Find the PDF of $Y = (X - \frac{1}{\theta})^2$

Question

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 $\theta$.

Answer 1

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 $f(y) = \frac{1}{2\sqrt{y}}f(\frac{1}{\theta}+\sqrt{y}) +\frac{1}{2\sqrt{y}} f(-\sqrt{y}+\frac{1}{\theta})$

$f(y) = \frac{1}{2\sqrt{y}}\theta e^{-\theta(\sqrt{y}+\frac{1}{\theta})} + \frac{1}{2\sqrt{y}}\theta e^{-\theta(-\sqrt{y}+\frac{1}{\theta})} = \frac{1}{2\sqrt{y}}\theta e^{-1}(e^{-\theta\sqrt{y}} + e^{\theta\sqrt{y}})$

Answer 2

It is important to specify the domains of each function. To begin with, the density of $X$ is: $$ f_X(x)=\begin{cases} 0 & x<0 \\ \theta\,e^{-\theta\,x}& x\ge0\end{cases} $$ Note that $f_X(x)$ is not continuous at $0$, so one would expect that something should happen near $Y=\frac{1}{\theta^2}$. The distribution is thus: $$ F_X(x)=\begin{cases} 0 & x<0 \\ 1-e^{-\theta\,x}& x > 0\end{cases} $$ and, although defining the value $x=0$ would give the correct value, one must remember that $F_X$ cannot be differentiated there.

Now, the derivation in BAYMAX's answer is correct: $$ \begin{aligned} F_Y(y)&=P(Y\le y)=P\left(\left(X-\frac{1}{\theta}\right)^2\le y\right)=P\left(\frac{1}{\theta}-\sqrt{y}\le X\le \frac{1}{\theta}+\sqrt{y}\right) \\ &=F_X\left(\frac{1}{\theta}+\sqrt{y}\right)-F_X\left(\frac{1}{\theta} - \sqrt{y}\right) \end{aligned} $$ However, before computing the derivative, let us check whether the arguments of $F_X$ are positive. The first one is positive for sure, but for the second term different cases may occur:

• Let $0 < y < \frac{1}{\theta^2}$. Then both arguments are positive and we take the derivative on the second branch of $F_X$: $$ f_Y(y)=\frac{\theta\,e^{-1}}{2\,\sqrt{y}}\left(e^{-\theta\,\sqrt{y}}+e^{\theta\,\sqrt{y}}\right) $$
• Let $ y > \frac{1}{\theta^2}$. Then the second argument is negative and we just have to take into account the first one: $$ f_Y(y)=\frac{\theta\,e^{-1}}{2\,\sqrt{y}} \,e^{-\theta\,\sqrt{y}} $$

Finally, the density of $Y$ is: $$ f_Y(x)=\begin{cases} 0 & x<0 \\\\ \displaystyle \frac{\theta\,e^{-1}}{2\,\sqrt{y}}\left(e^{-\theta\,\sqrt{y}}+e^{\theta\,\sqrt{y}}\right) & 0 < y < \frac{1}{\theta^2} \\\\ \displaystyle \frac{\theta\,e^{-1}}{2\,\sqrt{y}} \,e^{-\theta\,\sqrt{y}} & y > \frac{1}{\theta^2}\end{cases} $$

NOTE 1: If we only take into account the second branch, then $\lim_{y\to\infty} f_Y(y)\ne 0$ and the required condition $\int_{-\infty}^\infty f_Y(y) dy=1$ would not hold.

NOTE 2: $f_Y(y)$ is not bounded near $y=0$ but this is not a problem, see e.g. Difference between Probability and Probability Density