Lévy maximal inequality

Let $X_1,\dots ,X_n$ be independent random variables taking values in a normed vector space, each symmetric in law: $X_i$ and $-X_i$ have the same probability distribution. With partial sums $S_k={\sum }_{i=1}^k{X}_i$, the Lévy maximal inequality isThe statement requires no finite moments. Symmetry is stronger than having zero expected value and is the key reflection hypothesis.

For a proof, let $\tau$ be the first crossing index. On $\{\tau =k\}$, the future sum $R=X_{k+1}+\cdots +X_n$ is independent of the past and symmetric. For any vector $s$ with $\Vert s\Vert >t$, the two vectors $s+r$ and $s-r$ cannot both have norm at most $t$: their average is $s$, contradicting the triangle inequality. Symmetry therefore gives conditional terminal escape probability at least $1/2$. Sum over the disjoint first-crossing events to obtain the inequality. This controls every intermediate partial sum using a single terminal tail.