Quant Interview Question - JMPC - Poisson Process
Question:
You are waiting for a bus at a bus station. The buses arrive at the station according to a Poisson process with an average arrival time of 10 minutes (λ=0.1/minλ=0.1/min). If the buses have been running for a long time and you arrive at the bus station at a random time, what is your expected waiting time? On average, how many minutes ago did the last bus leave?
1. Understanding the problem
We have:
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Buses arrive according to a Poisson process with rate λ=0.1 per minute.λ=0.1 per minute.
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So the average interarrival time is 1/λ=10 minutes.
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The process has been running for a long time (stationary).
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You arrive at a random time.
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Two questions:
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What is your expected waiting time until the next bus?
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On average, how many minutes ago did the last bus leave?
2. Key property of Poisson process (and renewal processes)
For a Poisson process, if you arrive at a random time, the time since the last bus (age) and the time until the next bus (excess life) have the same distribution in this case, due to memoryless property of exponential interarrival times.
Let X = interarrival time (exponential, mean 10 minutes).
For a Poisson process:
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Distribution of the forward waiting time is the same as the original interarrival time distribution: exponential with mean 10 minutes.
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But careful — this seems contradictory: If I pick a random time, the time until next bus is exponential with mean 10? Let’s check.
Actually, known result: For a Poisson process, the residual waiting time distribution seen by a random arrival is exponential with mean 1/λ. This is because of memorylessness.
What is a Poisson Process?
A Poisson process is a mathematical model for events that happen randomly, independently, and at a constant average rate over time.
Examples include:
Number of customers arriving at a store.
Number of calls received by a call center.
Number of accidents on a highway.
We usually denote the rate by:
λ = average rate of arrivals per unit time
(lambda)
Example: λ = 5 per hour means on average 5 events per hour.
Formal Definition:
A counting process {N(t), t ≥ 0} is a Poisson process with rate λ if:
N(0) = 0
It has independent increments
It has stationary increments
The number of events in interval of length t follows a Poisson distribution:
\(P(N(t) = k) = ( (ฮปt)^k / k! ) * e^{( โฮปt )}\)
Distribution of Events:
Number of events in time t is given by N(t) which follows Poisson distribution with parameter λt. This is denoted by the following expression:
N(t) ~ Poisson( λt )
Here's a numeric example to help you understand the concept:
A coffee shop receives customers at λ = 3 per hour.
What is the probability of receiving 5 customers in 2 hours?
Mean = λt = 3 × 2 = 6
So: P(N(2) = 5) = (6^5 / 5!) * e^(−6)
Interarrival time
The time between events (Tโ, Tโ, Tโ, …) are i.i.d exponential with parameter λ.
The PDF of interarrival time T is:
f(t) = λ e^(−λt), for t ≥ 0
E[T] = 1 / λ
Example:
If events happen at rate λ = 10 per minute,
average time between events is 1/10 minute = 6 seconds.
3. Expected waiting time until next bus
Let T = time until next bus from random arrival.
We know for a Poisson process, T∼Exp(λ), mean 1/λ.
Thus:
E[T] = 1/λ = 10 minutes.E[T] โ= 10 minutes.
So expected waiting time = 10 minutes.
4. Expected time since last bus
Let A = time since last bus (age).
For a Poisson process, by symmetry (and memoryless property of exponential distribution), the distribution of A is also exponential with mean 1/λ.
Thus:
E[A]=10 minutes.
