I went straight for the naive E[G]/2 and felt pretty good about it for a moment.
Model the bus arrival process as a renewal process where interarrival times are either x or 3x/2 with equal probability. Use the inspection paradox to compute the expected waiting time as the ratio of the second moment to twice the first moment of the interarrival time distribution.
Pro tip: Explicitly state that you are using the inspection paradox and that the expected waiting time is not simply half the mean interarrival time. This shows you understand the subtlety and can avoid the common trap.
Let the time between consecutive buses be T. Since each bus is independently late by 0 or x/2 with equal probability, T can be x (if both on time or both late) or 3x/2 (if one on time and one late). Compute the probabilities: P(T=x)=1/2, P(T=3x/2)=1/2.
Calculate E[T] = (1/2)x + (1/2)(3x/2) = 5x/4. Calculate E[T^2] = (1/2)x^2 + (1/2)(9x^2/4) = 13x^2/8.
For a renewal process, the expected waiting time for a randomly arriving passenger is E[T^2]/(2E[T]). Plug in the moments: (13x^2/8) / (2 * 5x/4) = (13x^2/8) / (5x/2) = (13x^2/8) * (2/(5x)) = 13x/20.
If time permits, confirm by conditioning on the state of the previous bus and using the memoryless property of the uniform arrival time. This can catch errors and demonstrate thoroughness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Didn't get to fully work this out under pressure.
First, clarify the original problem setup and the binary delay model, then explain how replacing it with a continuous uniform distribution changes the expected waiting time and variance. Use linearity of expectation and the properties of the uniform distribution to derive the new expected delay, and discuss the implications for the overall system.
Pro tip: Show that you can generalize the solution by considering the distribution of the sum of delays and applying the Central Limit Theorem for large numbers of buses, which demonstrates deeper statistical insight.
Restate the original problem: each bus is either on time (delay 0) or late by x/2 with some probability. Identify the expected delay and variance in that case.
State that now each bus's delay is uniformly distributed on [0, x/2]. Compute the mean and variance of a single bus's delay: mean = x/4, variance = (x/2)^2 / 12 = x^2 / 48.
For n buses, the expected total delay is n * (x/4) by linearity of expectation. Compare this to the original expected total delay (which was n * (x/2) * p, where p is the probability of being late).
The variance of the total delay is n * x^2 / 48. For large n, the total delay is approximately normally distributed by the CLT. Discuss how this affects the probability of exceeding a threshold.
Consider how the change affects the worst-case delay (now bounded by n*x/2 instead of possibly n*x/2 if all late) and the shape of the distribution. Mention that the uniform assumption may be more realistic for continuous delays.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the problem setup: a bus arrives either d minutes late with probability p or on time with probability 1-p, and you arrive at a random time. Then compute the expected wait by conditioning on the bus state: if on time, expected wait is x/2; if late, expected wait is x/2 + d. Combine using linearity of expectation to get E[wait] = x/2 + p*d.
Pro tip: Always state your assumptions explicitly (e.g., uniform arrival time, independence) and sanity-check the result with edge cases like p=0, p=1, or d=0. This shows rigor and prevents miscommunication.
Confirm that you arrive at a uniformly random time within the interval [0, x] and that the bus's lateness is independent of your arrival time. Define the bus's arrival time as either 0 (on time) or d (late).
If the bus is on time (prob 1-p), your wait is uniform on [0, x], so expected wait = x/2. If the bus is late by d (prob p), your wait is uniform on [d, x+d], so expected wait = x/2 + d.
Combine the conditional expectations: E[wait] = (1-p)*(x/2) + p*(x/2 + d) = x/2 + p*d.
Verify that when p=0, E[wait] = x/2; when p=1, E[wait] = x/2 + d; and when d=0, E[wait] = x/2. Discuss how the expected wait increases linearly with both p and d.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by defining the gap-based formula for expected wait time, then intuitively explain how irregular headways increase the probability of longer waits. Use a simple example with two buses to illustrate the convexity effect, emphasizing that the average wait is driven by the second moment of headways.
Pro tip: Connect the concept to the inspection paradox: riders are more likely to arrive during longer gaps, which disproportionately increases their expected wait. This shows deep understanding and practical relevance.
State that for a random rider, expected wait time is E[H^2] / (2 E[H]), where H is the headway. This formula assumes random arrival and independent headways.
Show that for a fixed mean E[H], increasing variance (irregularity) increases E[H^2], which in turn increases the expected wait. This is because the function is convex.
Use two buses with headways alternating between 0 and 2m (mean 1m) versus constant 1m. Compute expected waits: irregular gives 1m, regular gives 0.5m, demonstrating the increase.
Explain that riders are more likely to arrive during longer gaps, so their wait is biased toward longer headways, further increasing the average wait.
Summarize that even with the same mean headway, irregularity always increases expected wait, highlighting the importance of consistency in service frequency.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Memoryless property means the expected wait is just x regardless of when you show up.
First, clarify the scheduled setup: if buses arrive exactly every x minutes, the average wait is x/2. Then, for a Poisson process with rate 1/x, use the memoryless property to show the wait time is exponential with mean x. Conclude that the expected wait doubles from x/2 to x.
Pro tip: Emphasize that this is the inspection paradox: random arrivals make you more likely to land in a longer gap, increasing average wait. This shows you understand the practical implications beyond the math.
Assume buses arrive exactly every x minutes. A passenger arriving at a random time waits uniformly between 0 and x, so expected wait is x/2.
Buses arrive as a Poisson process with rate λ = 1/x, so interarrival times are exponential with mean x.
Due to the memoryless property, the waiting time from a random arrival until the next bus is also exponential with mean x.
The expected wait increases from x/2 to x, i.e., it doubles.
This is the inspection paradox: random arrivals are more likely to occur during longer gaps, raising the average wait.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.