If $X \sim E(\lambda)$, find the density function of $Y = X^2+1$.

Question

If $X \sim E(\lambda)$, find the density function of $Y = X^2+1$.

My try: $$F_Y(x) = P(X^2+1 < x) = \mathrm P(-\sqrt{x-1} < X < \sqrt{x-1}) = \mathrm P(X < \sqrt{x-1}) -\mathrm P(X < -\sqrt{x-1})$$ Now since $$F_X(x) = \begin{cases}\ 1-e^{-\lambda x}, & x \ge 0 \\ 0 , & x< 0\end{cases}$$ we have $\mathrm P(X < -\sqrt{x-1}) \equiv 0$. Hence $$F_Y(x) = \mathrm P(X < \sqrt{x-1}) = F_X(\sqrt{x-1}) = \begin{cases} 1-e^{-\lambda \sqrt{x-1}}, & \sqrt{x-1} \ge 0 \\ 0 , & \sqrt{x-1}< 0\end{cases}$$ or $$F_Y(x) = \begin{cases} 1- e^{-\lambda \sqrt{x-1}}, & x > 1 \\ 0 , & x \le 1\end{cases}$$

1. I'm not sure putting the last intervals for $x$ though. Is there a mistake in this solution?

2. Now, I'm going to differentiate $F_Y(x)$ to find the probability density function. Is there any shortcut/other approach to this?

Answer 1

\begin{align} F_Y(x) & = \Pr(X^2+1\le x) \\[8pt] & = \begin{cases} \Pr(X\le \sqrt{x-1} ) & \text{if } x\ge1, \\ 0 & \text{if } x<1 \end{cases} \\[8pt] & = \begin{cases} 1 - e^{-\lambda\sqrt{x-1}} & \text{if } x\ge 1 \\ 0 & \text{if } x<1 \quad \text{(not “if $\sqrt{x-1} < 0$”)} \end{cases} \end{align}

Answer 2

Alternatively: because $y = g(x) = x^2 +1$ is invertible in $[0, \infty] \to [1, \infty]$ we can apply

$$ f_Y(y)= \frac{f_X(x)}{|g'(x)|} \bigg|_{x=g^{-1}(y)}$$

Now, $f_X(x) = \lambda e^{-\lambda x}$, $|g'(x)|= 2x$ , $g^{-1}(y)=\sqrt{y-1}$, hence

$$f_Y(y) = \frac{\lambda \exp(-\lambda \sqrt{y-1})}{2 \sqrt{y-1}}$$

for $y\ge 1$.