# (PTE) 4.6.3. Riez's decomposition


We know that any submartingales can be decomposed into a martingale and a predictable sequence (Doob’s decomposition). Riez’s decomposition allows us to do the similar to uniformly integrable non-negative supermartingales.

## Riez’s decomposition

A supermartingale $(X_n)$ is a potential if it is non-negative and $EX_n \to 0$ a.s.

Two notable properties of potentials is that (i) $X_n \to 0$ a.s. and (ii) $(X_n)$ is uniformly integrable. (i) is from supermartingale convergence and Fatou’s lemma. (ii) follows from $E|X_n| \le \epsilon$ for large $n$ for all $\epsilon>0.$

For a non-negative uniformly integrable supermartingale $(X_n),$ there uniquely exist a uniformly integrable martingale $(M_n)$ and a potential $(V_n)$ so that $X_n = M_n + V_n.$

By supermartingale convergence, there exists $X_\infty$ such that $X_n \to X_\infty$ a.s. Let $M_n = E(X_\infty|\mathcal{F}_n)$ be a regular, thus uniformly integrable martingale. It is enough to show that $V_n := X_n-M_n$ is a potential. $$E(V_{n+1}|\mathcal{F}_n) = E(X_{n+1}|\mathcal{F}_n) - E(M_{n+1}|\mathcal{F}_n) \le X_n - M_n = V_n \text{ a.s.}$$ Thus $V_n$ is a supermartingale. $$E(X_\infty|\mathcal{F}_n) \le \liminf_m E(X_m|\mathcal{F}_n) \le X_n \text{ a.s.}$$ for all fixed $n.$ Thus $V_n \ge 0$ for all $n.$ Now by Levy's theorem, $$\lim_n V_n = X_\infty - \lim_n E(X_\infty|\mathcal{F}_n) = 0 \text{ a.s.}$$ Since $(X_n), (M_n)$ are uniformly integrable, $(V_n)$ is also. By Vitali's lemma $EV_n \to E\lim_n V_n = 0.$ Thus $V_n$ is a potential.
For the uniqueness part, let $M_n+V_n = M_n'+V_n',$ $M_n = E(\eta_1|\mathcal{F}_n)$ a.s. and $M_n' = E(\eta_2|\mathcal{F}_n)$ a.s. $$M_n-M_n' = V_n'-V_n = E(\eta_1|\mathcal{F}_n) - E(\eta_2|\mathcal{F}_n) \to 0 \text{ a.s.}$$ since $V_n, V_n'$ are potentials. By Levy's theorem this implies $E(\eta_1|\mathcal{F}_\infty) - E(\eta_2|\mathcal{F}_\infty) = 0$ a.s. \begin{aligned} M_n &= E\left( E(\eta_1|\mathcal{F}_\infty) | \mathcal{F}_n \right) \\ &= E\left( E(\eta_2|\mathcal{F}_\infty) | \mathcal{F}_n \right) \\ &= E(\eta_2|\mathcal{F}_n) = M_n' \text{ a.s.} \end{aligned} Equivalence of $V_n,V_n'$ directly follows.

Acknowledgement

This post series is based on the textbook Probability: Theory and Examples, 5th edition (Durrett, 2019) and the lecture at Seoul National University, Republic of Korea (instructor: Prof. Sangyeol Lee).