Consider a simple symmetric random walk $(S_n)_{n \geq 0}$ with $S_0 = 0$ on $\mathbb{Z}.$ We add the killing probability: At each step, with probability $\alpha$, the random walk is killed and the process stops. With probability $1-\alpha$, the random walk continues. Let
$$\tau_1 = \min \{n : S_n = 1\}.$$
Set $\tau_1 = \infty$ if the process is killed before the first time it reaches $1.$ I want to find $\mathbb{P}(\tau_1 < \infty).$ To me, this is the probability that the random walk reaches $1$ before it is killed. I am struggling with this problem (I am doing practice qualifying exams and this was a question on an earlier exam, so obviously, no solution is provided).
My first thought was to calculate
$$\sum_{n=0}^{\infty}\mathbb{P}(S_n = 1) (1- \alpha)^{2n+1}.$$
However, I now know this is completely wrong. I know that you can only get to $1$ in an odd number of steps but $\mathbb{P}(S_n=1)$ includes the times that the random walk goes above $1$ and then gets back to $1$ in $n$ steps, which I do not want. So, I am completely stuck. Does anyone have any ideas?
Thanks!
Best Answer
Let $p_k=P_{k}(\tau_1<\infty)$, where the subscript $k$ indicates the starting point. I assume the killing happens with probability $\alpha=1-\beta$ at each step before the walking. Then $p_0=\beta(1/2+p_{-1}/2)$. On the other hand, by independence of the walk from $-1$ until reaching $0$ and from $0$ until reaching $1$, we have $p_{-1}=p_0^2$. Thus $$p_0=\beta(1+p_0^2)/2\,.$$ The solutions of this equation for $p_0$ are $$\frac{1\pm\sqrt{1-\beta^2}}{\beta}\,,$$ and only one of these is in $[0,1]$, so $$p_0=\frac{1-\sqrt{1-\beta^2}}{\beta}=\frac{\beta}{1+\sqrt{1-\beta^2}}\,. $$