Rayleigh and Exponential distributions?

Question

• If $X$ is Rayleigh distributed random variable. What is the distribution of $|X|^2$?

• If $X$ is Exponential distributed random variable. What is the distribution of $|X|^2$?

Answer 1

We deal with the Rayleigh. The other calculation is very similar.

Let $X$ be Rayleigh with parameter $\sigma^2$. Then $X$ has density function $\frac{x}{\sigma^2}e^{-x^2/(2\sigma^2)}$ for $x\ge 0$, and $0$ elsewhere. Let $Y=X^2$. We find the density function $f_Y(y)$ of $Y$.

To do this, we first find the cdf $F_Y(y)$ of $Y$. For $y\gt 0$, we have $$F_Y(y)=\Pr(Y\le y)=\Pr(X^2\le y)=\Pr(X\le \sqrt{y})=\int_0^{\sqrt{y}} \frac{x}{\sigma^2}e^{-x^2/(2\sigma^2)}\,dx.$$ Integrate, say using Substitution. We find that for $y\gt 0$, $$F_Y(y)=1-e^{-y/(2\sigma^2)}.$$ For the density function, differentiate. For completeness, note that $f_Y(y)=0$ if $y\lt 0$.

Remark: We went through the cdf. It is quicker to use the method of transformations. We avoided it because at this stage going through the cdf makes what is going on clearer.

Answer 2

Many thanks to André Nicolas. I know now that $|X|^2$ is Exponential when $X$ is Rayleigh. Based on André Nicolas's answer, I will deal with the Exponential case. Please correct me if I did something wrong.

Let $X$ be Exponential with parameter $\lambda$. Then $X$ has density function $\lambda e^{-\lambda x}$ for $x\ge 0$, and $0$ elsewhere. Let $Y=X^2$. We find the density function $f_Y(y)$ of $Y$.

Let us find the cdf $F_Y(y)$ of $Y$. For $y\gt 0$, we have $$F_Y(y)=\Pr(Y\le y)=\Pr(X^2\le y)=\Pr(X\le \sqrt{y})=\int_0^{\sqrt{y}} \lambda e^{-\lambda x}\,dx.$$ Now integrate. We find that for $y\gt 0$, $$F_Y(y)=1-e^{-\lambda\sqrt{y}}.$$

This is I do not know which distribution it is.