Assume that $P(X_i = 1) =1/2, P(X_i =-1)= 1/4,\text{ and }P(X_i = 0)=1/4$.
Consider the random walk starting at 1 given by
$$S_n = 1 + X_1 + X_2 + \cdots + X_n$$ where $X_1,X_2, ...$ are i.i.d.
What is the probability that the random walk ever reaches $0$?
I have tried to solve this using Binomial Theorem, but I do not have an equal probability here!
I would appreciate it so much if you could help me solve this problem.
On
Let P1 be the probability it ever reaches 0, starting at 1, and P2 the probability it ever reaches 0, starting at 2.
To reach 0 from 2, it has to reach 1 from 2, then 0 from 1.
Reaching 1 from 2 is the same as reaching 0 from 1.
So the probability $P2=P1^2$
Now, start at 1, and make one move. The chances are 1/3 you are at 0 (success!) and 2/3 you are at 2. So $$P1=\frac13(1)+\frac23(P2)\\1-3P1+2P1^2=0\\(1-2P1)(1-P1)=0$$
So P1=1 or 1/2. :(
On
The probability of ever reaching zero does not depend on the probability to not move at all. Thus you can instead consider the walk with a $2/3$ chance to go right and a $1/3$ chance to go left. There is a nice general solution to this problem. Let $n \in \mathbb{N}$, then the probability to hit $0$ before hitting $n$ starting at $0<x<n$ satisfies:
$$u_n(x)=\frac{2}{3} u_n(x+1)+\frac{1}{3} u_n(x-1).$$
Adjoin this to the boundary conditions $u_n(0)=1,u_n(n)=0$. Then the recurrence relation can be solved explicitly. (Hint: the general solution is of the form $c_1 r_1^x + c_2 r_2^x$, where $r_1,r_2$ are the roots of $\frac{2}{3} r^2-r+\frac{1}{3}$.) Then the desired quantity is $u(1)=\lim_{n \to \infty} u_n(1)$.
Suppose it reaches $0$ in n steps. We can calculate the probability using the multinomial distribution. Out of n steps some k will be +1, k+1 will be -1, and n-2k-1 will be $0$.
Hence, $P(S_n=0) = \sum_{k=0}^{(n-1)/2} (1/2)^k (1/4)^{n-k}\frac{n!}{k! (k+1)! (n-2k-1)!}$