Year: 2011
Paper: 3
Question Number: 12
Course: UFM Statistics
Section: Probability Generating Functions
The percentages attempting larger numbers of questions were higher this year than formerly. More than 90% attempted at least five questions and there were 30% that didn't attempt at least six questions. About 25% made substantive attempts at more than six questions, of which a very small number indeed were high scoring candidates that had perhaps done extra questions (well) for fun, but mostly these were cases of candidates not being able to complete six good solutions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
The random variable $N$ takes positive integer values
and has pgf (probability generating function) $\G(t)$.
The random variables $X_i$, where $i=1$, $2$, $3$, $\ldots,$
are independently and identically
distributed, each with pgf ${H}(t)$. The random variables $X_i$
are also independent of $N$. The random variable $Y$ is defined by
\[
Y=
\sum_{i=1}^N X_i \;.
\]
Given that the pgf of $Y$ is $\G(H(t))$,
show that
\[
\E(Y) = \E(N)\E(X_i)
\text{ and }
\var(Y) = \var(N)\big(\E(X_i)\big)^2 + \E(N) \var(X_i)
\,.\]
A fair coin is tossed until a head occurs. The total number of tosses is
$N$. The coin is then tossed a further $N$ times and the total number of
heads in these $N$ tosses is $Y$. Find in this particular case
the pgf of $Y$, $\E(Y)$, $\var(Y)$ and $\P(Y=r)$.Recall that for a random variable $Z$ with pgf $F(t)$ we have $F(1) = 1$, $\E[Z] = F'(1)$ and $\E[Z^2] = F''(1) +F'(1)$ so
\begin{align*}
&& \E[Y] &= G'(H(1))H'(1) \\
&&&= G'(1)H'(1) \\
&&&= \E[N]\E[X_i] \\
\\
&& \E[Y^2] &= G''(H(1))(H'(1))^2+G'(H(1))H''(1) + G'(H(1))H'(1) \\
&&&= G''(1)(H'(1))^2+G'(1)H''(1) + G'(1)H'(1) \\
&&&= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N](\E[X_i^2]-\E[X_i]) + \E[N]\E[X_i] \\
&&&= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N]\E[X_i^2] \\
&& \var[Y] &= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N]\E[X_i^2] - (\E[N])^2(\E[X_i])^2\\
&&&= (\var[N]+(\E[N])^2-\E[N])(\E[X_i])^2 + \E[N](\var[X_i]+\E[X_i]^2) - (\E[N])^2(\E[X_i])^2\\
&&&= \var[N](\E[X_i])^2 + \E[N]\var[X_i]
\end{align*}
Notice that $N \sim Geo(\tfrac12)$ and $Y = \sum_{i=1}^N X_i$ where $X_i$ are Bernoulli. We have that $G(t) = \frac{\frac12}{1-\frac12z}$ and $H(t) = \frac12+\frac12p$ so the pgf of $Y$ is $G(H(t) = \frac{\frac12}{1 - \frac14-\frac14p} = \frac{2}{3-p}$.
\begin{align*}
&& \E[X_i] &= \frac12\\
&& \var[X_i] &= \frac14 \\
&& \E[N] &= 2 \\
&& \var[N] &= 2 \\
\\
&& \E[Y] &= 2 \cdot \frac12 = 1 \\
&& \var[Y] &= 2 \cdot \frac14 + 2 \frac14 = 1
\\
&& \mathbb{P}(Y=r) &= \tfrac23 \left ( \tfrac13 \right)^r
\end{align*}
This question ran a close second to number 11 for unpopularity, but reflected the same level of success. Most attempts followed the method of the question, and if they got off on the right foot, often got most of the way through. Some struggled with the algebra for the variance result, and a few tripped up on the standard pgf for the number of tosses to the first head. Strangely, having found the pgf for Y successfully, and used it or the results of the question for expectation and variance, the final probabilities were often wrong, and not merely from overlooking the initial case.