= Lévy maximal inequality
{c}
Let $X_1,\ldots,X_n$ 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> $S_k=\sum_{i=1}^kX_i$, the <Lévy maximal inequality> is
$$
\mathbb P\left(\max_{1\leq k\leq n}\lVert S_k\rVert>t\right)
\leq2\mathbb P(\lVert S_n\rVert>t),\qquad t\geq0.
$$
The 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 $\lVert s\rVert>t$, 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.
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