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{\bf Question}

A certain river floods every year.  Suppose that the low water
mark is set at 1 and the high water mark $X$ has the cdf

$$F(x)=P(X \leq x)=1-\displaystyle\frac{1}{x^2},\ 1 \leq x <
\infty.$$

\begin{description}
\item[(a)]
Find $f(x)$, the pdf of $X$.

\item[(b)]
If the low water mark is reset at zero and we use a unit of
measurement which is $\frac{1}{10}$ of that given previously, the
high-water mark becomes $Y=10(X-1)$.  Find the cdf of $Y$.

\item[(c)]
Find the pdf of $Y$ by differentiating the cdf as obtained in part
(b), and also using the transformation technique.
\end{description}


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{\bf Answer}

\begin{description}
\item[(a)]
$$f(x)=\left\{\begin{array}{ccrcccc} 0 & {\rm if} & -\infty & < &
x & < & 1\\\displaystyle \frac{2}{x^3} & {\rm if} & 1 & \leq & x &
< & \infty.
\end{array} \right.$$

\item[(b)]
The cdf of $Y$ is

\begin{eqnarray*}&{}& G(y) = P\{Y \leq y\}\\ & = & P\{10(X-1) \leq y\}\\& =
& P\{X \leq 1+\frac{y}{10}\}\\& = &
F\left(1+\frac{y}{10}\right)\\& = &
1-\frac{1}{\left({1+\frac{y}{10}}\right)^2},\ \ \ 0 \leq y<\infty.
\end{eqnarray*}


\item[(c)]
The pdf of $Y$ is \begin{eqnarray*} g(y) & = & \frac{dG(y)}{dy}\\
& = & 2 \cdot \frac{1}{\left(1+\frac{y}{10}\right)^3}
\frac{1}{10},\ \ \ 0 \leq y < \infty\\ & = & \frac{1}{5}
\frac{1}{\left(1+\frac{y}{10}\right)^3},\ \ \ 0 \leq y < \infty
\end{eqnarray*}

Using the transformation technique:

The range of $y$ is $0 \leq y < \infty$.

The transformation is increasing.

$\displaystyle x=1+\frac{y}{10}=s(y)$

$\displaystyle \frac{dx}{dy}=\frac{1}{10}$

\begin{eqnarray*} g(y) & = & f(s(y))\left|\frac{dx}{dy}\right|\\ &
= & \frac{2}{\left(1+\frac{y}{10}\right)^3} \cdot \frac{1}{10}\\ &
= & \frac{1}{5} \frac{1}{\left(1+\frac{y}{10}\right)^3},\ \ \ 0
\leq y < \infty \end{eqnarray*}
\end{description}

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