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{\bf Question}

If $X \sim {\rm exponential}(\beta)$ then find the pdf of
$Y=X^{\frac{1}{\gamma}}$.  The random variable $Y$ is known as the
Weibull random variable.  Using a list of distributions, write
down its mean and variance.


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{\bf Answer}

Here $f(x|\beta)=\ds \frac{1}{\beta}e^{\frac{-x}{\beta}},\ \ \
0&lt;x&lt;\infty$

$$Y=X^{\frac{1}{\gamma}}\ \ \ \gamma &gt;0$$

The transformation is increasing and continuous.

Therefore $x=y^{\gamma} \Rightarrow \ds\frac{dx}{dy}=\gamma \ds
y^{\gamma-1}$

Therefore the pdf of \begin{eqnarray*} Y = g(y) &amp; = &amp; \ds
\frac{1}{\beta} \cdot e^{-\frac{y^{\gamma}}{\beta}} \cdot
\left|\gamma y^{\gamma-1}\right|,\ \ \ 0&lt;y&lt;\infty\\ &amp; = &amp; \ds
\frac{\gamma}{\beta} \cdot y^{\gamma-1} \cdot
e^{-\frac{y^{\gamma}}{\beta}},\ \ \ 0&lt;y&lt;\infty. \end{eqnarray*}

$$E(Y)=\beta^{\frac{1}{\gamma}} \Gamma\left( 1+
\frac{1}{\gamma}\right)\ \ {\rm and\ var}(Y)=\beta^{
\frac{2}{\gamma}}\left\{\Gamma\left( 1+
\frac{2}{\gamma}\right)-\Gamma^2\left( 1+
\frac{1}{\gamma}\right)\right\}$$


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