Consider an 1D infinite lattice. The lattice is fully occupied except from a vacancy in the origin which undergoes simple diffusion (in countinuous time).
At position $n>0$ in the lattice there is a "tracer particle".
Estimate the typical number of hops that the tracer particle does in the interval $(0,t)$.
I.e. the tracer particles does an hop whenever the vacancy goes undergoes the transition $(n-1)\to(n)$ or $(n)\to(n-1)$.
This is part of exercise 4.1 of the Krapivsky book "a kinetic view of statistical physics" but this book does not explain method to solve this kind of problems. I tried to google the problem but I have not found anything which explains directly how to solve this.
My idea to solve this problem is to use the properties of the random walk in discrete time and then generalize the result for a continuous time...but that seems me not the most immediate way. Can anyone give me an hint or suggest a good reference?
There are two way to think about this problem. First, let $N_t$ be the number of jumps by the time $t$ (clearly you only need to consider $N_t\ge n$) and let $\bar n=n-1/2$. Then you are interested in $$EV(n)=E(\sum_{i=1}^{N_t}(\frac 1 2 P(S_{i-1}=n-1)+\frac 1 2P(S_{i-1}=n)))=E(\sum_{i=n}^{N_t}\frac 1 2\binom {i-1}{\lfloor\frac {i-n}2\rfloor})$$ Now you can try different approximations. Assuming $n$ is not tiny and $N_t$ is fairly larger than $n$, you can approximate $S_i$ by a normal to get $$EV(n)\approx E(\sum_{i=n-1}^{N_t-1}\frac 1{\sqrt{2\pi i}} e^{-\frac{\bar n^2}{2i}})\approx E\int_{n-1}^{N_t-1}\frac 1 {\sqrt{2\pi i}}e^{-\frac{\bar n^2}{2i}}\,di=E((\bar n\cdot erf(\frac {\bar n}{\sqrt{2i}})+\sqrt{\frac 2 \pi}\sqrt ie^{-\frac {\bar n ^2}{2i}})|_{n-1}^{N_t-1})$$ There might be better approximations/exact forms based on combinatorial identities. You can also approximate $N_t$ which is usually assumed to be a Poisson process with rate $1$. The simplest and most tractable one is $N_t=t$. A less simplistic assumption would be normal approximation for largish $t$.
A alternative approach would consider occupation probability $P_t(n)$ to get:
$$EV(n)=\int_0^t\frac 1 2 (P_t(n-1)+P_t(n))dt$$ For $t<n$, $P_n(t)$ is exponentially small and for large $t$, $P_n(t)$ is approximately normal for we can approximate
$$EV(n)\approx\int_n^t\frac 1 {\sqrt{2\pi u}}e^{\frac {-(n-1/2)^2}{2u}}\,du$$ as before. If you want to approximate your random walk by a Browninan motion, you need to find $$E(t,n)=E(U_t([n-1,n])+D_t([n,n-1]))$$ where $U_t(I), D_t(I)$ are the number of upcrossings and downcrossings by time $t$ of interval I.