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Conditional on ${\mathcal{F}}_n$, the increment is $-X_n/2$ with probability $1-X_n$, and $(1-X_n)/2$ with probability $X_n$. ThereforeTaking expectations using the tower property yieldsThis is the conditional variance that drives the Bernoulli polarization martingale toward its endpoints.

Conditional on ${\mathcal{F}}_n$, put $x=X_n$. The two possible values have conditional meanThe process is adapted and bounded, hence integrable. Consequently it is a martingale. Since ${\sup }_n\mathbb{E}|X_n|\le 1$, the martingale convergence theorem gives an almost surely finite limit $X_{\infty }$, necessarily in $[0,1]$.

Moreover $|X_n-X_{\infty }{|}^2\le 1$ and these squared differences tend to zero almost surely. The dominated convergence theorem gives $\mathbb{E}|X_n-X_{\infty }{|}^2\to 0$. Thus the Bernoulli polarization martingale converges almost surely and in $L^2$.

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