If $X\sim exp(\lambda)$ what is the PDF of $X^2$?

Question

If $X\sim exp(\lambda)$ what is the Probability density function of $X^2$?
I'd like to know how to calculate it, and what is the way...

Thank you!

Answer 1

We have $X\sim$ exponential distribution $(\lambda)$. The CDF of $X$ is $$ F_X(x)=\Pr[X\le x]=1-e^{\large-\lambda x}. $$ Let $Y=X^2$, then the CDF of $Y$ is $$ \begin{align} F_Y(y)&=\Pr[Y\le y]\\ &=\Pr\left[X^2\le y\right]\\ &=\Pr[X\le \sqrt{y}]\\ &=F_X(\sqrt{y})\\ &=1-e^{\large-\lambda \sqrt{y}}. \end{align} $$ Using the CDF of $y$, we can easily obtain that $Y\sim$ Weibull $\left(\frac{1}{2},\frac{1}{\lambda^2}\right)$. The PDF can also easily be found using $$ f_Y(y)=\frac{d}{dy}F_Y(y). $$

Answer 2

Hint: One way that you could find this would be to find the cumulative distribution function for $X^2$, then use it to find the probability density function.

To find the CDF for $X^2$, note that $X$ is non-negative; so $$ P(X^2\leq x)=\begin{cases}0 & \text{if }x<0\\P(X\leq\sqrt{x}) & \text{if }x\geq0\end{cases}. $$ (If we didn't have a non-negative random variable, we would have to be more careful here; in fact, we would get $$ P(X^2\leq x)=\begin{cases}0 & \text{if }x<0\\P(-\sqrt{x}\leq X\leq\sqrt{x}) & \text{if }x\geq0\end{cases}, $$ which then simplifies to our answer by noting that in that case, $P(-\sqrt{x}\leq X\leq\sqrt{x})=P(X\leq\sqrt{x})$ for $x\geq 0$.)

Once you've found the CDF, do you know how to find the density?

Answer 3

If $Y=X^2$ and $u$ is a bounded, measurable function, then $$ {\rm E}[u(Y)]=\int_0^\infty u(x^2)\lambda e^{-\lambda x}\,\mathrm dx. $$ Now, do a change of variables with $y=x^2$ to obtain something of the form $$ \int_0^\infty u(y)v(y)\,\mathrm dy $$ for some function $v$ (to be determined by you) and conclude that the density of $Y$ is $y\mapsto v(y)\mathbf{1}_{y>0}$.