Seems trivial until you actually have to justify every step.
Recognize that the arrival time is uniformly distributed over the 10-minute interval between buses, so the waiting time is also uniform on [0,10]. Compute the expected value of a uniform random variable on that interval, which is the midpoint, 5 minutes. State the assumptions clearly and note that the result is independent of the schedule's starting point.
Pro tip: Explicitly state that you assume a fixed schedule with no randomness in bus arrivals, and that your arrival time is uniformly distributed. This shows you understand the model and avoids ambiguity.
Let W be the waiting time until the next bus. Since you arrive at a random time with no knowledge of the schedule, W is the time from your arrival to the next bus.
Because buses arrive every 10 minutes, the time until the next bus is uniformly distributed between 0 and 10 minutes. This assumes your arrival time is uniformly random over the interval.
For a uniform distribution on [0,10], the expected value is the midpoint: (0+10)/2 = 5 minutes.
The expected waiting time is 5 minutes. Emphasize that this is a theoretical average and actual waiting times will vary.
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Start by explicitly defining a precise model for bus lateness, such as each bus independently being late by an exponential amount with probability 0.5, and justify why this is reasonable. Compute the expected waiting time under this model, then discuss how the answer changes if you assume a different distribution, like a uniform delay, highlighting the impact of distributional assumptions.
Pro tip: Always state your assumptions clearly and connect them to real-world scenarios; interviewers value candidates who can reason under ambiguity and adapt their models when new information is presented.
Choose a specific probability distribution for the delay when a bus is late (e.g., exponential with rate λ) and specify that each bus is independently late with probability 0.5. Justify the choice by relating it to typical bus arrival patterns or simplicity.
Derive the expected waiting time for a passenger arriving at a random time, considering that the bus may be on time (zero delay) or late (positive delay). Use the law of total expectation.
Introduce a different reasonable model, such as a uniform delay between 0 and 10 minutes when late, and compute or qualitatively describe how the expected wait changes. Highlight that the expected wait depends on the distribution's mean and variance.
Compare the models, noting how assumptions affect the result and which model might be more realistic. Mention that in practice, data would be used to estimate parameters.
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Start by clarifying the two scenarios: fixed 10-minute headway means a uniform wait between 0 and 10 minutes, averaging 5 minutes. For a Poisson process with mean headway 10 minutes, the arrival rate is 0.1 per minute, and by memorylessness the wait time is exponential with mean 1/λ = 10 minutes. Explain that memorylessness makes the expected wait equal to the mean interarrival time, not half of it, because the residual time distribution is the same as the full interarrival distribution.
Pro tip: Emphasize that the Poisson process is a model for random arrivals, and in practice bus arrivals are not perfectly Poisson due to scheduling; but for the interview, show you understand the mathematical distinction and can connect it to real-world implications like passenger wait times.
State that with a fixed 10-minute headway, wait times are uniformly distributed between 0 and 10 minutes, so the expected wait is 5 minutes.
Specify that a Poisson process with mean headway 10 minutes has rate λ = 1/10 per minute, and interarrival times are exponential with mean 10 minutes.
Explain that due to the memoryless property of the exponential distribution, the expected residual wait time given any arrival time is still 10 minutes, not 5.
Contrast the two results: fixed schedule gives 5 minutes, Poisson gives 10 minutes, highlighting that variability increases average wait.
Mention that this is a classic result (e.g., inspection paradox) and that real-world bus systems often have less variability than Poisson, so waits may be between 5 and 10 minutes.
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First, clarify the fixed-delay model: a bus that is late is delayed by exactly δ minutes, while on-time buses are not delayed. Then derive the expected wait as a function of δ, showing that it increases linearly with δ. Finally, evaluate at δ = 10 to give a concrete answer, noting any assumptions about bus arrival distribution.
Pro tip: Mention that the result depends on the probability of a bus being late; if that probability is p, the expected wait increases by p·δ. This shows you understand the underlying stochastic model and can generalize.
State the assumptions: buses arrive according to a known schedule, but with a fixed delay δ applied only to late buses. Clarify the probability p that a bus is late.
Express the expected wait as a function of δ, typically as a linear function: E[wait] = base_wait + p·δ, where base_wait is the expected wait without delays.
Discuss how the expected wait changes with δ: it increases linearly, with slope equal to the probability of a late bus. If p=1, the slope is 1; if p<1, the slope is less than 1.
Plug in δ=10 to compute the expected wait. If p is known, give a numerical answer; otherwise, express it in terms of p.
Mention that if δ is very large, the expected wait grows without bound. Also note that if the bus is never late (p=0), δ has no effect.
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They saved this for the end and I was already a bit fried.
Use the inspection paradox and renewal theory to show that expected wait depends on the second moment of the interarrival distribution. Since the mean is fixed at 10, minimizing the second moment (and thus variance) minimizes the expected wait. The deterministic (perfectly regular) schedule has zero variance, so it achieves the minimum.
Pro tip: Connect the result to real-world systems: this explains why scheduled buses with low variance reduce average wait, and why high-variance arrivals (e.g., Poisson) increase it. Mentioning the inspection paradox shows depth.
State that we consider a stationary arrival process with mean interarrival time E[X] = 10 minutes. A random passenger arrives at an arbitrary time and waits until the next arrival; we want to minimize E[W].
Explain that a random passenger is more likely to land in a longer interarrival interval. The expected length of the interval containing the passenger is E[X^2]/E[X], not E[X].
For a stationary renewal process, the expected waiting time is E[W] = E[X^2]/(2E[X]). Since E[X] = 10 is fixed, minimizing E[W] is equivalent to minimizing E[X^2].
By Jensen's inequality, E[X^2] ≥ (E[X])^2, with equality iff X is constant. Thus the minimum occurs when X = 10 deterministically, i.e., a perfectly regular schedule.
Therefore, among all stationary arrival processes with mean gap 10, the deterministic schedule minimizes the random passenger's expected wait. This is a classic result in renewal theory.
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