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\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\begin{document}
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{\bf Question}

\bigskip
The probability density function of a random variable $X$ is given
by:$$f(x)=\left\{ \begin{array}{ll}   0 & 0 < x < 1 \\   cx^{-2} &
x \ge 1 \end{array}\right.$$

\begin{description}
\item[(a)] Find the value of $c$;
\item[(b)] Find the distribution function $F(X)$;
\item[(c)] Find $p(X>3)$;
\item[(d)] Find the mode of this distribution;
\item[(e)] Find the mean and standard deviation of the
distribution;
\item[(f)] Find the median and interquartile range of the
distribution.
\end{description}

\vspace{.25in}

{\bf Answer}
\bigskip


$f(x)=\left\{ \begin{array}{ll}   0 & 0 < x < 1 \\   cx^{-2} &
x\ge 1 \end{array}\right.$

\begin{description}
\item[(a)]
$\ds \int_0^\infty f(x) \, dx = c\int_1^\infty x^{-2} \, dx =
c\left[\frac{x^{-1}}{-1}\right]_1^\infty = c[1-0] = c = 1$

(equals one because it is a p.d.f.) Hence $c = 1$.

\item[(b)]
$\ds F(x) = \int_{-\infty}^x f(y) \, dy = \int_1^x x^{-2} \, dx =
\left[\frac{x^-1}{-1}\right]_1^x = 1 - \frac{1}{x}$



\item[(c)]
$\ds p(X>3) = 1 - F(3) = 1 - \left(1 - \frac{1}{3}\right) =
\frac{1}{3} = \int_3^\infty x^{-2} \, dx$

\item[(d)]
$$\\$$
\begin{center}
$ \begin{array}{l}
\textrm{Mode occurs where }f(x) \textrm{ has a max.}\\
\textrm{Maximum at }x = 1\\
\textrm{Mode }x_{\mathrm{mode}} = 1
\end{array}
\ \ \ 
\begin{array}{c}
\epsfig{file=175-10-1.eps, width=30mm}
\end{array} $
\end{center}

\item[(e)]
Mean $\ds \mu = \int_1^\infty xf(x) \, dx = \int_1^\infty
\frac{1}{x} \, dx$

This integral diverges! The mean cannot be defined.

Standard deviation

$\ds \sigma^2 = \int_{-\infty}^\infty (x - \mu)^2 \times
\frac{1}{x^2} \, dx$

Similarly the standard deviation cannot be defined

[$\infty$ will be accepted in both cases]

\item[(f)]m
The median $m$ is defined so that $F(m) = \frac{1}{2}$

$\ds F(m) = 1 - \frac{1}{m} = \frac{1}{2} \Rightarrow m = 2$

Interquartile range between $q_1$ and $q_3$ with $F(q_1) =
\frac{1}{4}$, $F(q_3) = \frac{3}{4}$

$\ds 1 - \frac{1}{q_1} = \frac{1}{4} \Rightarrow q_1 =
\frac{4}{3}$

$\ds 1 - \frac{1}{q_3} = \frac{3}{4} \Rightarrow q_3 = 4$

So the interquartile range $\ds \{q_1, q_3\} = \left\{\frac{4}{3},
4\right\}$

Note that this is a peculiar probability density function.
\end{description}

\end{document}
