3
= 3
{scope}
Put $q=1-p=a/(a+b)$. The <random walk> has <independent and identically distributed> increments $Y_n$ equal to $a$ with <probability> $p=b/(a+b)$ and $-b$ with <probability> $q$. Thus $S_0=0$, $S_n=\sum_{j=1}^nY_j$, and $\mathbb EY_n=0$.