1 problem found
A densely populated circular island is divided into \(N\) concentric regions \(R_1\), \(R_2\), \(\ldots\,\), \(R_N\), such that the inner and outer radii of \(R_n\) are \(n-1\) km and \(n\) km, respectively. The average number of road accidents that occur in any one day in \(R_n\) is \(2-n/N\,\), independently of the number of accidents in any other region. Each day an observer selects a region at random, with a probability that is proportional to the area of the region, and records the number of road accidents, \(X\), that occur in it. Show that, in the long term, the average number of recorded accidents per day will be \[ 2-\frac16\left(1+\frac1N\right)\left(4-\frac1N\right)\;. \] [Note: \(\sum\limits_{n=1}^N n^2 = \frac16 N(N+1)(2N+1) \;\).] Show also that \[ \P(X=k) = \frac{\e^{-2}N^{-k-2}}{k!}\sum_{n=1}^N (2n-1)(2N-n)^k\e^{n/N} \;. \] Suppose now that \(N=3\) and that, on a particular day, two accidents were recorded. Show that the probability that \(R_2\) had been selected is \[ \frac{48}{48 + 45\e^{1/3} +25 \e^{-1/3}}\;. \]
Solution: The area of \(R_n\) is \(\pi(n^2 - (n-1)^2) = (2n-1)\pi\). The area of the whole region is \(\pi N^2\). \begin{align*} && \E[X] &= \E[\E[X | \text{choose region }n]] \\ &&&= \sum_{n=1}^N \left (2 - \frac{n}{N} \right) \cdot \frac{(2n-1)\pi}{N^2 \pi} \\ &&&= \sum_{n=1}^N \left (2\cdot \frac{(2n-1)\pi}{N^2 \pi} - \frac{n}{N}\cdot \frac{(2n-1)\pi}{N^2 \pi} \right) \\ &&&= 2 - \frac{1}{N^3} \sum_{n=1}^N (2n^2-n) \\ &&&= 2 - \frac{1}{N^3} \left (\frac{2N(N+1)(2N+1)}{6} - \frac{N(N+1)}{2} \right) \\ &&&= 2 - \frac{N+1}{6N^2} \left (2(2N+1)-3 \right) \\ &&&= 2 - \frac{N+1}{6N^2} (4N - 1) \\ &&&= 2 - \frac16 \left (1 + \frac1N \right) \left (4 - \frac1N \right) \end{align*} Modelling each region as \(Po(2 - n/N)\) we have \begin{align*} \mathbb{P}(X = k ) &= \sum_{n=1}^N \exp(-2 + n/N) \frac{(2-n/N)^k}{k!} \frac{2n-1}{N^2} \\ &= \frac{e^{-2}N^{-k-2}}{k!} \sum_{n=1}^N e^{n/N} (2N-n)^k(2n-1) \end{align*} as desired. Supposing \(N=3\) and two accidents then \begin{align*} \mathbb{P}(R_2 | X = 2) &= \frac{\frac{3}{9} e^{-4/3}\frac{(\frac43)^2}{2!}}{\mathbb{P}(X=2)} \\ &= \frac{\frac{3}{9} e^{-4/3} \frac{(\frac43)^2}{2!}}{\frac{1}{9} e^{-5/3} \frac{(\frac53)^2}{2!} + \frac{3}{9} e^{-4/3} \frac{(\frac43)^2}{2!} + \frac{5}{9} e^{-2/3} \frac{(\frac33)^2}{2!}} \\ &= \frac{3 \cdot 16}{25e^{-1/3} + 3 \cdot 16 + 5 \cdot 9e^{1/3}} \\ &= \frac{48}{25e^{-1/3} + 48 + 45e^{1/3}} \end{align*} as required.