= Solution
Treat each observed arrival label as a constraint on an unobserved walking duration. Departures are treated as fixed design information. Writing $c$ for the minutes available until work starts, the derived late-label thresholds have frequencies $20:1,21:3,25:3,27:1,32:2$; the early-label thresholds have frequencies $29:1,34:1,36:3,37:1,38:1,39:1,43:1,44:1$. These are <censoring> thresholds, \b[not measured journey times]. A late label means $T>c$, and an early label means $T<c$; equality has <probability> zero under the continuous model.
The paper specifies no <prior distribution>. One explicit proper model is
$$
\mu\sim\operatorname{Unif}(0,60),\qquad \sigma\sim\operatorname{Unif}(1,20),\qquad T_i\mid\mu,\sigma\overset{\rm iid}{\sim}N(\mu,\sigma^2),
$$
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> $\sigma$, not a uniform prior on the <variance> $\sigma^2$. 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 $n$ 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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