I am reading the proof of Theorem 5.2.7 (Reflection Principle) in Durrett's book, avaialable here (see page 278, which is page 286 of the PDF file).
The theorem statement is: Let $\xi_1,\xi_2,\dotsc$ be i.i.d. random variables, with a distribution that is symmetric about $0$. Let $S_n=\xi_1+\dotsc+\xi_n$ and $a>0$. Then $P(\sup_{m\leq n}S_m \leq a)\leq 2P(S_n\geq a)$.
Before the proof is given, Durrett gives an imprecise intuitive proof idea. Then he gives the actual proof, using the Strong Markov Property.
It seems to me that the intuitive proof can easily be turned into a rigorous one. Yet, Durrett proceeds to give a proof using a deeper tool (the Strong Markov Property).
Is the proof below correct?
$P(S_n \geq a)\geq \sum_{m=1}^n P(S_1<a,S_2<a,\dotsc,S_{m-1}<a,S_m\geq a)\cdot \underbrace{P(S_n-S_m\geq 0)}_{\geq 1/2}\geq \frac{1}{2} P(\sup_{m\leq n}S_m \leq a) $.
My goal here is to gain insight into the Strong Markov Property and to see how it helps to solve somewhat difficult problems. This is why I'd like to know if my easy proof is correct.
EDIT: The first inequality follows by partitioning the probability space according to the first time $S_i$ is at least $a$ (if it ever is), and considering the event that the sum of the remaining $\xi_j$ after this first time is nonnegative (originally I wrote the first inequality as an eqaulity, but I only need the inequality, and I think I was actually wrong to write it as an equality).
I think you are correct. If a more detailed explanation is needed - maybe the following lines may serve this purpose:
Let $A_1:=\{S_1 \geq a\}, A_{k+1}:=\{max_{1\leq j\leq k}(S_j) < a, S_{k+1}\geq a \}$ for $1 \leq k < n$.
Then the $A$-events are pairwise disjoint and their union is: $$\cup_{k=1}^n A_k = \{ max_{1\leq k \leq n}(S_k) \geq a\} $$
Further, if $A_k$ occurs, then $S_k \geq a$ holds. Thus if then also $S_n - S_k\geq 0$ holds, then $S_n \geq a$ is implied, i.e. $$A_k \cap \{S_n-S_k\geq 0\} \subset \{S_n \geq a\}$$
Putting all this together yields: $$P(S_n \geq a)\geq \sum_{k=1}^n P(A_k \cap \{S_n-S_k \geq 0\})= \sum_k P(\{S_n-S_k \geq 0\})P(A_k)\geq \frac{1}{2}P(max_{1\leq k \leq n}(S_k)\geq a)$$