find density function of Z = (X - 1)^2 when X is a exponential distribution

Question

let X be an exponential distribution with lambda as parameter and want to find the density function of $Z = (X - 1)^2$

I tried that: $F(Z) = P(Z < z) = P((X - 1)^2 <z)= P(1 - \sqrt{z} < x < 1 + \sqrt{z})$

for the next part, do I have to assume that $\sqrt{z}$ is greater than one or we have to determine for both cases?

of $\sqrt{z} > 1$: $F(Z) = P(X < \sqrt{z}+1)$ since exponential distribution have x always greater than 0.

if $\sqrt{z} < 1$: $F(Z) = P(1 - \sqrt{z}< x < 1 + \sqrt{z})$

and then find the pdf?

thank you!

Answer 1

For finding the PDF of Y, given PDF of X, where $Y=g(X)$, we can use the following formula:

$$f_Y(y) = f_X(g^{-1}(y))|\frac{d}{dy}g^{-1}(y)|$$

see this paper.

In this case, $g(x) = (x-1)^2$, which is decreasing for $x < 1$ and increasing for $x > 1$. $g^{-1}(x) = \sqrt{x} + 1$. And $Z = g(X)$. The rest should be easy.

Answer 2

As you have: $F_{\small Z}(z)=\mathsf P(1-\surd z\lt X\leqslant 1+\surd z)$

Which you have identified will have a piecewise partition at $z=1$

And thus: $\quad F_{\small Z}(z)~=(\mathrm e^{-\lambda(1-\surd z)}-\mathrm e^{-\lambda(1+\surd z)})\mathbf 1_{0\leqslant z\lt 1}+(1-\mathrm e^{-\lambda(1+\surd z)})\mathbf 1_{1\leqslant z}$

Now just find the derivative.

So: $\qquad\qquad f_{\small Z}(z)~ = \dfrac{\lambda(\mathrm e^{-\lambda(1+\surd z)}+\mathrm e^{-\lambda(1-\surd z)})}{2\surd z}\mathbf 1_{0\leqslant z\lt 1}+\dfrac{\lambda\mathrm e^{-\lambda(1+\surd z)}}{2\surd z}\mathbf 1_{1\leqslant z}$