Year: 2018
Paper: 3
Question Number: 13
Course: UFM Statistics
Section: Probability Generating Functions
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: 1484.0
Banger Comparisons: 1
The random variable $X$ takes only non-negative integer values and has probability generating function $\G(t)$.
Show that
\[
\P(X = 0 \text{ or } 2 \text{ or } 4 \text { or } 6 \ \ldots )
= \frac{1}{2}\big(\G\left(1\right)+\G\left(-1\right)\big).
\]
You are now given that $X$ has a Poisson distribution with mean $\lambda$. Show that
\[
\G(t) = \e^{-\lambda(1-t)}
\,.
\]
\begin{questionparts}
\item
The random variable $Y$ is defined by
\[
\P(Y=r)=
\begin{cases}
k\P(X=r) & \text{if $r=0, \ 2, \ 4, \ 6, \ \ldots$ \ }, \\[2mm]
0& \text{otherwise},
\end{cases}
\]
where $k$ is an appropriate constant.
Show that the probability generating function of $Y$ is $\dfrac{\cosh\lambda t}{\cosh\lambda}\,$.
Deduce that $\E(Y) < \lambda$ for $\lambda > 0\,$.
\item The random variable $Z$ is defined by
\[\P(Z=r)=
\begin{cases}
c \P(X=r) &
\text{if $r = 0, \ 4, \ 8, \ 12, \ \ldots \ $}, \\[2mm]
0& \text{otherwise,}
\end{cases}
\]
where $c$ is an appropriate constant.
Is $\E(Z) < \lambda$ for all positive values of $\lambda\,$?
\end{questionparts}\begin{align*}
&&G_X(t) &= \mathbb{E}(t^N) \\
&&&= \sum_{k=0}^{\infty} \mathbb{P}(X = k) t^k \\
\Rightarrow && G_X(1) &= \sum_{k=0}^{\infty} \mathbb{P}(X = k) \\
\Rightarrow && G_X(-1) &= \sum_{k=0}^{\infty} (-1)^k\mathbb{P}(X = k) \\
\Rightarrow && \frac12 (G_X(1) + G_X(-1) &= \sum_{k=0}^{\infty} \frac12 (1 + (-1)^k) \mathbb{P}(X = k) \\
&&&= \sum_{k=0}^{\infty} \mathbb{P}(X =2k)
\end{align*}
\begin{questionparts}
\item \begin{align*}
1 &= \sum_r \mathbb{P}(Y = r) \\
&= \sum_{k=0}^\infty k \cdot \mathbb{P}(X = 2k) \\
&= k \cdot \frac12 \l e^{-\lambda(1-1) } + e^{-\lambda(1+1) }\r \\
&= \frac{k}{2}(1+e^{-2\lambda})
\end{align*}
Therefore $k = \frac{2}{1+e^{-2\lambda}} = e^{\lambda} \frac{1}{\cosh \lambda}$
\begin{align*}
&& G_X(t) + G_X(-t) &= \sum_{k=0}^\infty \mathbb{P}(X = k)t^k(1^k + (-1)^k) \\
&&&= \sum_{k=0}^\infty \mathbb{P}(X = k)t^k(1^k + (-1)^k) \\
&&&= 2\sum_{k=0}^\infty \mathbb{P}(X = 2k)t^{2k} \\
&&&= 2\sum_{k=0}^\infty \frac{1}{k}\mathbb{P}(Y = 2k)t^{2k} \\
&&&= \frac{2}{k}G_Y(t) \\
\Rightarrow && G_Y(t) &= k \cdot \frac{G_X(t) + G_X(-t)}{2} \\
&&&= k\frac{e^{-\lambda(1-t)} + e^{-\lambda(1+t)}}{2} \\
&&&= \frac{e^\lambda}{\cosh \lambda} \frac{e^{-\lambda} (e^{\lambda t}+e^{-\lambda t}) }{2} \\
&&&= \frac{\cosh \lambda t}{\cosh \lambda}
\end{align*}
Since $\mathbb{E}(Y) = G_Y'(1)$ and
\begin{align*}
&& G_Y'(t) &= \frac{\lambda \sinh \lambda t}{\cosh \lambda t} \\
\Rightarrow && G_Y'(1) &= \lambda \tanh \lambda \\
&&&< \lambda
\end{align*}
since $\tanh x < 1$
\item \begin{align*}
&& \frac14 \l G_X(t) + G_X(it) +G_X(-t) + G_X(-it) \r &= \sum_{k=0}^\infty \mathbb{P}(X=k)t^k (1 + i^k + (-1)^k + (-i)^k) \\
&&&= \sum_{k=0}^\infty \mathbb{P}(X = 4k)t^{4k} \\
&&&= \frac{G_Z(t)}{c}
\end{align*}
Since $G_Z(1) = 1$ we must have $c = \frac1{\frac14 \l G_X(1) + G_X(i) +G_X(-1) + G_X(-i) \r}$
\begin{align*}
&& c &= \frac{4e^{\lambda}}{e^{\lambda} + e^{-\lambda} + e^{i\lambda} + e^{-i\lambda}} \\
&&&= \frac{2e^{\lambda}}{\cosh \lambda + \cos \lambda} \\
&& G_Z(t) &= c \cdot \frac14 \l e^{-\lambda(1-t)}+e^{-\lambda(1-it)}+e^{-\lambda(1+t)}+e^{-\lambda(1+it)} \r \\
&&&= \frac{ce^{-\lambda t}}{4} \l 2\cosh \lambda t + 2 \cos \lambda t\r \\
&&&= \frac{\cosh \lambda t + \cos \lambda t}{\cosh \lambda + \cos \lambda}
\end{align*}
We are interested in $G_Z'(1)$ so:
\begin{align*}
&& G_Z'(t) &= \frac{\lambda (\sinh \lambda t - \sin \lambda t)}{\cosh \lambda + \cos \lambda }
\end{align*}
Considering various values of $\lambda$, it makes sense to look at $\lambda = \pi$ (since $\cos \lambda = -1$ and the denominator will be small). From this we can see:
\begin{align*}
G'_Z(1) &= \frac{\pi (\sinh \pi-0)}{\cosh \pi-1} \\
&= \frac{\pi}{\tanh \frac{\pi}{2}} > \pi
\end{align*}
So $\mathbb{E}(Z)$ is larger than $\lambda$ for $\lambda = \pi$ (and probably many others)
\end{questionparts}
As already mentioned, this was the joint least popular question, although its success rate was only very marginally less than that for question 2. The first result of the stem was generally well done with clear explanation. Likewise, the pgf for the Poisson distribution was generally well calculated, although a few candidates merely quoted the result from the formula booklet and then performed some trivial rearrangements which, of course, did not satisfy what was required. In part (i), candidates recognised how to apply the stem to calculate, and then either applied it further to express the pgf as or directly identified the cosh power series. Similarly, most justified the result for the expectation correctly. For part (ii), neatly gave the pgf of (not asked but required to progress) from the previous result, although frequently, candidates worked from or directly from power series again. Occasional attempts to use the same method again (i.e. considering) failed. Candidates generally failed to complete the last stage, either not attempting it, or providing poor justifications of the existence of counterexamples as → ∞ rather than simply furnishing an explicit counterexample.