Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 27K (Applied Probability)

This is Wald's identity. Expand the stopped sum and use $\{M\ge i\}$, which depends only on earlier interarrivals and is independent of ${\xi }_i$: $E{\sum }_{i=1}^M{\xi }_i={\sum }_{i}E[1_{M\ge i}]E{\xi }_i={\lambda }^{-1}EM$.

Apply part (a) to the stopping time $N_t+1$ to obtain $E{T}_{N_t+1}={\lambda }^{-1}E(N_t+1)$.

The strong law gives $T_n/n\to 1/\lambda$. Since $T_{N_t}\le t<T_{N_t+1}$ and $N_t\to \infty$, sandwiching yields $N_t/t\to \lambda$ almost surely.

If ${\xi }_1\le C$, then $t<T_{N_t+1}\le t+C$. Taking expectations and using (b) yields $\lambda t-1<E{N}_t\le \lambda (t+C)-1$. Divide by $t$.