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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?

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2 Answers 2

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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 $y\to0^+$.)

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however, when computing its expected value, you have to multiply by $y$ and $y f_{\rm Y}(y)$ doesn't seem to converge for all values of $\alpha$ or am I getting something wrong? –  ACAC Jul 14 '11 at 0:20
    
True, $E(Y)$ is finite if and only if $a>2$. One can prove this using the density $f_Y$ (as you suggest) or the function $y\mapsto P(Y\le y)$ or, more directly, using the density $f_X$ of $X$ since $E(Y)=E(X^{-a})$ is the integral of $x^{-a}$ times $f_X(x)$. –  Did Jul 14 '11 at 8:16

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 $0$).

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The expression given above for $f_Y (x)$ indeed integrates to $1$ -- confirmed numerically (using Wims function calculator). –  Shai Covo Jul 13 '11 at 1:40

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