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{\bf Question}

Let $X$ have pdf $f(x)=\frac{1}{2}(1+x),\ \ -1&lt;x&lt;1.$ Find the pdf
of $Y=X^2$.

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{\bf Answer}

$f(x)=\frac{1}{2}(1+x),\ \ \ -1&lt;x&lt;1.$

The transformation is $y=x^2$.  It is decreasing in $-1&lt;x \leq 0$
and increasing in $0&lt;x&lt;1$.

Also $0&lt;y&lt;1$ and $x=\pm \sqrt y$

Therefore $\displaystyle \left|\frac{dx}{dy}\right|=\frac{1}{2
\sqrt y}$

The pdf of $Y$ is \begin{eqnarray*} g(y) &amp; = &amp; \frac{1}{2}(1-\sqrt
y) \cdot \frac{1}{2\sqrt y} +  \frac{1}{2}(1+\sqrt y) \cdot
\frac{1}{2\sqrt y},\ \ \ 0&lt;y&lt;1\\ &amp; = &amp; \frac{1}{4\sqrt y}.2,\ \ \
0&lt;y&lt;1\\ &amp; = &amp; \frac{1}{2\sqrt y},\ \ \ 0&lt;y&lt;1. \end{eqnarray*}


Alternative:

\begin{eqnarray*} F(x) &amp; = &amp; \int_{-1}^x f(t) \,dt\\ &amp; = &amp;
\frac{1}{2} \int_{-1}^x (1+t) \,dt\\ &amp; = &amp; \frac{1}{2}
\left[t+\frac{t^2}{2}\right]_{-1}^x\\ &amp; = &amp; \frac{1}{2}
\left\{x+\frac{x^2}{2}+1-\frac{1}{2}\right\}\\ &amp; = &amp;
\frac{1}{2}\left\{x+\frac{x^2}{2}+\frac{1}{2}\right\}
\end{eqnarray*}

\begin{eqnarray*} G(y) &amp; = &amp; P(Y \leq y)\\ &amp; = &amp; P\{x^2 \leq y\}\\
&amp; = &amp; P\{-\sqrt y \leq x \leq \sqrt y\}\\ &amp; = &amp; F(\sqrt
y)-F(-\sqrt y)\\ &amp; = &amp; \frac{1}{2}\left\{\sqrt y
+\frac{y}{2}+\frac{1}{2}+\sqrt y -
\frac{y}{2}-\frac{1}{2}\right\}\\ &amp; = &amp; \sqrt y\end{eqnarray*}

Therefore the pdf of $Y$ is $\displaystyle g(y)=\frac{dG(y)}{dy} =
\frac{1}{2\sqrt y},\ \ \ 0&lt;y&lt;1.$

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