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{\bf Question}

Suppose that X has a uniform distribution on the interval (0,1).
Show that the pdf of $Y=(8X)^{\frac{1}{3}}$ is given by

$$f(x)=\left\{ \begin{array} {ll} \frac{3}{8}y^2, & {\rm for}\
0<y<2;\\ 0, & {\rm otherwise} \end{array} \right.$$


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{\bf Answer}

The transformation is $$ y  = (8x)^{\frac{1}{3}} =
2x^{\frac{1}{3}} $$

Therefore the range of $y$ is $0<y<2$.

Also $\displaystyle x=\frac{y^3}{8}$.

Therefore $\displaystyle \frac{dx}{dy}=\frac{1}{8} \cdot 3 \cdot
y^2.$

Therefore the pdf of $Y$ if \begin{eqnarray*} g(y) & = & 1\cdot
\left|\frac{3}{8}y^2\right|,\ \ \ 0<y<2\\ & = & \frac{3}{8}y^2,\ \
\ 0<y<2. \end{eqnarray*}


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