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Let $0<\alpha <1$, $X_0=x\in [0,1]$, and let $B_{n+1}$ be a binary random variable satisfying $\mathbb{E}[B_{n+1}\mid {\mathcal{F}}_n]=X_n$. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable and $B_{n+1}$ is ${\mathcal{F}}_{n+1}$-measurable. The updateremains in $[0,1]$ and defines a bounded martingale. For example, take independent uniform random variables $U_{n+1}$ independent of the current filtration and set $B_{n+1}=1_{\{U_{n+1}\le X_n\}}$.

The binary identity $B_{n+1}^2=B_{n+1}$ givesand henceThe L2-bounded martingale convergence theorem gives a terminal random variable $X_{\infty }$ with almost sure convergence and convergence in L2. Iteration followed by the bounded convergence theorem yieldsThus $X_{\infty }$ takes only $0$ and $1$, and preservation of its expected value gives a Bernoulli distribution with parameter $x$. The terminal conditional expectation representation gives the exact mean squared errorThe finite-time paths can stay strictly inside $(0,1)$ while their limit lies entirely at the endpoints. The positive conditional variance in the interior makes permanent convergence to an interior value impossible.

With the natural filtration ${\mathcal{F}}_n=\sigma (Z_1,\dots ,Z_n)$, the sums form a martingale. Put $v_n={\sum }_{k\le n}{a}_k^2$. Independence and the characteristic function of a standard normal random variable giveIf $M_n\to M_{\infty }$ almost surely with a finite limit, dominated convergence gives convergence of these characteristic functions to $\mathbb{E}{e}^{iu{M}_{\infty }}$. If $v_n\to \infty$, that limit would be $1$ at zero and $0$ at every nonzero $u$. A characteristic function is continuous at zero: apply dominated convergence to $e^{iu{M}_{\infty }}$ as $u\to 0$. This contradiction provesThis is also sufficient by the L2-bounded martingale convergence theorem, since $\mathbb{E}{M}_n^2=v_n$. Thus the almost sure convergence criterion for an independent Gaussian series is an equivalence.

The martingale property gives $\mathbb{E}[X_n\mid {\mathcal{F}}_m]=X_m$. Both $X_m$ and $X_n$ are square-integrable, so their product is integrable by the Cauchy-Schwarz inequality. Using the ${\mathcal{F}}_m$-measurability of $X_m$ in conditional expectation givesThe pull-out step for the possibly unbounded $X_m$ can be justified by first truncating it, then passing to the limit in the L2 norm and using the Cauchy-Schwarz inequality. Expanding the square now yieldsThe left side is nonnegative. Thus $a_n=\mathbb{E}[X_n^2]$ is nondecreasing and bounded above by $M$, and so $a_n\to a\le M$. For $m\le n$,By symmetry of the norm difference this proves that $(X_n)$ is a Cauchy sequence in the L2 space. The completeness established in part (a) gives an $X_{\infty }\in L^2$ withThis is the L2-bounded martingale convergence theorem, derived from martingale-difference orthogonality and completeness. The same sequence also converges almost surely by part (b), since $\mathbb{E}|X_n|\le \sqrt{M}$; uniqueness of a limit in probability identifies that limit with this $X_{\infty }$.

An elementary predictable process with stopping-time intervals has the form $H={\sum }_{j=0}^{r-1}{h}_j{1}_{({\tau }_j,{\tau }_{j+1}]}$, where the stopping times are ordered and $h_j$ is bounded and ${\mathcal{F}}_{{\tau }_j}$-measurable. Deterministic endpoints give the usual simple predictable process. Define the zero-starting stochastic integral byCommon refinement and telescoping show that this definition is independent of the representation. Use $M_{\infty }$ for an infinite endpoint. The L2-bounded martingale convergence theorem gives convergence of $M_t$ in $L^2$ to $M_{\infty }$ and $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$.

Put $D_j=M_{{\tau }_{j+1}}-M_{{\tau }_j}$. The optional sampling theorem gives $\mathbb{E}[D_j\mid {\mathcal{F}}_{{\tau }_j}]=0$. For $j<k$, the variable $h_j{D}_j{h}_k$ is ${\mathcal{F}}_{{\tau }_k}$-measurable; hence $\mathbb{E}[h_j{D}_j{h}_k{D}_k]=0$. Applying the same theorem to the assumed martingale $M^2-[M]$ givesConsequently the Itô isometry follows:The unbounded stopping times cause no optional-sampling gap. The $L^2$ terminal representation makes $M$ uniformly integrable and makes the family of stopped $M^2$ uniformly integrable, by conditional Jensen applied to $M_{\infty }^2$. Also $\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2<\infty$, from monotone convergence theorem and the assumed square-compensator identity. Thus $M^2-[M]$ has the needed uniform integrability, and truncating stopping times and passing in $L^1$ is legitimate. Bounded coefficients are part of the simple-integrand convention; more general coefficients require the displayed square-integrability condition.

The L2-bounded martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$ converges both almost surely and with convergence in L2 to a square-integrable random variable. Its hypotheses hold because $0<X_n<1$. Thus there is $X_{\infty }\in [0,1]$ almost surely with both kinds of convergence. Alternatively, the martingale convergence theorem gives the almost sure convergence, and the bound $|X_n-X_{\infty }{|}^2\le 1$ gives convergence in L2 by the dominated convergence theorem.^[ref]

The calculation of the conditional expectation of the square is especially informative. The martingale property makes the cross term vanish in the square expansion, soSubtracting this from $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$ givesTaking expected values and iterating provesBy bounded convergence theorem, $\mathbb{E}[X_{\infty }(1-X_{\infty })]=0$. The random variable inside this expected value is nonnegative, so it is zero almost surely; on $[0,1]$ this forces $X_{\infty }\in \{0,1\}$. Again by bounded convergence theorem, $\mathbb{E}{X}_{\infty }={\lim }_n\mathbb{E}{X}_n=x$. HenceThis is the Bernoulli polarization martingale with update size $1/2$. It also has an exact mean squared error: the terminal representation $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ and the equality case for conditional second moments give