Solution

ID: past-exam-of-the-msc-in-advanced-computer-science-course-of-the-university-of-oxford/2021/bayesian-statistical-probabilistic-programming/3/a/solution

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Past exam of the MSc in advanced computer science course of the University of Oxford / 2021 / Bayesian statistical probabilistic programming / 3 / a / Solution by Codex 0 2026-10-25
Treat each observed arrival label as a constraint on an unobserved walking duration. Departures are treated as fixed design information. Writing for the minutes available until work starts, the derived late-label thresholds have frequencies ; the early-label thresholds have frequencies . These are censoring thresholds, not measured journey times. A late label means , and an early label means ; equality has probability zero under the continuous model.
The paper specifies no prior distribution. One explicit proper model is
(1)
with independent priors; all times are in minutes. These ranges are a transparent modelling choice, not information deduced from the observations. In particular, this is a prior on the standard deviation , not a uniform prior on the variance . The normal distribution permits negative durations; it is the mathematical approximation requested here, rather than a claim that such durations are physically possible. A truncated model would be a different model.
The following Python implements hard conditioning by rejection sampling. None marks a rejected trace; an accepted Boolean answers whether all five future arrivals are late. The same parameters are shared across past and future days.
import math
import random

LATE = {20: 1, 21: 3, 25: 3, 27: 1, 32: 2}
EARLY = {29: 1, 34: 1, 36: 3, 37: 1, 38: 1, 39: 1, 43: 1, 44: 1}
RECORDS = [(c, True) for c, n in LATE.items() for _ in range(n)]
RECORDS += [(c, False) for c, n in EARLY.items() for _ in range(n)]

def journey_trace(rng):
    mu = rng.uniform(0.0, 60.0)
    sigma = rng.uniform(1.0, 20.0)
    # Twenty latent past durations and five future durations.
    past = [rng.gauss(mu, sigma) for _ in RECORDS]
    future = [rng.gauss(mu, sigma) for _ in range(5)]
    if not all(t > c if late else t < c
               for t, (c, late) in zip(past, RECORDS)):
        return None
    return all(t > 30.0 for t in future)

def accepted_average(trace, n, rng):
    if n <= 0:
        raise ValueError("n must be positive")
    successes = accepted = 0
    while accepted < n:
        result = trace(rng)
        if result is not None:
            accepted += 1
            successes += int(result)
    return successes / n
For example, accepted_average(journey_trace, n, random.Random(2021)) estimates the desired conditional probability from accepted traces. Accepting False is essential: treating a false answer as a rejected trace would condition on the desired conclusion. The model deliberately emphasizes clear semantics; rejection can be inefficient because all twenty past-label constraints must hold together.
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