My question concerns an excercise given by my probability theory teacher.
Consider a simple random walk, i.e., we have iid sequence $U_1, U_2,\dots$ with $\mathbb P(U_i=-1)=\mathbb P(U_i=1)=\frac{1}{2}$.
Define $X_0=0$ and for $n\geq 0$: $X_n=\sum_{i=1}^nU_i$.
Define for $k>0$, $q_k=\mathbb P$(the walker leaves the interval $(-1,k)$ in position $k$)
Show that $q_k=\frac{1}{2}((1-q_{k-1})q_k+q_{k-1}) $.
I think I have an answer: if the walker goes to 1 (probability $\frac{1}{2}$) and we condition on this position we see that he has probability $q_{k-1}$ to leave the interval $(0,k)$ in position $k$, and he has probability $1-q_{k-1}$ to not do this and go back to zero. Here he essentially starts over, and he has probability $q_k$ to leave the interval $(-1,k)$ in position $k$.
However, how do I make this more formal instead of mere telling why this is so?