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{\bf Question}

Let $X$ have pdf $f(x)=42x^5(1-x),\ \ 0<x<1.$  Find the pdf of
$Y=X^3$.  Show that the pdf integrates to 1.


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{\bf Answer}

The range of $y$ is $0<y<1$.

Also $\displaystyle x=y^{\frac{1}{3}}$

Therefore $\displaystyle
\frac{dx}{dy}=\frac{1}{3}y^{-\frac{2}{3}}$

Therefore the pdf of $Y$ is \begin{eqnarray*} g(y) & = & 42
y^{\frac{5}{3}} \left(1-y^{\frac{1}{3}}\right) \cdot
\left|\frac{1}{3}y^{-\frac{2}{3}}\right|,\ \ \ 0<y<1.\\ & = &
14y\left(1-y^{\frac{1}{3}}\right),\ \ \ 0<y<1 \end{eqnarray*}

\begin{eqnarray*} \int_0^1 g(y) \,dy & = & 14\int_0^1
\left(y-y^{\frac{4}{3}}\right) \,dy\\ & = &
14\left(\frac{1}{2}-\frac{1}{7/3}\right)\\ & = &
14\left(\frac{1}{2}-\frac{3}{7}\right)\\ & = & 14 \cdot
\frac{7-6}{14}\\ & = & 1 \end{eqnarray*}

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