Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 26J (Applied Probability)

For independent identically distributed positive holding times $S_i$, define $T_0=0$, $T_n={\sum }_{i=1}^n{S}_i$ and the renewal count $X_t=\max \{n:T_n\le t\}$. With $0<\mu =E{S}_1<\infty$, the renewal strong law is $X_t/t\to 1/\mu$ almost surely. The elementary renewal theorem is $m(t)/t\to 1/\mu$, where $m(t)=E{X}_t$. More generally the rate is zero when the mean holding time is infinite, under the usual nonexplosion assumptions.

One complete cycle consists of ten stops of mean one minute and ten journeys of mean four minutes, so its mean duration is fifty minutes. Over the eight-hour shift, the elementary renewal approximation gives $480/50=9.6$ completed circuits on average.

For the end-of-shift calculation use the long-run occupation probabilities of the twenty successive phases. Each stop phase has probability $1/50$, and each travel phase probability $4/50$. Alighting stop $j$ is therefore selected with probability $1/50+4/50=1/10$, from either stopping there or travelling toward it. If that stop is the station, no further stops are passed. If it is stop $j\ge 2$, the intermediate stops are $j+1,\dots ,10$, numbering $10-j$. Thus the requested approximate mean isThe approximation treats eight hours as long compared with a cycle and ignores the initial-phase transient. Exponential holding times justify the twenty-phase Markov description; independent cycle times justify the renewal rate.