Square of Exponential Random Variable

Question

I want to find the pdf of the square of an exponential distribution. Let $X \sim \exp(\lambda)$ and $Y=X^2$. I want $f_Y(y)$.

I start from the $\operatorname{CDF}$ of $Y$: $$ P(Y\leq y)=P(X^2\leq y) \\ =P(X\leq y/X) \\ =\int_0^\infty\left( \int_0^{y/X} \lambda e^{-\lambda x} dx \right)\lambda e^{-\lambda x} dx\\ = \int_0^\infty \left(e^{\lambda y / x} - 1 \right) e^{-\lambda y / x} \lambda e^{\lambda x} dx$$

which after a bit of solving gives me: $$= \lambda \int_0^\infty e^{\lambda x}-e^{\lambda x - \frac{\lambda y}{x}} dx $$

At this point the calculus starts to get a bit tricky, and this is where I get lost.

Am I on the right track and I just need to muscle through the calculation, or is there an easier way to approach this?

Answer 1

The step $$ P(X\leq y/X) =\int_0^\infty\left( \int_0^{y/X} \lambda e^{-\lambda x} dx \right)\lambda e^{-\lambda x} dx$$ is not valid for several reasons:

• there are two integrals depending on $x$
• there is a capital $X$ while the left hand side is a number.

Instead, use $P(X^2\leqslant y)=\mathbb P(X\leqslant\sqrt{y})$ for $y\geqslant 0$, which is valid since $X$ is positive.