We have a symetric random walking. Random variables $\xi_i: P(\xi_i = 1) = \frac{1}{2}, P(\xi_i = -1) = \frac{1}{2}, \forall i$. Random walking $S_n = \sum_{k=0}^n \xi_k, S_0 = 0$. Denote $\alpha_p = min(n : |S_n| = p)$. I need to calculate $E\alpha_p$. I did the following: if $p = 1$ obviously that $E\alpha_1 = 1$. Assume that $p = 2$, it's not difficult to verify that $P(\alpha_2 = 2) = \frac{1}{2}, P(\alpha_2 = 4) = \frac{1}{2}, P(\alpha_2 = 6) = \frac{1}{2}$ and so on. We have that $E\alpha_2 = \sum_{k=1}^\infty \frac{2k}{2} = \infty$. Also, it's obviously that $\forall p_1, p_2 : p_2 > p_1 \rightarrow E\alpha_{p_1} < E\alpha_{p_2}$. Thereby $E\alpha_p = \begin{cases}\ 1, & \text{p = 1} \\ \infty, &p > 1 \end{cases}$
Could you please tell me if I have solved everything that is written correctly? (this result seems a bit paradoxical).
Your $P(\alpha_2=4)$ calculation is not right. It is actually $1/4$. Likewise, $P(\alpha_2=6)=1/8$ and so on. Computing the expectation, we get $$E(\alpha_2)=4$$
Before computing expectation, you can already know that your probabilities are not correct because they don’t add up to 1 over all k.
Actually, finding the $E(\alpha_p)$ by finding probabilities over all paths is quite tedious (okay for p values until 4, then a pain). If someone can point out a simple way to do that, I am all ears.
An easier way is to use recurrence relationships. To solve in the case of when the barriers are at $-p$ and $p$, let us denote by $E_n$, the expected lifespan if you begin at location $n$. We will finally want to find $E_0$.
Note three things $$E_{-p}=0$$ $$E_{p}=0$$ And the recurrence $$E_n=1+\frac{E_{n-1}}{2}+ \frac{E_{n+1}}{2}$$ The intuition behind this is that if we start from position $n$, we end up at $n-1$ or $n+1$ after 1 step with probability of $\frac{1}{2}$ each and each of the $E$ terms on the RHS of the equation give the expected lifespan from the new location.
For the recurrence, we try a solution $E_n=a+bn +cn^2$. Plugging that into the recurrence equation and the two boundary conditions, we get $$a=p^2, b=0, c=-1$$
Or $E_n=p^2-n^2$ or $E_0=p^2$. So we have $$E(\alpha_p)=E_0=p^2$$