# The probability that a 2d continuous time random walk avoids the origin

I am trying to find a reference of a proof of a continuous time version of a result of Dvoretzky and Erdos from their paper "Some problems on random walk in space" that says the probability $\gamma_2(n)$ that a two dimensional random walk does not return to the origin after first leaving it up until time n is asymptotically $$\gamma_2(n)=\frac{\pi}{\log n}+O\left(\frac{\log\log n}{\log^2 n}\right).$$

Explicitly, I am looking to prove a result of the following kind. Let $\tilde{S}_t$ be a continuous time random walk on the two dimensional integer lattice then $$\mathbb{P}[\tilde{S}_s\neq 0 \text{ for }0<s\leq t]\approx\frac{\pi}{\log t}\text{ as }t\to\infty.$$ I feel the proof should be an easy adaption with integrals in place of sums etc. but it is alluding.

Thanks in advance for any help.

• What exactly is a statement? Commented Mar 7, 2015 at 17:35
• Does not this probability equal to 0?! Amazing. Commented Mar 7, 2015 at 17:55
• A brownian motion has probability $0$ to return to exactly $0$. It will return arbitrarily close to $0$ arbitrarily often, but you have to add an allowable distance to $0$ in the mix to phrase your problem correctly. Commented Mar 7, 2015 at 22:35
• It's a continuous time random walk on the 2 dimensional lattice. Random walks on the 2 dimensional integer lattice are point recurrent. I'm not discussing Brownian motions. Commented Mar 8, 2015 at 12:45
• @YemonChoi The usual rate one exponential waiting time. That is, this is the Poissonization of the two dimensional discrete time simple symmetric random walk. Commented Mar 8, 2015 at 18:14

Let $X_n$ be a discrete-time random walk on the integer lattice, and let $N(t)$ be a rate-1 Poisson process independent of $X_n$. Then your continuous-time random walk $S_t$ has the same distribution (as a process) as $X_{N(t)}$.

Let $T = \min\{n \ge 1 : X_n = 0\}$ be the return time for $X_n$, so that $P(T > n) = \gamma_2(n)$. So the probability $p(t)$ that $S(t)$ does not return to $0$ by time $t$ is $p(t) = P(T > N(t))$. But $T$ is independent of $N(t)$ so by conditioning on $N(t)$ we have $p(t) = E[\gamma_2(N(t))]$.

Now recall the law of large numbers says that $N(t)/t \to 1$ a.s.

So let us write \begin{align*} p(t) \log(t) &= E\left[\gamma_2(N(t))\log(t)\right] \\ &= E\left[\gamma_2(N(t)) \log(N(t)) \cdot \frac{\log(t)}{\log(N(t))}\right] \\ &= E\left[\gamma_2(N(t)) \log(N(t)) \cdot \frac{\log(t)}{\log(t) + \log(N(t)/t)}\right] \end{align*} Now the first factor inside the expectation converges to $\pi$ almost surely and boundedly, and the second factor converges to 1 almost surely.

There is a little more work to do to pass the limit inside the expectation since we need a little control over the second factor. I've run out of time to think about this right now, but hopefully you'll be able to fill it in.

• The above answer is off by a factor of 2. More precisely, $\gamma_2(n)$ is off by a factor of 2. It is not difficult to see that $\gamma_2(n)\sim 2\pi/\log(n)$, where $\log$ is the natural logarithm.
– user83290
Commented Nov 26, 2015 at 9:25