Determine probability density function

Question

If the p.d.f. of a random variable $X$ is $f(x) = \begin{cases} \exp{(-x)} & x > 0, \\ 0 & x \leq 0, \end{cases}$

how can I determine the p.d.f. of $Y= X^{1/2}$?

Answer 1

Hint: It's easier to reason with the CDF. The CDF of $X$ is $$F_X(x) = 1 - \exp(-x)$$ Then find the CDF of $Y$: $$F_Y(y) = \mathbb P[Y \leq y] = \mathbb P[\sqrt X \leq y] = \mathbb P[X \leq y^2] = \ldots$$ Finally, differentiate $F_Y$ to find the PDF of $Y$.

Answer 2

Hint: It's easier to use the chain rule and not bother with the CDF.

$$\begin{align}f_{\surd X}(y) &= \left\lvert\dfrac{\mathop{\rm d} y^2 }{\mathop{\rm d} y}\right\rvert\cdot f_X(y^2) \end{align}$$