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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?

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  • $\begingroup$ By the way, in order to use this method to compute $p_0$, you have to start with the assumption that $p_k < 1$ for some $k$ (or prove it). Otherwise, there's nothing in the calculation that allows you to differentiate $p_0 = 1$ from $p_0 = 2-2p$. $\endgroup$ Commented May 3, 2023 at 22:01

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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.

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