Partial sums of independent centered variables are a martingale

To justify the finite $L^1$ operations independently of the affected supplier proof, augment every finite disjoint simple display by the complement with coefficient $0$. Pairwise intersections of two augmented displays partition the whole space and carry equal coefficients; finite additivity and $0\cdot (+\mathrm{\infty })=0$ prove representation independence. Common refinements give simple addition and monotonicity; scalar zero is direct and positive scalars are termwise. Supremum over simple minorants, increasing simple approximation and the sets $\{f_j\ge cs\}$ for $0<c<1$ give monotone convergence and nonnegative additivity. Positive/negative and real/imaginary decompositions give finite $L^1$ linearity. This also repairs the integral base beneath the event identities defining the conditional classes used below. The generated sigma-algebras are nested because their generator families are nested. The finite sum $S_n$ is ${\mathcal{F}}_n$-measurable and $E\mathrm{\mid }{S}_n\mathrm{\mid }\le \mathrm{\mid }c\mathrm{\mid }+{\sum }_{k=1}^{n}E\mathrm{\mid }{Y}_k\mathrm{\mid }<\mathrm{\infty }$. Independence means independence of the sigma-algebras $\sigma (Y_k)$ Independent random elements. Group the first $n$ of these separately from $\sigma (Y_{n+1})$. Then $Y_{n+1}$ is independent of ${\mathcal{F}}_n$, so $E[Y_{n+1}\mid {\mathcal{F}}_n]=E{Y}_{n+1}=0$. At $n=0$ independence of the trivial sigma-algebra follows directly from its two events.

Conditioning the finite identity $S_{n+1}=S_n+Y_{n+1}$ gives $E[S_{n+1}\mid {\mathcal{F}}_n]=S_n+0=S_n$. Hence this is a martingale Martingale submartingale and supermartingale. Each $S_k$ for $k\le n$ is measurable for $\sigma (Y_1,\dots ,Y_n)$, while $Y_k=S_k-S_{k-1}$ is measurable for $\sigma (S_0,\dots ,S_n)$. Minimality in both directions proves equality of these sigma-algebras, including the trivial initial one, as required by Natural filtration of a process. AC is inherited solely from the conditional classes; the independent sequence was given.

For a concrete model let $\mathrm{\Omega }=\{-1,1{\}}^2$ with each point of mass $1\mathrm{/}4$, $Y_1(u,v)=u$, $Y_2(u,v)=2v$, and $Y_k=0$ for $k\ge 3$. The two coordinate events have product probabilities because their intersections have cardinality the product of their cardinalities; adding constant variables preserves this identity. Here $S_0=c$, $S_1=c+u$, and $S_n=c+u+2v$ for $n\ge 2$. Averaging the two values in each fixed-u fibre gives $c+u$, and the first average is $c$. This displays the calculation without an identical-distribution assumption.