Rayleigh and Exponential distributions? 
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*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 $|X|^2$?
 A: 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.  
A: 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.
