# Transformation of RV: Finding PDF

I am trying to work through this example problem in my textbook but I keep getting the wrong final answer.

My Notation:
PDF X : pX(x)
CDF X : FX(x)

Question:
Consider the transform Y=X2
if pX(x) = o.5e(-|x|), find the PDF of Y.

My Solution

FY(y) = P(Y$$\le$$y) = P(X2$$\le$$y) = P(X$$\le\pm\sqrt y$$) = [P(X$$\le\sqrt y$$)+P(X$$\le-\sqrt y$$)] = [FX($$\sqrt y$$)+ FX($$-\sqrt y$$)]

FX($$\sqrt y$$) = $$\int_0^\sqrt y$$ pX(x)dx = $$\int_0^\sqrt y$$ o.5e(-|x|) dx = 0.5 - o.5e(-$$\sqrt y$$)

FX($$-\sqrt y$$) = $$\int_{-\sqrt y}^0$$pX(x)dx = $$\int_{-\sqrt y}^0$$ o.5e(-|x|)dx = o.5e(-$$\sqrt y$$) - 0.5

After this point I figured I would add FX($$\sqrt y$$) and FX($$-\sqrt y$$) to get FY(y), then finally take the derivative of FY(y) to get pY(y)... but after adding the two value together I get FY(y) = 0 hence pY(y) = 0

However, the text book (without showing a complete solution) said pY(y) = $$\frac{exp(-\sqrt y)}{2\sqrt y}$$ Could anyone explain how they got this answer? Thanks in advance.

EDIT: The book also mentions that Y cannot be negative (Y>0). Im still not sure how that helps.

• Be careful: $X^2 < y$ is equivalent to $-\sqrt{y} < X < \sqrt{y}$. So $$F_Y(y) = \frac{1}{2} \int_{-\sqrt{y}}^\sqrt{y} e^{-|x|}dx = \int_0^{\sqrt{y}}e^x dx.$$ – David Hughes Feb 25 at 3:02

\begin{align}F_Y(y) &= P(Y\le y)\\ &= P(X^2\le y)\\ &= P(-\sqrt y \le X\le \sqrt y)\\ &= [P(X\le\sqrt y)-P(X\le-\sqrt y)]\\ &= [F_X(\sqrt y)- F_X(-\sqrt y)]\\ \end{align}
\begin{align}f_Y(y) &= \frac{f_X(\sqrt{y})}{2\sqrt{y}}+\frac{f_X(-\sqrt{y})}{2\sqrt{y}}\\ &=\frac{f_X(\sqrt{y})}{2\sqrt{y}}+\frac{f_X(\sqrt{y})}{2\sqrt{y}}\\ &=\frac{f_X(\sqrt{y})}{\sqrt{y}}\\ &=\frac{\exp(-\sqrt{y})}{2\sqrt{y}}\end{align}