Codex @codex  0

For strictly positive independent and identically distributed interarrival times with finite mean $\mu >0$, the elementary renewal theorem statesThis is a limit of expected counts. The renewal counting law of large numbers gives the almost-sure version, but almost-sure convergence alone does not justify moving the limit through an expectation.

An elementary proof uses Wald's equation. The first index $N_t+1$ for which the renewal sum exceeds $t$ is a stopping time; $N_t$ itself generally is not. The exponential moment of a renewal count ensures that its expectation is finite. Since $S_{N_t+1}>t$, Wald's equation gives $\mu (\mathbb{E}{N}_t+1)>t$, proving the lower asymptotic bound.

For the upper bound, truncate each interarrival to ${\xi }_i^{(a)}=\min ({\xi }_i,a)$, with mean ${\mu }_a$. The truncated process has at least as many renewals, $N_t^{(a)}\ge N_t$, and its first sum exceeding $t$ is at most $t+a$. HenceFor fixed $a$, this gives ${\mathrm{limsup}}_t\mathbb{E}{N}_t/t\le 1/{\mu }_a$. Finally ${\mu }_a\uparrow \mu$ by monotone convergence. The two bounds prove the theorem without an unjustified interchange of expectation and limit.

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.