Poisson Bus Waiting Time

Arrival rate: λ = 0.1 buses/minute

Mean interarrival time = 1 / λ = 1 / 0.1 = 10 minutes

Imagine arriving at a random time

🚌 Last Bus

👤 You Arrive

🚌 Next Bus

← Time since last bus

Time until next bus →

1. How long until the next bus?

For a Poisson process, the waiting time until the next arrival follows an exponential distribution.

W ~ Exp(λ)

The mean of an exponential random variable is:

E[W] = 1 / λ

E[W] = 1 / 0.1 = 10 minutes

Therefore, even if you arrive at a completely random time, your expected wait is still:

Expected time until next bus

10 minutes

2. How long ago did the last bus depart?

A stationary Poisson process looks the same if we look forward or backward in time.

Therefore, the time since the previous bus also follows an exponential distribution:

A ~ Exp(λ)

E[A] = 1 / λ = 10 minutes

Expected time since last bus

10 minutes

The surprising symmetry

🚌 ← 👤

Looking backward

E[A] = 10 min

👤 → 🚌

Looking forward

E[W] = 10 min

⚠️ A subtle point: the inspection paradox

You might now think:

Last bus ← 10 min → YOU ← 10 min → Next bus

So the bus-to-bus interval containing you has expected length:

E[A + W] = E[A] + E[W] = 10 + 10 = 20 minutes

But an ordinary bus-to-bus interval has mean only 10 minutes.

      Why?

      Because when you arrive at a random time, you are more likely
      to land inside a long bus gap than a short one.

      A 20-minute gap gives you twice as many opportunities to arrive
      inside it as a 10-minute gap.

      This phenomenon is called the
      inspection paradox.

Final Answer

1. Expected wait until next bus

10 min

2. Expected time since last bus