\frac{x^2}{2\sigma^2}}, x\geq 0  $$
let $Y=X^{-\alpha}$, with $\alpha \geq 1$.The issue, when computing the pdf of $Y$ is that there is a  singularity with $x=0$, for which $1/x$ diverges. How do I deal with this? Is there any known result I can use?
 A: Perhaps you overlooked the contribution from the exponent; the density of $Y$ should approach $0$ as $x \downarrow 0$ (and in any case should integrate to $1$).
EDIT: For any $x > 0$,
$$
{\rm P}(Y \leq x) = {\rm P}(X^{-\alpha} \leq x) = {\rm P}(X^\alpha   \ge 1/x) = 1 - {\rm P}(X^\alpha   \le x^{-1}) = 
1 - {\rm P}(X \le x^{ - 1/\alpha } ).
$$
Hence the distribution function of $Y$ is given, for $x > 0$, by
$$
F_Y (x) = 1 - F_X (x^{ - 1/\alpha }),
$$
where $F_X$ is the distribution function of $X$.
Hence the pdf of $Y$ is given, for $x > 0$, by
$$
f_Y (x) = f_X (x^{ - 1/\alpha }) \alpha^{-1} x^{-1/\alpha - 1},
$$
where $f_X$ is the pdf of $X$.
Thus,
$$
f_Y (x) = \frac{{x^{ - 1/\alpha } }}{{\sigma ^2 }}\exp \bigg( - \frac{{x^{ - 2/\alpha } }}{{2\sigma ^2 }}\bigg)\alpha ^{ - 1} x^{ - 1/\alpha  - 1} = \frac{1}{{\alpha \sigma ^2 }}x^{ - 2/\alpha  - 1} \exp \bigg( - \frac{{x^{ - 2/\alpha } }}{{2\sigma ^2 }}\bigg).
$$
While $\int_0^\infty  {x^{ - 2/\alpha  - 1} \,dx}  = \infty $ due to the singularity at $0$, the contribution from the exponential term obviously implies that $\lim _{x \to 0^ +  } f_Y (x) = 0$ (hence $f_Y$ does not have a singularity at $0$).
A: The distribution of $X$ is characterized by the fact that,  for every $x\ge0$, 
$$
P(X\ge x)=\exp(-ux^2),\quad u=1/(2\sigma^2).
$$
Now, for every positive $y$, $[Y\le y]=[X^{-a}\le y]=[X\ge y^{-1/a}]$ hence 
$$
P(Y\le y)=\exp(-u/y^{2/a}).
$$
The probability density function is the derivative of the cumulative distribution function, hence
$$
f_Y(y)=\frac{2u\exp(-u/y^{2/a})}{ay^{1+2/a}}=\frac{\exp(-1/(2\sigma^2y^{2/a}))}{a\sigma^2y^{1+2/a}}.
$$
One sees that $f_Y(y)\to0$ when $y\to0^+$ and when $y\to+\infty$. 
(But recall that the density of a nonnegative random variable may be unbounded, for example one may have $f(y)\to+\infty$ when $y\to0^+$.)
