Bus-Waiting and the Memoryless Property

The Puzzle

Buses arrive according to a Poisson process with rate $\lambda = 0.1$ per minute. That means the mean interarrival time is:

$\frac{1}{\lambda} = 10 \text{ minutes}$

You show up at a random time.

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Questions:

1. What is your expected waiting time until the next bus?
2. On average, how long ago did the last bus depart?

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Key Concept: The Memoryless Property

In a Poisson process, the interarrival times are exponentially distributed. The exponential distribution has the memoryless property:

$P(T > s + t \mid T > s) = P(T > t)$

This implies that given no event has occurred so far, the distribution of the waiting time until the next event remains the same.

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Forward Recurrence Time (Next Bus)

Since arrivals follow a Poisson process, the time until the next bus from a random observation point is distributed:

$\text{Waiting time} \sim \text{Exp}(\lambda) \Rightarrow \mathbb{E}[\text{wait}] = \frac{1}{\lambda} = 10 \text{ minutes}$

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Backward Recurrence Time (Last Bus)

Likewise, the time since the last bus is also distributed as:

$\text{Time since last bus} \sim \text{Exp}(\lambda) \Rightarrow \mathbb{E}[\text{elapsed time}] = \frac{1}{\lambda} = 10 \text{ minutes}$

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Final Answers

1. Expected waiting time until next bus: $\boxed{10}$ minutes
2. Expected time since last bus: $\boxed{10}$ minutes

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Reference

• [1] Exponential distribution and the memoryless property