Year: 2012
Paper: 2
Question Number: 13
Course: UFM Statistics
Section: Poisson Distribution
There were just over 1000 entries for paper II this year, almost exactly the same number as last year. Overall, the paper was found marginally easier than its predecessor, which means that it was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50 candidates scored 100 marks or more, with more than 400 gaining at least half marks on the paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40 marks. It was pleasing to note that the advice of recent years, encouraging students not to make attempts at lots of early parts to questions but rather to spend their time getting to grips with the six that can count towards their paper total, was more obviously being heeded in 2012 than I can recall being the case previously. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were the most popular questions, although each drew only around 800 "hits" – fewer than usual. Questions 3 – 5 & 8 were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most successfully attempted question on the paper – with very little interest shown in the rest of Sections B or C.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
In this question, you may assume that $\displaystyle \int_0^\infty \!\!\! \e^{-x^2/2} \d x = \sqrt{\tfrac12 \pi}\,$.
The number of supermarkets situated in any given region can be modelled by a Poisson random variable, where the mean is $k$ times the area of the given region.
Find the probability that there are no supermarkets within a circle of radius $y$.
The random variable $Y$ denotes the distance between a randomly chosen point in the region and the nearest supermarket. Write down $\P(Y < y)$ and hence show that the probability density function of $Y$ is $\displaystyle 2\pi y k \e^{-\pi k y^2}$ for $y\ge0$. Find $\E(Y)$ and show that $\var(Y) = \dfrac{4-\pi}{4\pi k}$.
A circle radius $y$ has a number of supermarkets $X$ where $X \sim Po(k \pi y^2)$.
\[ \mathbb{P}(X = 0) = e^{-k\pi y^2} \frac{1}{0!} = e^{-k\pi y^2} \]
The probability $\mathbb{P}(Y < y) = 1-\mathbb{P}(Y \geq y) = 1-e^{-k\pi y^2}$, and in particular $f_Y(y) = 2k\pi y e^{-k\pi y^2}$ (by differentiating).
\begin{align*}
&& \mathbb{E}(Y) &= \int_0^\infty yf_Y(y) \d y \\
&&&= \int_0^\infty 2\pi y^2 k e^{-\pi k y^2} \d y \\
\sigma^2 = \frac{1}{2k\pi}:&&&= \pi k \sqrt{2 \pi}\sigma \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sigma }y^2 e^{-\frac12 \cdot 2\pi k y^2} \d y \\
&&&=\pi k \sqrt{2 \pi}\sigma \mathbb{E}\left (N(0, \sigma^2)^2 \right) \\
&&&= \pi k \sqrt{2 \pi}\sigma\sigma^2 \\
&&&= \pi k \sqrt{2 \pi} \frac{1}{(2k\pi)^{3/2}} \\
&&&= \frac{1}{2\sqrt{k}}
\end{align*}
\begin{align*}
&& \mathbb{E}(Y^2) &= \int_0^\infty y^2f_Y(y) \d y \\
&&&= \int_0^\infty 2\pi y^3 k e^{-\pi k y^2} \d y \\
&&&= \int_0^{\infty}y^2 2y \pi k e^{-\pi k y^2} \d y \\ \\
&&&= \left [-y^2 e^{-\pi k y^2}\right]_0^{\infty}+\int_0^\infty 2ye^{-\pi k y^2} \d y \\
&&&= \left [-\frac{1}{\pi k}e^{-\pi k y^2} \right]_0^{\infty} \\
&&&= \frac{1}{\pi k} \\
\Rightarrow && \textrm{Var}(Y) &= \mathbb{E}(Y^2) - \left [ \mathbb{E}(Y)\right]^2 \\
&&&= \frac{1}{\pi k} - \frac{1}{4k} \\
&&&= \frac{4 - \pi}{4\pi k}
\end{align*}
This question drew very little interest from candidates. Most attempts gained the first couple of answers and then differentiated to find the pdf of Y. In the attempts to find E(Y) and E(Y²), most candidates rightly attempted integration by parts, although some coefficients went astray when either making a substitution or comparing the integrals with the corresponding ones of the standard normal distribution. Slightly surprisingly, it was relatively common to find E(Y²) correct but the variance incorrect, as candidates failed to make this modest extra step without error.