Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 27L (Applied Probability)

PutThis is the renewal process generated by the interarrival times. Since the variables have a density and positive finite mean, $S_n\to \infty$ almost surely, so $N_t$ is finite and tends to infinity with $t$. The strong law givesOn this probability-one event,Dividing by $N_t$ and using $N_t\to \infty$ givesBoth outside expressions tend to $1/\lambda$, soThis is the renewal counting law of large numbers.

The size-biased distribution associated with ${\xi }_1$ is defined byfor every bounded measurable $h$. Thus, if ${\xi }_1$ has density $f$, then ${\hat{\xi }}_1$ has density $\lambda xf(x)$.

Now suppose that ${\xi }_1$ is exponential with rate $\lambda$. The renewal epochs form a Poisson process. Writeso that $L(t)=A_t+R_t$. Independent stationary increments and the exponential waiting-time law show that $R_t$ is exponential with rate $\lambda$, independent of the history up to $t$ and hence of $A_t$. For fixed $a\ge 0$ and $t>a$,Consequently $A_t$ converges in distribution to another rate-$\lambda$ exponential variable, independent of $R_t$. Therefore $L(t)$ converges to the sum of two independent rate-$\lambda$ exponentials, whose density isThe size-biased density of ${\xi }_1$ is likewiseHence the exponential renewal interval limit is

Let $F$ be the distribution function of ${\xi }_1$, write $\overline{F}=1-F$, and fix $x>0$. SetConditioning on the first interarrival givesIndeed, if ${\xi }_1>t$, the first interval contains $t$ and must also have length at least $x$; if ${\xi }_1\le t$, the process starts afresh at time ${\xi }_1$.

Put $h=\overline{F}(x)$ and $d(t)=q(t)-h$. ThenwhereIn particular $g\ge 0$. Iterating the renewal equation yieldsFor fixed $t$, the last term tends to zero in absolute value because $d$ is bounded andIt follows thatThus the stochastic length bias of a renewal interval givesfor every $x,t>0$.