Year: 2018
Paper: 3
Question Number: 12
Course: UFM Statistics
Section: Bivariate data
The total entry was a record number, an increase of over 6% on 2017. Only question 1 was attempted by more than 90%, although question 2 was attempted by very nearly 90%. Every question was attempted by a significant number of candidates with even the two least popular questions being attempted by 9%. More than 92% restricted themselves to attempting no more than 7 questions with very few indeed attempting more than 8. As has been normal in the past, apart from a handful of very strong candidates, those attempting six questions scored better than those attempting more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
A random process generates, independently, $n$
numbers each of which is drawn from a uniform (rectangular) distribution on the interval 0 to 1.
The random variable $Y_k$ is defined to be the $k$th smallest number (so there are $k-1$ smaller numbers).
\begin{questionparts}
\item Show that, for $0\le y\le1\,$,
\[
{\rm P}\big(Y_k\le y) =\sum^{n}_{m=k}\binom{n}{m}y^{m}\left(1-y\right)^{n-m}
.
\tag{$*$}
\]
\item
Show that
\[
m\binom n m = n \binom {n-1}{m-1}
\]
and obtain a similar expression for
$\displaystyle (n-m) \, \binom n m\,$.
Starting from $(*)$, show that the probability density function of $Y_k$ is
\[ n\binom{ n-1}{k-1} y^{k-1}\left(1-y\right)^{ n-k} \,.\]
Deduce an expression for
$\displaystyle \int_0^1 y^{k-1}(1-y)^{n-k} \, \d y \,$.
\item
Find $\E(Y_k) $ in terms of $n$ and $k$.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& \mathbb{P}(Y_k \leq y) &= \sum_{j=k}^n\mathbb{P}(\text{exactly }j \text{ values less than }y) \\
&&&= \sum_{j=k}^m \binom{m}{j} y^j(1-y)^{n-j}
\end{align*}
\item This is the number of ways to choose a committee of $m$ people with the chair from those $m$ people. This can be done in two ways. First: choose the committee in $\binom{n}{m}$ ways and choose the chair in $m$ ways so $m \binom{n}{m}$. Alternatively, choose the chain in $n$ ways and choose the remaining $m-1$ committee members in $\binom{n-1}{m-1}$ ways. Therefore $m \binom{n}{m} = n \binom{n-1}{m-1}$
\begin{align*}
(n-m) \binom{n}{m} &= (n-m) \binom{n}{n-m} \\
&= n \binom{n-1}{n-m-1} \\
&= n \binom{n-1}{m}
\end{align*}
\begin{align*}
f_{Y_k}(y) &= \frac{\d }{\d y} \l \sum^{n}_{m=k}\binom{n}{m}y^{m}\left(1-y\right)^{n-m} \r \\
&= \sum^{n}_{m=k} \l \binom{n}{m}my^{m-1}\left(1-y\right)^{n-m} -\binom{n}{m}(n-m)y^{m}\left(1-y\right)^{n-m-1} \r \\
&= \sum^{n}_{m=k} \l n \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n \binom{n-1}{m} y^{m}\left(1-y\right)^{n-m-1} \r \\
&= n\sum^{n}_{m=k} \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n\sum^{n+1}_{m=k+1} \binom{n-1}{m-1} y^{m-1}\left(1-y\right)^{n-m} \\
&= n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k}
\end{align*}
\begin{align*}
&&1 &= \int_0^1 f_{Y_k}(y) \d y \\
&&&= \int_0^1 n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k} \d y \\
&&&= n \binom{n-1}{k-1} \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\
\Rightarrow && \frac{1}{n \binom{n-1}{k-1}} &= \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\
\end{align*}
\item \begin{align*}
&& \mathbb{E}(Y_k) &= \int_0^1 y f_{Y_k}(y) \d y \\
&&&= \int_0^1 n \binom{n-1}{k-1} y^{k}(1-y)^{n-k} \\
&&&= n \binom{n-1}{k-1}\int_0^1 y^{k}(1-y)^{n-k} \d y \\
&&&= n \binom{n-1}{k-1}\int_0^1 y^{k+1-1}(1-y)^{n+1-(k+1)} \d y \\
&&&= n \binom{n-1}{k-1} \frac{1}{(n+1) \binom{n}{k}}\\
&&&= \frac{n}{n+1} \cdot \frac{k}{n} \\
&&&= \frac{k}{n+1}
\end{align*}
\end{questionparts}
Although it was only attempted by 14% of the candidates, it was moderately successfully done, just slightly less so than question 8 but still with a mean over 9/20. Part (i) was frequently poorly justified with candidates often attempting to describe the given expression without connecting it to the actual problem. However, binomial coefficients and factorials were generally successfully manipulated in part (ii), although care had to be taken to choose a form for the second expression which would be useful later. Most realised that it was necessary to differentiate the cdf and apply the previous result, though some failed to take care of the details. The deduction of the integral at the end of part (ii) was generally well done, and the majority correctly spotted that this a constant multiple of the integral of the pdf. Part (iii) was largely well done by those reaching this stage, either by recognising that this was of the form of the previous part or by direct integration by parts. A common mistake in this last part was to forget the constant term of the pdf when calculating the expectation.