I am stuck on this question,
"In a symmetric random walk of a particle on integer number line, starting from origin, what is the probability that it hits $-9$ before $+6$"
Here symmetric implies: P(going left at each step) = 1/2 and ditto for going right.
I could not think of a method for this, I just know that the expected number of steps to reach an integer $k$ can be defined recursively as:
Let $f(k)$ denote the expected number of steps starting from $k$,
$f(0)$ = $\frac{1}{2}$$(1+f(-1))$ + $\frac{1}{2}$$(1+f(1))$
In general,
$f(k)$ = $\frac{1}{2}$$(1+f(k-1))$ + $\frac{1}{2}$$(1+f(k+1))$
But how do I relate this or without relating this to the required probability.
Please help me with this. Thanks in advance!
PS: Please excuse me for the poor LaTeX syntax, I'm new to it.
Best Answer
You are asking yourself "how many steps?" when the question is "is position $-9$ reached before $+6$?".
Instead, consider $g(k)$ as this probability starting at $k$. You now have $$g(k) = \tfrac12 g(k-1)+\tfrac12 g(k+1)$$ and the boundary values $$g(-9)=1,g(+6)=0$$ which you should be able to solve
If you were really interested in the number of steps until you stop, the boundary values are $f(-9)=0,f(+6)=0$ and the solution is a little more complicated (you get a quadratic function of $k$, with $f(0)=54$ and it not a coincidence that $54=6\times 9$)