TITLE: Hitting time for two out of three random walk particles
QUESTION [10 upvotes]: I'm imagining a simple random walk on $\mathbb{Z}$ with three independent particles (maybe add laziness so they don't jump over each other). Suppose the particles are initially placed at, say, $-10$, $0$ and $10$.  The distance between any two particles evolves like a random walk, so the expected hitting time for those two specific particles is infinite.  Now, if we require only that any two of the three particles hit, is the expected hitting time still infinite?

REPLY [3 votes]: Expanding the comment above to a complete proof of Noam Elkies' formula:
Let $\tau$ be the first meeting time for independent simple random walks $X_t, Y_t$ and  $Z_t$ started at the even points $x<y<z$. Write $f(x,y,z)= (z-y)(y-x)$.  Claim: $ E(\tau)=f(x,y,z)$.
Proof: let $F_n=f(X_{\tau \wedge n},Y_{\tau \wedge n},Z_{\tau \wedge n})$ and $M_n:=F_n+{\tau \wedge n}$. It is easy to verify that $M_n$ is a Martingale so 
$f(x,y,z)=M_0=E(F_n+\tau\wedge n)$ for any $n$. 
Therefore   $E(\tau) \le f(x,y,z)$.  It remains to verify that $\lim_n E(F_n)=0$. Clearly $F_n \to 0$ a.s. so it suffices to verify that $\max_n F_n$ is integrable.
For this we will use the linear martingale $L_n=(Z_n-X_n)/2$ and the quadratic martingale $Q_n=F_n+ 2 L_n^2$. The first is a lazy SRW on the integers. The AMGM inequality gives $4f(x,y,z)\le (z-x)^2$,
so $E (\max_{n \le t} F_n) \le E(\max_{n \le t} L_n^2) \le 4 E(L_t^2)$ by Doob’s $L^2$ maximal inequality.  But $4E(L_t^2) \le 2E(Q_t)=2Q_0$ for every $t$. We conclude that $ E (\max_n F_n) \le 2Q_0$.