Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 16

The arrival phase has a uniform distribution over a cycle of length $R$. The no-delay phases occupy length $R-r$, soIndividual phase endpoints have probability zero. The model neglects clearance time and releases waiting vehicles together when the stop interval ends; the degenerate case $r=R$ is interpreted as release at successive cycle boundaries.

For a stop-phase arrival at time $t$, the delay is $s=r-t$. An interval of delays of width $ds$ therefore corresponds to an interval of arrival phases of the same width, because $|dt/ds|=1$. Its unconditional probability is $ds/R$. Hence the positive-delay part has densityThis ordinary density integrates to $r/R$, not to one: the remaining probability mass is at $s=0$. Thus $S$ has a uniform distribution with an atom at zero, with mixing weight $q=r/R$. Conditional on a positive delay, the density would instead be $1/r$ on $(0,r)$.

The graph of the positive-delay probability density function is a horizontal segment at height $1/R$ from $s=0$ to $r$, followed by zero from $r$ to $R$. Values at individual endpoints do not affect probabilities. The separate atom $1-r/R$ at zero is not a finite-height part of this density graph.

The companion cumulative distribution function in the figure is $F_S(s)=0$ for $s<0$, $F_S(s)=1-r/R+s/R$ for $0\le s<r$, and $F_S(s)=1$ for $s\ge r$. Its jump at zero displays the discrete part without confusing density height with probability mass.

The probability of a positive delay is the integral of its density:Adding the separate no-delay event givesThis is the normalization of a probability distribution with both discrete and continuous parts.

The atom at zero contributes zero to the expected value, while all positive delays must retain their unconditional density. ThereforeThis mean is $r/R$ times the mean $r/2$ conditional on encountering a stop; using $r/2$ alone would omit the vehicles passing without delay.

The second moment isSubtracting the squared expected value gives the variance and standard deviation:The variance is nonnegative because $0<r\le R$. Both $\mu$ and $\sigma$ have units of time. At $r=R$ the result reduces to $R/\sqrt{12}$, the standard deviation of a uniform distribution over one full cycle.

Condition throughout on the given arrival time $t$ of A, with $0\le t<r$. A departs at time $r$. B also departs at $r$ exactly when it arrives before that release, so $t+{\tau }_0\le r$, or $0<{\tau }_0\le r-t$. Since ${\tau }_0$ is uniform on an interval of length $R$,The original timeline below shows the three disjoint ranges of ${\tau }_0$. The same idealized simultaneous-release convention as in part (a) is used.

If B arrives during the remaining pass phase, then $r<t+{\tau }_0<R$, or $r-t<{\tau }_0<R-t$. It passes immediately, soBecause $r-t>0$, this is strictly between zero and ${\tau }_0$. The phase interval has length $(R-t)-(r-t)=R-r$, giving

The remaining arrivals occur in the next stop phase: $R\le t+{\tau }_0\le R+t$, or $R-t\le {\tau }_0\le R$. B waits until $R+r$, while A departed at $r$, giving ${\tau }_1=R\ge {\tau }_0$. This interval has length $t$, henceEndpoint assignments do not affect probabilities. The three outcomes are exhaustive and satisfyThese are conditional probabilities for the stated fixed $t$; there is no additional averaging over A's arrival phase in the question.