Finding density function of reciprocal of Rayleigh random variable

Question

I have a Rayleigh distributed random variable $X$. How can I find pdf of another random variable $Y= \frac{1}{X}$.

Using some knowedge of transformations I found that pdf of $Y$, $f_Y(y) = \frac{-1}{x^2}f_X(\frac{1}{x^2})$, but I am not sure whether I am correct.

Answer 1

Hint.

The PDF is the derivative of CDF. By the definition of CDF $$ P(Y\le y)=P(X\ge 1/y)=1-P(X<1/y)\quad y>0 $$

Now use the Rayleigh distribution assumption.

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For your edited question, you make two mistake on the right: the variable should by $y$, not $x$; inside the density, you should use $\frac1y$.