Independent zero-mean random variables maximum inequality

Question

Let $X_1,X_2,\ldots$ be i.i.d . zero-mean random variables with symmetric PDF's and $S_k = \sum_{i=1}^k X_i.$ Rhen for any $\varepsilon \in \mathbb{R}$ $$ \mathbb{P}(\max_{1\leq k\leq n} S_k>\varepsilon) \leq 2 \mathbb{P}(S_n>\varepsilon) $$ I can just observe that these PDF's has to be symmetric around mean, thus around zero.

Answer 1

Assume $\epsilon > 0$ and $S_0 := 0$. By symmetry, the proof will be the same for $-\epsilon$. The idea is to break up $\{S_n > \epsilon\}$ into disjoint events when $S_k$ first passes $\epsilon$. Define the events $$ A_k := \{S_k > \epsilon,\ S_i \le \epsilon\ \forall i < k\} $$ which are the disjoint events that $k$ is the first time we exceed $\epsilon$. Notice that $$ P(\max_{1\le k \le n} S_k > \epsilon) = \sum_{k=1}^n P(A_k) $$ since the maximum exceeds $\epsilon$ if and only if at least one of the $S_k$ exceeds $\epsilon$ by time $n$. Since we want to stay above $\epsilon$ at the final time, multiply both sides by $P(S_n - S_k \ge 0)$. By symmetry, $P(S_n - S_k \ge 0) \ge 1/2$, so $$ \frac{1}{2}P(\max_{1\le k \le n} S_k > \epsilon) \le \sum_{k=1}^n P(A_k)P(S_n - S_k \ge 0) $$ $S_n - S_k$ depends only on $X_{k+1},\ldots,X_n$, while $A_k$ depends only on $X_1,\ldots,X_k$. The $X_i$ are i.i.d. so in fact $$ P(A_k)P(S_n - S_k \ge 0) = P(A_k \cap \{S_n - S_k \ge 0\}) = P(S_k > \epsilon,\ S_i \le \epsilon\ \forall i < k,\ S_n - S_k \ge 0) $$ If $S_k > \epsilon$ and $S_n-S_k \ge 0$, then $S_n > \epsilon$. Moreover, $A_k \cap \{S_n - S_k \ge 0\}$ are still disjoint sets. Thus, $$ \frac{1}{2}P(\max_{1\le k \le n} S_k > \epsilon) \le \sum_{k=1}^n P(A_k \cap \{S_n - S_k \ge 0\}) \le P(S_n > \epsilon) $$ and therefore, $P(\max_{1\le k \le n}S_k > \epsilon) \le 2P(S_n > \epsilon)$. Note that this proof does not require existence of a PDF.