Find $P(\eta_t=m)$, $m=0,1,2,\dots,$

Question

Let $\epsilon_t$, $t=1,2,\dots$ independent random variables with $P(\epsilon_t=1)=p$ and $P(\epsilon_t=-1)=1-p$. If $\eta_0=0,\eta_t=\eta_{t-1}+\epsilon_t$ , $t=1,2,\dots$ where $\eta_t$ is markov chain.Find $P(\eta_t=m)$, $m=0,1,2,\dots,$

I know that $\epsilon_t=\eta_t-\eta_{t-1}$ then $$P(\epsilon_t=1)=P(\eta_t-\eta_{t-1}=1)=p$$ $$P(\epsilon_t=-1)=P(\eta_t-\eta_{t-1}=-1)=1-p$$

this seems to be related to the binomial, where we step forward with probability $p$ and step back with probability $1-p$

can anyone help?

Answer 1

Yes. A Drunkards Walk follows a similar principle to Binomial Distribution; but rather than successes (and failures) in a certain number of trials, you are counting steps forward and backwards.

$\mathsf P(\eta_t=m)$ is the probability that in $t$ steps of $\pm 1$ units each, you will have moved $m$ units in total; consisting of $x$ steps forward, and $y$ steps backwards in some order.   What is the probability of this happening?

Hint 1: Use simultaneous equations to solve $x,y$ in terms of $m,t$.

Hint 2: How many ways can the $x$ and $y$ events be arranged?

Hint 3: Or to simplify, the number of moves forward $x$ is binomially distributed.