Year: 2013
Paper: 3
Question Number: 13
Course: UFM Statistics
Section: Cumulative distribution functions
With the number of candidates submitting scripts up by some 8% from last year, and whilst inevitably some questions were more popular than others, namely the first two, 7 then 4 and 5 to a lesser extent, all questions on the paper were attempted by a significant number of candidates. About a sixth of candidates gave in answers to more than six questions, but the extra questions were invariably scoring negligible marks. Two fifths of the candidates gave in answers to six questions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
\begin{questionparts}
\item
The continuous random variable $X$ satisfies
$0\le X\le 1$, and has probability density function
$\f(x)$ and cumulative distribution
function $\F(x)$. The
greatest value of $\f(x)$ is $M$, so that $0\le \f(x) \le M$.
\begin{enumerate}
\item
Show that $0\le \F(x) \le Mx$ for $0\le x\le1$.
\item
For any function $\g(x)$, show that
\[
\int_0^1 2 \g(x) \F(x) \f(x) \d x = \g(1) -
\int_0^1 \g'(x) \big( \F(x)\big)^2 \d x
\,.
\]
\end{enumerate}
\item
The
continuous random variable $Y$
satisfies
$0\le Y\le 1$, and has probability density function
$k \F(y) \f(y)$, where $\f$ and $\F$ are as above.
\begin{enumerate}
\item
Determine the
value of the constant $k$.
\item
Show that
\[
1+ \frac{nM}{n+1}\mu_{n+1} - \frac{nM}{n+1}
\le \E(Y^n) \le 2M\mu_{n+1}\,,
\]
where $\mu_{n+1} = \E(X^{n+1})$ and $n\ge0$.
\item
Hence show that, for $n\ge 1$,
\[
\mu _n \ge \frac{n}{(n+1)M} -\frac{n-1}{n+1}
\,.\]
\end{enumerate}
\end{questionparts}\begin{questionparts}
\item \begin{enumerate}
\item $\,$ \begin{align*}
&& 0 &\leq f(t) &\leq M \\
\Rightarrow && \int_0^x 0 \d t &\leq \int_0^x f(t) \d t & \leq \int_0^x M \d x \\
\Rightarrow && 0 &\leq F(x) &\leq Mx
\end{align*}
\item $\,$ \begin{align*}
&& \int_0^1 2g(x)F(x)f(x) \d x &= \left [ g(x) F(x)^2 \right] - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \\
&&&= g(1) - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x
\end{align*}
\end{enumerate}
\item \begin{enumerate}
\item $\,$ \begin{align*}
&& 1 &= \int_0^1 kF(y)f(y) \d y \\
&&&= k\left [ \frac12 F(y)^2\right]_0^1 \\
&&&= \frac{k}{2} \\
\Rightarrow && k &= 2
\end{align*}
\item $\,$ \begin{align*}
\E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\
&\geq \int_0^1 y^n 2My f(y) \d y \\
&= 2M\int_0^1 y^{n+1} f(y) \d y \\
&= 2M \E[X^{n+1}] = 2M\mu_{n+1} \\
\\
\E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\
&= 1 - \int_0^1 ny^{n-1} F(y)^2 \d y \\
&\geq 1 - \int_0^1 ny^{n-1}My F(y) \d y \\
&= 1 - M\int_0^1 ny^n F(y) \d y \\
&= 1 - M[\frac{n}{n+1}y^{n+1} F(y)]_0^1 + M\int_0^1\frac{n}{n+1} y^{n+1} f(y) \d y \\
&= 1 - \frac{nM}{n+1} + \frac{nM}{n+1} \mu_{n+1}
\end{align*}
\item Since $\E[Y^{n-1}] \geq 0$ we must have
\begin{align*}
&& 2M\mu_n \geq 1 + \frac{(n-1)M}{n}\mu_n - \frac{(n-1)M}{n} \\
\Rightarrow && \mu_n \left (2M + \frac{(n-1)M}{n} \right) \geq 1 - \frac{(n-1)M}{n} \\
\Rightarrow && \mu_n \frac{3Mn-M}{n} & \geq \frac{n-(n-1)M}{n} \\
\Rightarrow && \mu_n & \geq \frac{n-(n-1)M}{3Mn-M}
\end{align*}
\end{enumerate}
\end{questionparts}
The number attempting this was very similar to that attempting question 3 with the same level of success as question 11. In general, candidates attempted both parts of (a) correctly, and then likewise part (i) of (b) then stopped. However, part (b)(ii) tripped up many. Some successfully dealt with part (iii) without having managed (ii).