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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.

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

Differentiating, we get

\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}

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