I have the following problem. Consider a 1D random walk with step size L. At each step one can move to the left, to the right, or stay in the same spot, all with equal probability.
I understand that of course the expected value of my displacement after N steps is still zero, just as in the usual unbiased random walk in 1D without pauses, since pauses don't break the symmetry in any way.
I would expect though that the root mean square displacement after N steps is different in this case, not just the usual $L\sqrt{N}$ one gets for the 1D unbiased random walk without pauses.
The problem is that I don't know how to calculate the RMS displacement for this walk with pauses, can anyone tell me how it is done and find it? Thank you. (If you could also derive the probability distribution of the displacement after N steps, for large N, for such a walk that would be a plus, and much appreciated.)
Best Answer
To find the expected squared displacement, condition it on the number of steps that you don't move, say $S_N$, which is Binomial($N,1/3$) distributed and use the total expectation formula. The conditional expected squared displacement, conditioned on $S_N=n$, is just like an ordinary random walk on $N-n$ steps, so its expected squared displacement is $L^2(N-n)$. Now
$$E[X_N^2]=\sum_{n=0}^N E[X_N^2 \mid S_N=n] P[S_N=s]=L^2(N-E[S_N])=L^2(2N/3).$$