= Solution
The random-walk form of the <Skorokhod embedding theorem> says the following. If $S_n=Y_1+\cdots+Y_n$, where the $Y_i$ are independent and identically distributed with
$$
\mathbb EY_i=0,
\qquad
\mathbb EY_i^2=\sigma^2<\infty,
$$
then on a space carrying a Brownian motion $B$ there are stopping times
$$
0=T_0\leq T_1\leq\cdots
$$
such that $(B_{T_n})_{n\geq0}$ has the same law as $(S_n)_{n\geq0}$, and the increments $T_n-T_{n-1}$ are independent and identically distributed with mean $\sigma^2$.
To prove the one-step statement, first note that every centered distribution is a mixture of centered two-point distributions. Indeed, match the equal-mass size-biased measures $x\,\mathbb P(Y\in dx)$ on $(0,\infty)$ and $|x|\,\mathbb P(Y\in dx)$ on $(-\infty,0)$. This produces a random pair $(L,R)$ of positive numbers such that, conditionally on $(L,R)$, $Y$ has values $-L,R$ with probabilities
$$
\frac R{L+R},\qquad\frac L{L+R},
$$
and $\mathbb E[LR]=\mathbb EY^2=\sigma^2$. Include the atom at zero by taking the stopping time zero.
Choose $(L,R)$ independently of $B_t$ and stop Brownian motion on first leaving $(-L,R)$. The <Brownian exit from an interval> formulas give the displayed two-point probabilities and conditional mean stopping time $LR$. Thus $B_T$ has the law of $Y$ and $\mathbb ET=\sigma^2$.
Starting from $T_0=0$, repeat this construction after each $T_{n-1}$. The <Strong Markov property> makes the new Brownian increments independent copies of the first embedding, proving the <Skorokhod embedding of a centered random walk>.
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