Find the PDF of $Y = (X - \frac{1}{\theta})^2$ Let $X$ be a random variable with probability density function 
$f(x) =  \theta e^{(-\theta x)}$ for $x \geq 0$, otherwise $0$, when $\theta > 0$.
Let  $Y  = (X  - \frac{1}{\theta})^2$. Find pdf of $Y$. 
I'm just not sure how to incorporate $\theta$.  

Let $Y = (X -\frac{1}{\theta})^{2}$ we want the  PDF of $Y$,
So Lets find the CDF of Y that is $F_{Y}(y)$ and to find the PDF we will differentiate $F_{Y}(y)$ so that $f(y) = F'_{Y}(y)$.
$F_{Y}(y) = P(Y \leq y) = P((X -\frac{1}{\theta})^{2} \leq y) = P(-\sqrt{y}+\frac{1}{\theta} \leq X  \leq \frac{1}{\theta}+\sqrt{y})$
$= F_{X}(\frac{1}{\theta}+\sqrt{y}) - F_{X}(-\sqrt{y}+\frac{1}{\theta})$
So $F_{Y} = F_{X}(\frac{1}{\theta}+\sqrt{y}) - F_{X}(-\sqrt{y}+\frac{1}{\theta})$
to obtain th PDF we differentiate the CDF
so differentiating we obtain
$f(y) = \frac{1}{2\sqrt{y}}f(\frac{1}{\theta}+\sqrt{y}) +\frac{1}{2\sqrt{y}} f(-\sqrt{y}+\frac{1}{\theta})$
$f(y) = \frac{1}{2\sqrt{y}}\theta e^{-\theta(\sqrt{y}+\frac{1}{\theta})} + \frac{1}{2\sqrt{y}}\theta e^{-\theta(-\sqrt{y}+\frac{1}{\theta})} = \frac{1}{2\sqrt{y}}\theta e^{-1}(e^{-\theta\sqrt{y}} + e^{\theta\sqrt{y}})$
