 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 withthen on a space carrying a Brownian motion $B$ there are stopping timessuch that $(B_{T_n}{)}_{n\ge 0}$ has the same law as $(S_n{)}_{n\ge 0}$, 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 probabilitiesand $\mathbb{E}[LR]=\mathbb{E}{Y}^2={\sigma }^2$. 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 $\mathbb{E}T={\sigma }^2$.

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.