Year: 2022
Paper: 3
Question Number: 11
Course: LFM Stats And Pure
Section: Binomial Distribution
One question was attempted by well over 90% of the candidates two others by about 90%, and a fourth by over 80%. Two questions were attempted by about half the candidates and a further three questions by about a third of the candidates. Even the other three received attempts from a sixth of the candidates or more, meaning that even the least popular questions were markedly more popular than their counterparts in previous years. Nearly 90% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A fair coin is tossed $N$ times and the random variable $X$ records the number of heads. The mean deviation, $\delta$, of $X$ is defined by
\[ \delta = \mathrm{E}\big(|X - \mu|\big) \]
where $\mu$ is the mean of $X$.
\begin{questionparts}
\item Let $N = 2n$ where $n$ is a positive integer.
\begin{enumerate}
\item Show that $\mathrm{P}(X \leqslant n-1) = \frac{1}{2}\big(1 - \mathrm{P}(X=n)\big)$.
\item Show that
\[ \delta = \sum_{r=0}^{n-1}(n-r)\binom{2n}{r}\frac{1}{2^{2n-1}}\,. \]
\item Show that for $r > 0$,
\[ r\binom{2n}{r} = 2n\binom{2n-1}{r-1}\,. \]
Hence show that
\[ \delta = \frac{n}{2^{2n}}\binom{2n}{n}\,. \]
\end{enumerate}
\item Find a similar expression for $\delta$ in the case $N = 2n+1$.
\end{questionparts}\begin{questionparts}
\begin{enumerate}
\item First notice that $\mathbb{P}(X \leq n-1) = \mathbb{P}(X \geq n + 1)$ so
\begin{align*}
&& 1 & = \mathbb{P}(X \leq n-1) + \mathbb{P}(X = n) + \mathbb{P}(X \geq n + 1) \\
&&&= 2 \mathbb{P}(X \leq n-1) + \mathbb{P}(X = n) \\
\Rightarrow && \mathbb{P}(X \leq n-1) &= \tfrac12(1 - \mathbb{P}(X = n))
\end{align*}
\item $\,$ \begin{align*}
&& \delta &= \E[|X-\mu|] \\
&&&= \E[|X - N|] \\
&&&= \sum_{r=0}^{n-1} (n - r) \mathbb{P}(X = r) + \sum_{r=n+1}^{2n} (r-n) \mathbb{P}(X=r) \\
&&&= \sum_{r=0}^{n-1} (n - r) \mathbb{P}(X = r) + \sum_{r=0}^{n-1} (n-r) \mathbb{P}(X=r) \\
&&&= \sum_{r=0}^{n-1} 2(n-r) \frac{1}{2^{2n}} \binom{2n}{r}\\
&&&= \sum_{r=0}^{n-1} (n-r) \frac{1}{2^{2n-1}} \binom{2n}{r}\\
\end{align*}
\item Notice that \begin{align*}
&& r \binom{2n}{r} &= \frac{r(2n)!}{r!(2n-r)!} \\
&&&= \frac{2n \cdot (2n-1)!}{(r-1)!(2n-1-(r-1))!} \\
&&&= 2n \binom{2n-1}{r-1}
\end{align*}
So \begin{align*}
&& \delta &= \sum_{r=0}^{n-1} (n-r) \frac{1}{2^{2n-1}} \binom{2n}{r} \\
&&&= 2n\mathbb{P}(X \leq n-1) - \frac{1}{2^{2n-1}} \sum_{r=0}^{n-1} r \binom{2n}{r} \\
&&&= 2n \mathbb{P}(X \leq n-1) - \frac{1}{2^{2n-1}} \sum_{r=1}^{n-1} 2n \binom{2n-1}{r-1} \\
&&&= 2n \mathbb{P}(X \leq n-1) - \frac{1}{2^{2n-1}} 2n \sum_{r=0}^{n-2} \binom{2n-1}{r} \\
&&&= 2n \mathbb{P}(X \leq n-1) - \frac{1}{2^{2n-1}} 2n \left (\frac12 2^{2n-1} - \binom{2n-1}{n-1} \right)\\
&&&= 2n \frac12 (1 - \mathbb{P}(X=n) ) - n + \frac{2n}{2^{2n-1}} \binom{2n-1}{n-1} \\
&&&= -\frac{n}{2^{2n}} \binom{2n}{n} + \frac{2n}{2^{2n}}\binom{2n}{n} \\
&&&= \frac{n}{2^{2n}} \binom{2n}{n}
\end{align*}
\end{enumerate}
\item When $N = 2n+1$, $\mu = n +\frac12$ and so
\begin{align*}
&& \delta &= \E[|X-\mu|] \\
&&&= \sum_{i=0}^n (n + \tfrac12 - i) \frac{1}{2^{2n+1}} \binom{2n+1}{i} + \sum_{i=n+1}^{2n+1} (i-n - \tfrac12) \frac{1}{2^{2n+1}} \binom{2n+1}{i} \\
&&&= 2\sum_{i=0}^n (n + \tfrac12 - i) \frac{1}{2^{2n+1}} \binom{2n+1}{i} \\
&&&= \frac{(2n +1)}{2^{2n+1}}\sum_{i=0}^n \binom{2n+1}i - \frac{2}{2^{2n+1}}\sum_{i=0}^n i \binom{2n+1}{i} \\
&&&= \frac{(2n +1)}{2^{2n+1}}\sum_{i=0}^n \binom{2n+1}i - \frac{2}{2^{2n+1}}\sum_{i=1}^n (2n+1) \binom{2n}{i-1} \\
&&&= \frac{(2n +1)}{2^{2n+1}}2^{2n} - \frac{2(2n+1)}{2^{2n+1}}\sum_{i=0}^{n-1} \binom{2n}{i} \\
&&&= \frac{(2n +1)}{2} - \frac{2(2n+1)}{2^{2n+1}} \frac12\left (2^{2n} - \binom{2n}{n} \right) \\
&&&= \frac{2n+1}{2^{2n+1}}\binom{2n}{n}
\end{align*}
\end{questionparts}
A quarter of the candidates attempted this, scoring a mean of one third marks. Of these, about a quarter made little or no progress. However, there were also several very good attempts achieving most or all of the marks available. In part (i)(a) the majority of candidates noticed the symmetry of the distribution and were therefore able to answer this part well, although errors such as omitting the binomial coefficients and forgetting that X could take the value 0 were made in some cases. In part (i)(b) most candidates were able to see that the modulus sign in the sum effectively meant that the calculation of δ should be split into two sums. However, in several cases candidates simply observed that the given result followed from the two sums by symmetry without sufficient justification to earn the marks. Almost all candidates who attempted part (i)(c) were able to show the first result by applying the definition of C(2n,r) and then cancelling terms. A small number of candidates argued the result by viewing the two expressions as representing different ways of counting the same total number of things. For the next part of (i)(c) the majority of candidates split the sum and then applied the previous result to the second term. Many candidates, however, did not pay sufficient attention to the case where r = 0 and ended up with an incorrect term in the sum. Many candidates jumped straight to the given answer at this point and therefore did not show sufficient detail to earn the remaining marks for this part. Many of the candidates who progressed further with this part dealt with the two sums separately, but some used the fact that C(2n,r) = C(2n-1,r-1) + C(2n-1,r) to rearrange into a sum of differences, most of which then cancelled out. Candidates who had completed part (i)(c) well were able to apply the same methods to the case in part (ii) and this part was generally completed well, although a small number of candidates failed to notice that the expression for the mean in terms of n had changed.