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.
 A: 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.
