Linear variance bound for a predictably weighted martingale transform

A linear variance bound for martingale transforms of a Markov chain. For bounded measurable $g$ and $u$, set

Then under the chain law from any initial distribution,

A martingale transform, not just a martingale. The weight $u(X_k)$ is predictable: it is measurable with respect to $\sigma(X_0,\dots,X_k)$, the same $\sigma$-algebra against which the increment $g(X_{k+1}) - (Pg)(X_k)$ has zero conditional mean. Hence $\mathbb E[D_k \mid \sigma(X_0,\dots,X_k)] = u(X_k)\cdot 0 = 0$: multiplying by a predictable factor preserves the martingale-difference property, and with it the orthogonality $\mathbb E[D_jD_k]=0$ for $j\ne k$. Expanding the square leaves only the diagonal, giving the linear-in-$n$ bound. Taking $u \equiv 1$ recovers the plain martingale case.

Why the weighted form is needed. In the central limit theorem for a Markov chain one must control the quadratic variation $\frac1n\sum_{k<n} D_k^2$ of the martingale approximation, and expanding

produces exactly a martingale transform with the predictable weight $u = Pg$, alongside two terms that are functions of a single coordinate. The bound above makes the transform term $O(n)$ in $L^2$, hence $o(n)$ after dividing by $n$, so the quadratic variation converges to a deterministic limit without any appeal to an ergodic theorem.