Let $S_0 := 0, S_1, \cdots$ be random integers. Define the increment $(S_i - S_{i - 1})_{i = 1} ^\infty$ as i.i.d. symmetric random variables and $T_1 := \min\{ n > 0: S_n = 0 \}$. We ask if $k$ is even, what is $\mathbb{P}\{ T_1 = k \}$.
I am trying to understand an equality in the proof:
It is claimed that $\mathbb{P}\{ S_1 = -1 \text{ and }S_{k - 1} = 1 \} = \frac{1}{2}\mathbb{P}\{ S_{k - 2} = 2 \}$.
My understanding is the following graph:
We are shifting the coordinate system (coordinate system in red $S_{t'}$). Then I think what I see here is the following equality:
$$\mathbb{P}\{ S_1 = -1 \text{ and }S_{k - 1} = 1 \} = \mathbb{P}\{ S_{k - 2} = 2 \}$$
But this is off by a constant multiple of $\frac{1}{2}$ on the right hand side when comparing to the claim. How is the $\frac{1}{2}$ appearing in the equality?
Note that we had to assume that $S_1 = -1$ in order to do this shifting of the coordinate system. Therefore, the probability that $S_{k-1} = 1$ is equal to the probability that $S_{k-2} = 2$ but only under the assumption that $S_1 = -1$. The $\tfrac{1}{2}$ should fall out from there.
EDIT: Here's a less handwavey answer. Given two events $A$ and $B$, one can define the probability of $A$ given that $B$ has already happened as $$\mathbb{P}(A \text{ given that } B) = \frac{\mathbb{P}(A \text{ and }B)}{\mathbb{P}(B)}.$$ Read more about this here (Conditional Probability). In our situation, we can let $A$ be the event that $S_{k-1}=1$ and let $B$ be the event that $S_1 = -1$ to get
$$\mathbb{P}(S_1 = -1 \text{ and } S_{k-1}=1) =\mathbb{P}(S_1 = -1)\mathbb{P}(S_{k-1}=1 \text{ given that } S_1 = -1).$$
What you are calculating with your graph is actually the probability $\mathbb{P}(S_{k-1}=1 \text{ given that } S_1 = -1)$ since you are assuming that $S_1 = -1$ already. Thus, you must multiply by the probability that $S_1 = 1$ which is $\frac{1}{2}$. Hope this is more clear!