Consider a process $X_t$, given by $X_0 = 1$ and $X_t = X_{t-1} + B_t$ where
\begin{align} B_t = \left\{ \begin{array}{lr} 1, \ \ \text{with probability } \frac{p}{2} \\ 0, \ \ \text{with probability } \frac{1}{2}\\ -1, \ \ \text{with probability } \frac{q}{2} \end{array} \right\} \end{align}
and $p+q=1$
Consider the event $E_k = \{X_s = 0 \text{ for some } s>0, X_0 = k\}$. Explain why $\mathbb{P}(E_2) = \mathbb{P}(E_1)^2$.
So, I think this is a random walk, but it's not a simple random walk as $B_t$ can take 3 values (correct me if I'm wrong!).
A hint I have been given for this question is to use the same concept that is used to show this in a simple random walk, but this requires the assumption that given the current state $X_s$, the next state $X_{s+1}$ is independent from the history of the random walk. But, in my course we haven't seen this result for anything but a simple random walk and we have only seen simple random walks so I'm a bit stumped with this...
If I were to assume this could I write
$$ \begin{align} \mathbb{P}(E_2) &= \mathbb{P}(\exists 0 < s_1 < s_2 \text{ s.th } X_{s_1} = 1 | X_0 = 2) \\ &= \mathbb{P}(X_{s_2} = 0 | X_{s_1} = 1) \mathbb{P}(X_{s_1} = 1 | X_0 = 2) \\ &= \mathbb{P}(E_1)^2 \end{align} $$
Where the last equality is because $$ \begin{align} \mathbb{P}(E_1) = \mathbb{P}(X_{s_i} = k-i | X_{s_{i-1}} = k - i + 1) \end{align} $$
Is this correct? And, if so why? If anyone could point me to a good resource on this, introductory stochastic processes I would really appreciate it. I'm struggling to find anything that isn't too advanced for me and I really want to get my head around these basics before moving on to Markov Chains.