Problem source

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 source

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.