Year: 2014
Paper: 3
Question Number: 13
Course: UFM Statistics
Section: Probability Generating Functions
A 10% increase in the number of candidates and the popularity of all questions ensured that all questions had a good number of attempts, though the first two questions were very much the most popular. Every question received at least one absolutely correct solution. In most cases when candidates submitted more than six solutions, the extra ones were rarely substantial attempts. Five sixths gave in at least six attempts.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
I play a game which has repeated rounds. Before the
first round, my score is $0$. Each round can have three outcomes:
\begin{enumerate}
\item my score is unchanged and the game ends;
\item my score is unchanged and I continue to the next round;
\item my score is increased by one and I continue to the next round.
\end{enumerate}
The probabilities of these outcomes are $a$, $b$
and $c$, respectively (the same in each round),
where $a+b+c=1$ and $abc\ne0$. The random
variable $N$ represents my score at the end of a randomly chosen
game.
Let $G(t)$ be the probability generating function of
$N$.
\begin{questionparts}
\item
Suppose in the first round, the game ends.
Show that the probability generating
function conditional on this happening is 1.
\item
Suppose in the first round,
the game continues to the next round with no change in
score.
Show that the probability generating function conditional on this happening
is $G(t)$.
\item By comparing the coefficients of $t^n$, show that
$
G(t) = a + bG(t) + ctG(t)\,.
$
Deduce that,
for $n\ge0$,
\[
P(N=n) = \frac{ac^n}{(1-b)^{n+1}}\,.
\]
\item Show further that, for $n\ge0$,
\[
P(N=n) = \frac{\mu^n}{(1+\mu)^{n+1}}\,,
\]
where $\mu=\E(N)$.
\end{questionparts}\begin{questionparts}
\item If the game ends in the first round then the score is exactly $0$ and the pgf is $1\cdot x^0 = 1$
\item If the game moves onto the next round with no change in the first round then it's as if nothing happened, therefore the pgf is the original pgf $G(t)$
\item If the game moves into the next round with the score increased by one, then the pgf is $tG(t)$ since all the scores are increased by $1$. Therefore
\begin{align*}
&& G(t) &= \E[t^N] \\
&&&= \E[\E[t^N | \text{first round}]] \\
&&&= a + bG(t) + ctG(t) \\
\Rightarrow && G(t)(1-b-ct) = a \\
\Rightarrow && G(t) &= \frac{a}{(1-b)-ct} \\
&&&= \frac{a}{(1-b)} \frac{1}{1- \left(\frac{c}{1-b}\right)t} \\
&&&= \sum_{n=0}^\infty \frac{a}{1-b} \frac{c^n}{(1-b)^n} t^n\\
&&&= \sum_{n=0}^{\infty} \frac{ac^n}{(1-b)^{n+1}}t^n
\end{align*}
Therefore by comparing coefficients, $\mathbb{P}(N=n) = \frac{ac^n}{(1-b)^{n+1}}$
\item $\,$ \begin{align*}
&& \E[N] &= G'(1) \\
&&&= \frac{ac}{((1-b)-c)^2} \\
&&&= \frac{ac}{a^2} = \frac{c}{a} \\
\\
&& \frac{\mu^n}{(1+\mu)^{n+1}} &= \frac{c^na^{-n}}{(a+c)^{n+1}a^{-n-1}} \\
&&&= \frac{ac^n}{(a+c)^{n+1}}\\
&&&= \frac{ac^n}{(1-b)^{n+1}}\\
&&&= \mathbb{P}(N=n)
\end{align*} as required
\end{questionparts}
Just a handful of candidates more attempted this than question 12, but scoring marginally less with one quarter marks. A small number did just part (i), but otherwise candidates tended to either score zero or nearly all of the marks. There was some very shaky logic in finding the first result of part (iii) and then failing to deduce, as required, the probability result. Quite often, the working for the required value in part (iv), whilst usually correct, was extremely convoluted.