= Expected occupation before interval exit for a simple symmetric random walk
{title2=$G_{a,b}(i,j)=\frac{2((i\wedge j)-a)(b-(i\vee j))}{b-a}$}
Let $a<i<b$ and $a<j<b$ be integers. A <simple symmetric random walk> started at $i$ exits the integer interval at $\tau=\inf\{n\geq0:S_n\in\{a,b\}\}$. Its expected occupation of $j$ before exit, including its initial state, is
$$
G_{a,b}(i,j)
=\mathbb E_i\sum_{k=0}^{\tau-1}\mathbf1_{\{S_k=j\}}
=\frac{2((i\wedge j)-a)(b-(i\vee j))}{b-a}.
$$
A block of $b-a$ consecutive positive steps forces exit, so the <geometric tail bound from a uniform escape probability> proves $\tau<\infty$ <almost surely>. <Optional stopping> at the <bounded stopping time> $n\wedge\tau$, followed by the <bounded convergence theorem> for $S_{n\wedge\tau}\in[a,b]$, gives the <gambler's ruin> probabilities
$$
\mathbb P_i(S_\tau=a)=\frac{b-i}{b-a},\qquad
\mathbb P_i(S_\tau=b)=\frac{i-a}{b-a}.
$$
Translate the <absolute value martingale of a simple symmetric random walk> from zero to $j$. Bounded-time <optional stopping> yields
$$
\mathbb E_i\sum_{k<n\wedge\tau}\mathbf1_{\{S_k=j\}}
=\mathbb E_i|S_{n\wedge\tau}-j|-|i-j|.
$$
Use <monotone convergence> on the increasing visit count and the <bounded convergence theorem> on the right. Substituting the exit <probabilities> gives
$$
G_{a,b}(i,j)
=\frac{(b-i)(j-a)+(i-a)(b-j)}{b-a}-|i-j|,
$$
which simplifies to the displayed expression by considering $i\leq j$ and $i\geq j$. In particular, $G_{a,b}(0,0)=-2ab/(b-a)$ when $a<0<b$.
The proof establishes finite <expected value> of the visit count. Therefore each translated stopped absolute-value <martingale> is dominated in absolute value by a constant plus that total count, an <integrable random variable>; its family of values is <uniformly integrable>. This supplies the limit justification without assuming <uniform integrability> in advance.
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