## Probability sheet 4, number 9

I would like to highly recommend working out number 9 on probability sheet 4, simply because it’s so amusing.

It reads like this:

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Passengers arrive at a bus stop at rate 1 per minute. Find the distribution of the number

of passengers boarding a typical bus in two cases: (a) buses arrive regularly every 10

minutes; (b) buses arrive as a Poisson process with rate 1 per 10 minutes. Which one has

higher variance?

I arrive at the bus stop at 2pm. Find the distribution of the number of other passengers

boarding the same bus as me in the two cases above.

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Most of the problem is straightforward, and I won’t go over it here. For example, the mean number of passengers boarding will be 10 in either case, but, naturally, the variance will be higher in case (b). But what’s amusing is the second part, where we assume you arrive at the bus at some given time, and calculate the mean number of *other* passengers boarding with you. For case (a), you find 10 as before, but for case (b), the mean turns out to be 20!

This situation is sometimes referred to as the ‘inspector’s paradox’. That is, if you’re an inspector trying to check up on the mean number of passengers boarding at a given stop by arriving at 2 PM for a number of days to take the bus, you will tend to find a larger mean than the true mean for the average bus, at least in the model (b). You should ask yourself why this happens.

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