Year: 2023
Paper: 3
Question Number: 11
Course: UFM Statistics
Section: Poisson Distribution
No solution available for this problem.
The total entry was a marginal increase on that of 2022 (by just over 1%). Two questions were attempted by more than 90% of candidates, another two by 80%, and another two by about two thirds. The least popular questions were attempted by more than a sixth of candidates. All the questions were perfectly answered by at least three candidates (but mostly more than this), with one being perfectly answered by eighty candidates. Very nearly 90% of candidates attempted no more than 7 questions. One general comment regarding all the questions is that candidates need to make sure that they read the question carefully, paying particular attention to command words such as "hence" and "show that".
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Show that
\[\sum_{k=1}^{\infty} \frac{k+1}{k!}\, x^k = (x+1)\mathrm{e}^x - 1\,.\]
In the remainder of this question, $n$ is a fixed positive integer.
\begin{questionparts}
\item Random variable $Y$ has a Poisson distribution with mean $n$. One observation of $Y$ is taken. Random variable $D$ is defined as follows. If the observed value of $Y$ is zero then $D = 0$. If the observed value of $Y$ is $k$, where $k \geqslant 1$, then a fair $k$-sided die (with sides numbered $1$ to $k$) is rolled once and $D$ is the number shown on the die.
\begin{enumerate}
\item Write down $\mathrm{P}(D = 0)$.
\item Show, from the definition of the expectation of a random variable, that
\[\mathrm{E}(D) = \sum_{d=1}^{\infty} \left[ d \sum_{k=d}^{\infty} \left( \frac{1}{k} \cdot \frac{n^k}{k!}\, \mathrm{e}^{-n} \right) \right].\]
Show further that
\[\mathrm{E}(D) = \sum_{k=1}^{\infty} \left( \frac{1}{k} \cdot \frac{n^k}{k!}\, \mathrm{e}^{-n} \sum_{d=1}^{k} d \right).\]
\item Show that $\mathrm{E}(D) = \frac{1}{2}(n + 1 - \mathrm{e}^{-n})$.
\end{enumerate}
\item Random variables $X_1, X_2, \ldots, X_n$ all have Poisson distributions. For each $k \in \{1, 2, \ldots, n\}$, the mean of $X_k$ is $k$.
A fair $n$-sided die, with sides numbered $1$ to $n$, is rolled. When $k$ is the number shown, one observation of $X_k$ is recorded. Let $Z$ be the number recorded.
\begin{enumerate}
\item Find $\mathrm{P}(Z = 0)$.
\item Show that $\mathrm{E}(Z) > \mathrm{E}(D)$.
\end{enumerate}
\end{questionparts}
Although it was unpopular, those that attempted this scored better than on other Applied questions and, indeed, five of the Pure questions. Most candidates recognised that the expression on the left in the stem could be separated into two sums, one of which would produce eˣ − 1 while the other would produce xeˣ. A small number of candidates falsely seemed to assume that since Σ(xᵏ/k!) = eˣ it immediately follows that Σ(k+1)(xᵏ/k!) = (x+1)eˣ. Some candidates chose to write out the sums without using summation notation, which made some parts of the justification more difficult to express clearly. In part (i) most candidates were able to work out the value of P(D = 0) and a large number were able to give some justification of the first formula for E(D). Many however did not manage to justify fully how all parts of the required expression were deduced and in a question in which the answer to be reached is known, it is important that solutions clearly express the steps that are involved. In particular, many candidates simply stated that the sum over values of k ran from d to infinity without any comment that this is because there had to be at least d sides on the die. Similarly, many candidates failed to express the clear reasoning required to show the second form of E(D). Candidates were generally successful in using the formula for E(D) to complete part (i)(c) and the majority recognised the significance of the result in the stem to the work here. Many of the candidates who attempted part (ii) were able to make good progress, although there were some who failed to understand the sequence in which the events take place in this second situation. Most were able to find an initial expression for the value of P(Z = 0) and the majority recognised that this was a sum of a geometric series. However, several candidates calculated the sum to infinity instead of the required sum of n terms and some of those who correctly calculated the sum of n terms then made errors when dealing with the powers in the simplification. Many of the candidates who attempted to calculate E(Z) were able to reach a correct form, but relatively few recognised that changing the order of summation (as in part (i)) would again help to simplify the expression. Those who found the correct expression were generally able to justify that E(Z) > E(D), although some did not comment on the fact that the exponential term must be positive as part of their justification.