Let $\Omega = \{w = (x_1, \dots, x_N) | \; x_i \in \{-1, 1\}\}, \;X_k(w) = x_k, \;S_n(w) = \sum_{k=1}^n X_k(w), \; S_0(w) = 0.$ After having proven a few theorems about $S_n$, in our lecture about probability theory and statistics, we stated the following theorem:
Let $L(w) = \max \{0 \leq n \leq 2N |\; S_n(w) = 0\}$. The distribution of $L$ is the so called arcsine-distribution: $$P(L = 2n) = P(S_{2n} = 0) \cdot P (S_{2N-2n} = 0) = 2^{-2N} \binom{2n}{n} \binom{2N-2n}{N-n}.$$
I was really confused after reading this theorem. First of all, the definition of $L$ seems rather odd to me. We take $L$ as the maximum $0 \leq n \leq 2N$ such that $S_n(w) = 0$. How could possibly $n > N$ hold? After all, in our original definition of $\Omega$, we only look at finite periods of length $N$, or am I completely mistaken? I thought maybe that was a typo, so I looked at the actual theorem. It is a explicit formula for $P(L=2n)$. But this implies $2n \leq N$ and therefore $2N-2n \geq N$ but we are also looking at the term $P(S_{2N-2n} = 0)$. For these two terms to make sense, we must have $2n = N$. This would make the theorem rather useless, in my opinion.
Where did I go completely wrong?
The formula for $P[L=2n]$ holds for every $0\leqslant n\leqslant N$. In this range the binomial coefficients on the RHS are (well defined and) positive. The sum of the values of the RHS from $n=0$ to $n=N$ is $1$. Finally, for every $n$, $P[L=2n+1]=0$.