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The mean increment is $\mathbb{E}{X}_1=-6/7+2/7=-4/7$, so $M_n=S_n+(4/7)n$ is a martingale. Applying the bounded optional stopping theorem at $n\wedge T$ givesThe terminal sums are bounded by the exit-state bounds from (b), while $n\wedge T$ increases to the integrable stopping time $T$. The dominated convergence theorem on the left and the monotone convergence theorem on the right therefore proveThis is also Wald's equation here. In fact, the stopped identity already implies $\mathbb{E}(n\wedge T)\le 7m/4$ from $S_{n\wedge T}\ge -m$, giving another direct finite-expectation justification.

Let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ be the natural filtration. An integer-valued random variable $M$ is a stopping time with respect to $(X_i)$ whenThus whether one has stopped by time $n$ can be decided from the first $n$ observations, without seeing future variables. Equivalently, $\{M\ge i\}\in {\mathcal{F}}_{i-1}$.

Write the stopped sum asThe event $\{M\ge i\}$ depends only on $X_1,\dots ,X_{i-1}$, so its indicator and $X_i$ are independent random variables. Since $\mathbb{E}|X_1|<\infty$ and $\mathbb{E}M<\infty$, Tonelli theorem givesWe may consequently exchange expectation and summation. Using the tail-sum formula for $M$ proves Wald's equation:

For the renewal process, put $S_n=X_1+\cdots +X_n$ and $N_t=\max \{n:S_n\le t\}$. The random variable $N_t+1$ is a stopping time. Since $S_{N_t+1}>t$, Wald's equation givesand hence

For $c>0$, set $Y_i=X_i\wedge c$ and ${\mu }_c=\mathbb{E}{Y}_i$. Wald's equation still applies to the stopping time $N_t+1$ and the independent variables $Y_i$. Because $S_{N_t}\le t$,ThereforeBy the monotone convergence theorem, ${\mu }_c\uparrow \mu$ as $c\to \infty$. Combining the bounds proves the elementary renewal theorem

For a fixed word over $q$ equally likely symbols, the waiting time for a word in independent uniform symbols has mean $\sum q^k$, where $k$ runs through the lengths of the borders of a word, including the full word. The word lava' has no proper nonempty border, so $$ \boxed{\mathbb E\tau_{\mathtt{lava}}=100^4=100\,000\,000}. $$ The word aa' has borders of lengths $1$ and $2$, soFor a direct check, let $E_0$ be the expected remaining time with no trailing a' and $E_1$ with one trailing a'. Thenwhich gives $E_0=10100$.