Bus Waiting Paradox

Expectation · Medium · Free problem

Buses arrive at a stop according to a Poisson process with rate $\lambda$, so the average time between buses is

/\lambda$. You show up at the bus stop at a uniformly random time.

  1. What is the expected time until the next bus arrives?
  1. Explain why this is NOT
    /(2\lambda)$ (half the average inter-arrival time), even though that seems like the natural guess.
  1. Generalize: for a renewal process with inter-arrival times having mean $\mu$ and variance $\sigma^2$, what is the expected remaining time?

Hints

  1. Think about which gaps you are more likely to land in -- are you sampling gaps uniformly, or is there a bias toward longer ones?
  2. The exponential distribution has a special property: $P(X > s + t \mid X > s) = P(X > t)$. What does this imply about the remaining time?
  3. For the general case, use the renewal theory result $E[W] = E[X^2]/(2\mu)$ and verify that it reduces to
    /\lambda$ for exponential inter-arrivals.

Worked Solution

How to Think About It: Your gut says: "The average gap between buses is

/\lambda$. I arrive at a random point in a random gap, so I should wait half the gap on average -- that is
/(2\lambda)$." That reasoning is correct for deterministic arrivals (buses every
/\lambda$ minutes exactly). But Poisson arrivals are random, and here is the catch: you are more likely to arrive during a long gap than a short one. This is called length-biased sampling or the inspection paradox. The gaps you "see" are not a representative sample of all gaps -- they are biased toward the longer ones. For exponential inter-arrivals, the memoryless property gives you a clean answer: the remaining time has the same distribution as a fresh inter-arrival time.

Quick Estimate: Suppose $\lambda = 1$ bus per minute. The naive guess is $E[W] = 0.5$ minutes. But the exponential distribution is memoryless: $P(X > s + t \mid X > s) = P(X > t)$. So no matter when you arrive within the current gap, the remaining time is $\text{Exp}(1)$ with mean

$. The answer is $E[W] = 1/\lambda = 1$ minute -- double the naive guess.

Approach: We derive the result two ways: via the memoryless property (Poisson-specific) and via the general renewal theory formula.

Formal Solution:

*Method 1 -- Memoryless Property:*

For a Poisson process, inter-arrival times are i.i.d. $\text{Exp}(\lambda)$. Suppose you arrive at time $s$ within the current inter-arrival interval. The remaining time $R$ satisfies:

$P(R > t) = P(X > s + t \mid X > s) = P(X > t) = e^{-\lambda t}$

So $R \sim \text{Exp}(\lambda)$ regardless of $s$, and $E[R] = 1/\lambda$.

*Method 2 -- General Renewal Theory (Inspection Paradox):*

For a renewal process with inter-arrival distribution having mean $\mu$ and second moment $E[X^2]$, the expected remaining time when you arrive at a random time is:

$E[W] = \frac{E[X^2]}{2\mu} = \frac{\mu}{2} + \frac{\sigma^2}{2\mu}$

For the exponential with $\mu = 1/\lambda$ and $\sigma^2 = 1/\lambda^2$:

$E[W] = \frac{1}{2\lambda} + \frac{1/\lambda^2}{2/\lambda} = \frac{1}{2\lambda} + \frac{1}{2\lambda} = \frac{1}{\lambda}$

The formula $E[W] = \mu/2 + \sigma^2/(2\mu)$ shows that $E[W] \geq \mu/2$ always, with equality only when $\sigma^2 = 0$ (deterministic arrivals). More variance in the gaps means longer expected waits.

Answer: For a Poisson process with rate $\lambda$, the expected wait is $E[W] = 1/\lambda$, not

/(2\lambda)$, due to the memoryless property of the exponential distribution. For a general renewal process, $E[W] = (\mu^2 + \sigma^2)/(2\mu)$, which always exceeds $\mu/2$ unless arrivals are deterministic.

Intuition

The inspection paradox is one of those results that feels wrong until you internalize the sampling bias. When you arrive at a random time, you are not picking a random inter-arrival gap -- you are picking a random moment in time, and longer gaps occupy more moments. This is length-biased sampling. It is the same reason that the average class size experienced by a student is larger than the average class size reported by the registrar, and why the average bus you ride feels more crowded than the average bus on the route.

In quant work, this shows up whenever you sample a process at a random time. If you look at the spread on a random tick, you are more likely to catch a wide-spread period (because wide spreads last longer). If you measure the age of a position at a random audit time, you will overestimate typical holding periods. The general principle: any time you sample a duration-weighted quantity, you get a biased view, and the bias is proportional to the variance of the underlying distribution.

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