Square of a martingale minus quadratic compensator

The generated sigma-algebras are nested, and finite sums are measurable by Arithmetic and lattice operations preserve measurability whenever they are defined. Cauchy–Schwarz with the constant one gives $E\mathrm{\mid }{Y}_k\mathrm{\mid }\le (E{Y}_k^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Repeated use of $(a+b{)}^2\le 2a^2+2b^2$ shows that every finite sum $S_n$ has finite second moment, and hence finite first moment. Group independence and [F2] give $E[Y_{n+1}\mid {\mathcal{F}}_n]=0$, while the known $S_n$ conditions to itself. Thus $E[S_{n+1}\mid {\mathcal{F}}_n]=S_n$, proving the martingale property Martingale submartingale and supermartingale.

The measurable variable $Y_k^2$ is integrable and its Borel events belong to $\sigma (Y_k)$, independent of ${\mathcal{F}}_{k-1}$. Thus $E[Y_k^2\mid {\mathcal{F}}_{k-1}]=E{Y}_k^2={\sigma }_k^2$. Since $S_k-S_{k-1}=Y_k$, the bracket formula gives $⟨S{⟩}_n={\sum }_{k=1}^n{\sigma }_k^2$. The square-minus-bracket theorem now gives the asserted martingale. AC is inherited from these conditional classes and the bracket construction.

To see why the optional sum differs, take $\mathrm{\Omega }=\{-1,0,1\}$ and $P=\frac{1}{4}{\delta }_{-1}+\frac{1}{2}{\delta }_0+\frac{1}{4}{\delta }_1$. This is a measure by [F7]–[F8], and its total mass is one. Set $Y_1(\omega )=\omega$ and $Y_k=0$ for $k\ge 2$. The family $(Y_k{)}_{k\ge 1}$ is independent because all but one member have only probability-zero or probability-one events. Here $E{Y}_1=0$ and $E{Y}_1^2=1\mathrm{/}2$. Thus $[S{]}_1=Y_1^2$ takes values $0,1$ with probabilities $1\mathrm{/}2,1\mathrm{/}2$, whereas $⟨S{⟩}_1=1\mathrm{/}2$ everywhere. The compensated square takes values $-1\mathrm{/}2,1\mathrm{/}2$ of equal mass and thereafter stays fixed.