Consider a particle that moves along the set of integers in the following manner. If it is presently at i, then it next moves to i+1 with probability p and to i-1 with probability 1-p. Starting at 0, let a denote the probability that it ever reaches 1.
show that $$ a=p+(1-p)a^2$$ I think problem is just modification of random walk problems
I found the ''a '' as follows but ı can not convert it to above equation. Can someone help to find my mistakes or above equation ? $$ a=f_{01}=\sum_{m=1}^\infty f_{01} (m)=p+(1-p)p^2+2*(1-p)^2p^3+3(1-p)^3p^4..... $$
This is a non-symetric, 1-dimensional random walk on $\mathbb{Z}$.
Firts notice that $a$ is the probability of ever reaching $k+1$ whenever starting from $k$.
Starting from $0$ you can reach $1$ in two distinct ways. First, you can go straight to $1$, this has probability $p$. Or, you can first go to $-1$ (with probability $1-p$), then eventually reach $0$ (with probability $a$, see above with $k=-1$) and then, starting in $0$, eventually reaching $1$ (again with probability $a$). As the steps are independent you may multiply the probabilities and write $$ a=p+(1-p)\,a^2. $$