Year: 2009
Paper: 3
Question Number: 12
Course: UFM Statistics
Section: Probability Generating Functions
The vast majority of candidates (in excess of 95%) attempted at least five questions, and nearly a quarter attempted more than six questions, though very few doing so achieved high scores (about 2%). Most attempting more than six questions were submitting fragmentary answers, which, as the rubric informed candidates, earned little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item
Albert tosses a fair coin $k$ times, where $k$ is a given positive integer. The number of heads he gets is $X_1$. He then tosses the coin $X_1$ times, getting $X_2$ heads. He then tosses the coin $X_2$ times, getting $X_3$ heads. The random variables $X_4$, $X_5$, $\ldots$ are defined similarly.
Write down $\E(X_1)$.
By considering $\E(X_2 \; \big\vert \; X_1 = x_1)$, or otherwise, show that $\E(X_2) = \frac14 k$.
Find $\displaystyle \sum_{i=1}^\infty \E(X_i)$.
\item Bertha has $k$ fair coins. She tosses the first coin until she gets a tail. The number of heads she gets before the first tail is $Y_1$. She then tosses the second coin until she gets a tail and the number of heads she gets with this coin before the first tail is $Y_2$. The random variables $Y_3, Y_4, \ldots\;$, $Y_k$ are defined similarly, and $Y= \sum\limits_{i=1}^k Y_i\,$.
Obtain the probability generating function of $Y$, and use it to find $\E(Y)$, $\var(Y)$ and $\P(Y=r)$.
\end{questionparts}\begin{questionparts}
\item $X_1 \sim B(k, \tfrac12)$ so $\E[X_1] = \frac{k}{2}$
Note that $X_2 | X_1 = x_1 \sim B(x_1, \tfrac12)$ so $\E[X_2 | X_1 = x_1) = \frac{x_1}{2}$ or $\E[X_2 | X_1] = \frac12 X_1$.
Therefore by the tower law, $\E[\E[X_2|X_1]] = \E[\frac12 X_1] = \frac14k$
Notice also that $\E[X_n] = \frac1{2^n} k$ and so
\begin{align*}
&& \sum_{i=1}^\infty \E[X_i] &= \sum_{i=1}^{\infty} \frac1{2^i} k \\
&&&= \frac{\frac12 k}{1-\frac12} = k
\end{align*}
\item Note that $Y_1 \sim Geo(\tfrac12)-1$ which has generating function $\E[t^{Y_1}] = \E[t^{G-1}] = \frac{\frac12 t}{1-(1-\frac12)t}\frac1{t} = \frac{\frac12}{1-\frac12t}$.
Notice that
\begin{align*}
&& \E \left [ t^Y \right] &= \E \left [ t^{\sum_{i=1}^kY_i} \right] \\
&&&= \prod_{i=1}^k \E[t^{Y_i}] \\
&&&= \frac{1}{(2-t)^k}
\end{align*}
Therefore $\E[Y] = G'(1) = k(2-1)^{-(k+1)} = k$
$\E[Y^2] = (tG'(t))'|_{t=1} = k(k+1)(2-1)^{-(k+2)}+k(2-1)^{-(k+1)} = k^2+2k$ so $\var[Y] = k^2+2k - k^2 =2 k$.
Finally $\mathbb{P}(Y=r) = \binom{k+r-1}{k} \frac{1}{2^{r+k}}$
\end{questionparts}
[Note: this second distribution is a negative binomial distribution]
About a tenth of the candidates attempted this, usually earning quarter marks. Quite often the conditional bit in part (i) threw them, so they were 3 marks down before they got into the question. 80% of the attempts did not obtain or use the pgf as required. A small number of candidates really knew their stuff and did it very well.