One-sided maximal inequality for symmetric independent sums

Symmetric real random variables: A real random variable $X$ is symmetric if its law as defined in def-law-or-distribution-of-a-random-element equals the law of $-X$. Equivalently, $\mathbb{P}(X\in B)=\mathbb{P}(X\in -B)$ for every Borel $B\subseteq \mathbb{R}$, where $-B=\{-b:b\in B\}$. No existence of an expectation is assumed in this definition. In particular atoms, including an atom at zero, are allowed.

Disjoint groups of an independent sigma-algebra family remain independent: Let $({\mathcal{F}}_i{)}_{i\in I}$ be an independent family of sigma-algebras on a probability space, and let $J_0,\dots ,J_{m-1}\subseteq I$ be pairwise disjoint index sets. For each $r<m$, define ${\mathcal{G}}_r:=\sigma \left({\bigcup }_{i\in J_r}{\mathcal{F}}_i\right)\mathrm{.}$ Then the sigma-algebras ${\mathcal{G}}_0,\dots ,{\mathcal{G}}_{m-1}$ are independent.

Measurable coordinatewise functions preserve independence: Let $(X_i{)}_{i\in I}$ be an independent family of random elements $X_i:(\mathrm{\Omega },\mathcal{F},\mathbb{P})\to (S_i,{\mathrm{\Sigma }}_i)$. For each $i$, let $g_i:(S_i,{\mathrm{\Sigma }}_i)\to (T_i,{\mathcal{T}}_i)$ be measurable. Then the family $(g_i\circ X_i{)}_{i\in I}$ is independent.

Independent random elements have product joint law: Let $n\ge 1$, and let $X_i:(\mathrm{\Omega },\mathcal{F},\mathbb{P})\to (S_i,{\mathrm{\Sigma }}_i)$ for $i<n$ be independent random elements. Define $X=(X_0,\dots ,X_{n-1}):\mathrm{\Omega }\to {\prod }_{i<n}{S}_i\mathrm{.}$ Then $X$ is a random element of $({\prod }_{i<n}{S}_i,{⨂}_{i<n}{\mathrm{\Sigma }}_i)$, and its law is the finite product of the marginal laws: ${\mathbb{P}}_X={⨂}_{i<n}{\mathbb{P}}_{X_i}\mathrm{.}$