Product martingale from independent mean one factors

The generated finite-history sigma-algebras form a filtration; finite products are adapted. Apply factorization to the Borel functions $t\mapsto \mathrm{\mid }t\mathrm{\mid }$ to get $E\mathrm{\mid }{M}_n\mathrm{\mid }={\prod }_{k=1}^{n}E\mathrm{\mid }{Y}_k\mathrm{\mid }<\mathrm{\infty }$ for $n\ge 1$; at $n=0$ it equals one. Group independence separates $\sigma (Y_{n+1})$ from the finite past, so $E[Y_{n+1}\mid {\mathcal{F}}_n]=E{Y}_{n+1}=1$ (also for the trivial past at zero).

The variable $M_n$ is finite and ${\mathcal{F}}_n$-measurable. Both $Y_{n+1}$ and $M_n{Y}_{n+1}=M_{n+1}$ are integrable by step 1.1. The unbounded taking-out clause therefore gives $E[M_{n+1}\mid {\mathcal{F}}_n]=M_{n}E[Y_{n+1}\mid {\mathcal{F}}_n]=M_n$. This proves the martingale assertion Martingale submartingale and supermartingale. AC is inherited from CE; neither positivity nor identical distribution of factors was used. If one adjoins the deterministic factor $Y_0=1$, the displayed filtration is exactly the natural filtration of the resulting zero-based factor process Natural filtration of a process, not necessarily that of the products.

For example take $\mathrm{\Omega }=\{-1,3{\}}^2$ with four equal masses, let $Y_1,Y_2$ be its coordinates and $Y_k=1$ for $k\ge 3$. Each first factor has mean $(-1+3)\mathrm{/}2=1$ and absolute mean $2$; coordinate rectangle counting proves independence. The four values of $M_2$ are $1,-3,-3,9$, so $E{M}_2=1$ and $E\mathrm{\mid }{M}_2\mathrm{\mid }=4=2\cdot 2$. Conditional on $Y_1=-1$ the product averages $(1-3)\mathrm{/}2=-1$, and conditional on $Y_1=3$ it averages $(-3+9)\mathrm{/}2=3$.