Recitation: The Absent-Minded Professor

What this lecture covers

This recitation problem uses a professor who double-books two student appointments to practice exponential random variables. It asks for the expected total time from when the first student arrives until the second student leaves, given that appointment lengths are exponential and the second student arrives five minutes late.

It reinforces the memoryless property of the exponential distribution and the total expectation theorem, both introduced earlier in the course, by applying them to a scenario with two possible timelines depending on whether the first appointment finishes before or after the second student arrives.

Key ideas

• Exponential random variable: a continuous, non-negative random variable with rate lambda, commonly used to model durations; its CDF is 1 - e^(-lambda*t) and its mean is 1/lambda.
• Memoryless property: if you observe an exponential random variable partway through, the remaining time behaves exactly like a fresh exponential with the same rate, independent of time already elapsed.
• Splitting into scenarios: the total time depends on whether the first student finishes within five minutes (before the second arrives) or takes longer than five minutes (so the second student waits).
• Total expectation theorem: the overall expected time is found by weighting the expected time in each scenario by the probability of that scenario.
• Naive averaging can mislead: simply adding the two individual expected appointment lengths gives 60 minutes, but the actual answer is slightly higher because of the idle gap that can occur when the first student finishes early.

Walkthrough

Setting up the two appointments (0:00)

Two students, T1 and T2, each take an independent exponential amount of time with mean 30 minutes. The first arrives on time, and the second arrives exactly five minutes later. The goal is the expected time X from when the first student arrives until the second student leaves.

The two possible timelines (3:03)

If the first student's appointment lasts more than five minutes, the second student is already waiting when the first leaves, so the two appointments run back to back after the five-minute mark. If the first student finishes within five minutes, the professor sits idle until the second student arrives, adding a gap that a naive 60-minute estimate misses.

Applying the memoryless property (5:12)

In the case where the first appointment runs past five minutes, the memoryless property says the remaining time behaves like a brand-new exponential with mean 30, so the expected total time in that case is 5 (guaranteed) + 30 (remainder of the first appointment) + 30 (second appointment) = 65 minutes. In the case where the first appointment finishes within five minutes, the total time is simply 5 + 30 = 35 minutes, since the second student's appointment starts fresh once they arrive.

Combining with total expectation (10:18)

Weighting the 35-minute and 65-minute scenarios by their respective probabilities, 1 - e^(-5/30) and e^(-5/30), and summing gives an expected total time of approximately 60.4 minutes — a little more than the naive 60-minute guess, because of the wasted idle time when the first student leaves early.

Before you watch

• Know the exponential distribution's CDF and mean, and be comfortable with its memoryless property.
• Review the total expectation theorem for combining expectations across mutually exclusive scenarios.
• Being familiar with splitting a problem into cases based on a conditioning event will make the setup easier to follow.

Check your understanding

1. Why does the naive calculation of adding the two mean appointment lengths underestimate the true expected time?
2. How does the memoryless property justify treating the remainder of the first appointment as a fresh exponential random variable?
3. What are the two scenarios the problem is split into, and what event defines the boundary between them?
4. Why does an idle gap only occur in one of the two scenarios?