Finding a probability density function for $Y=X^2$

Question

Question: Let $X$ have an exponential distribution with parameter $λ$. Let $Y = X^2$. Compute $f_Y(t)$.

My answer: I know that $f_X(x)=\lambda e^{-\lambda x}$ for $x \geq 0$.

I found that $F_Y(t)=P(Y \leq t)=P(X^2 \leq t)=P(X \leq \sqrt{t})$ so that $f_X(x)= F'_X(\sqrt{t})= f_X(\sqrt{t})\frac{d}{dy}(\sqrt{t})=f_X(\sqrt{t}) \frac{1}{2\sqrt{t}}$.

Now, $f_X(\sqrt{t})= \lambda e^{-\lambda \sqrt{t}}$.

Thus, $f_Y(t)= \frac{ \lambda}{2\sqrt{t}}e^{-\lambda\sqrt{t}}$ for $t \geq 0$.

I am unsure if what I have done is correct or if there is an easier way to tackle such questions.

Answer 1

Yes it's correct; just be careful to justify that

$$P(X^2\le t) = P(X \le \sqrt t)$$

is valid only because $X \ge 0$. I think this is the easiest way.