Consider a biased random walk on $\mathbb{Z}$ with $P(x,x+1)=p>\frac{1}{2}$. Let $T_0$ be the first hitting time of $0$.
I'm trying to find the probability $p_0=P_0(T_0<\infty)$. In words, it is the probability of the random walk returning to $0$.
My attempt: I'm using the hint given in the first comment here. Let $p_x:= P_x(T_0<\infty)$. Then using the Markov property after one step, we get:
$$
\begin{align*}
p_0 &= p\cdot p_{1} + (1-p)\cdot p_{-1} \\\\
&= p\cdot p_{1} + (1-p) \\\\
&= p\cdot (p\cdot p_{2}+ (1-p)) + (1-p) \\\\
&= p^2\cdot p_{2} + p(1-p) + (1-p) \\\\
&= \sum_{i=0}^{\infty} p^i(1-p) \\\\
&= (1-p)\cdot \frac{1}{1-p}
\end{align*}
$$
The above is clearly not right and the answer is supposed to be $2(1-p)$ and it would be great to know what I did wrong above?
Best Answer
This step: $$p^2\cdot p_{2} + p(1-p) + (1-p) = \sum_{i=0}^{\infty} p^i(1-p)$$ is false.
In particular, if you continued your method, you'd have $$p^2 \cdot p_2 + p(1-p) + (1-p) = p^3\cdot p_3 + p^2(1-p)\cdot p_1 + p(1-p) + (1-p)$$ That is, you have seemingly just assumed $p_1=1$ when you wrote the infinite series. Certainly, if $p_{-1}=p_1=1$, then $p_0=1$, but that's really all you've shown.