I’ve been thinking a lot about (discrete-time) martingales lately, as one might deduce from some of my recent posts. In particular, I’ve been trying to understand the Burkholder-Davis-Gundy inequalities for martingale transforms. This attempt has led me to the Doob and Krickeberg decomposition theorems which have powerful statements, yet are not too difficult to prove.

We first state and prove the Doob decomposition.

Theorem 1.A stochastic process adapted to a filtration , such that , can be (a.s.) uniquely writtenwhere is a martingale and is a predictable process with for all . . If is a submartingale (supermartingale), then a.s.

*Proof. *Suppose has such a Doob decomposition. Then, for ,

where we use the martingale property of and the predictability of . Note that if is a submartingale, then it is obvious from that , and analogously for the supermartingale case. We use this identity to define by setting and

It is tautological that is -measurable for all , where is the trivial -algebra. That follows from the hypothesis that and the triangle inequality.

We define . Clearly, is adapted to the filtration. To see that it defines a martingale, observe that for ,

We now prove the uniqueness claim. Suppose is another Doob decomposition. Our first identity shows that the predictable process is uniquely determined, hence a.s. for all . Hence,

Recall that a stochastic process adapted to a filtration is a semi-martingale if , where is martingale and such that , and . An immediate consequence of the Doob decomposition is that sub- and supermartingales are both semi-martingales.

We now prove the Krickeberg decomposition.

Theorem 2.Let be a~~submartingale~~martingale. Then there is a representation of , where and are nonnegative martingales, if and only if .

*Proof. *First, note that since is a martingale, we have by Jensen’s inequality that

so that

Thus, is an increasing sequence, and therefore it suffices to show that by the monotone convergence theorem.

If such a representation exists, then since and for all , we have that

for all .

Now suppose that . By definition of the Lebesgue integral, there exist nonnegative functions such that . Let denote the filtration with respect to which is adapted.

Since the mappings and define convex functions, it follows from the conditional version of Jensen’s inequality that and are nonnegative submartingales. Hence, for all

By induction, it follows that the sequence is a nondecreasing sequence of functions. Since and exists, we see from the monotone convergence theorem that

exist for all . We complete by observing that

We can generalize Theorem 2 to the case where is a submartingale.

Theorem 3.Suppose is a submartingale which is bounded in (i.e. ). Then we can write , where is a martingale and is a nonnegative supermartingale.

*Proof.* Let . If , then by repeated conditioning

by definition of submartingale. Thus exists a.s. as increasing (possibly infinite) limit and . Since is -measurable for all and pointwise limits of measurable functions are again measurable, we have that is -measurable.By the monotone convergence theorem,

by hypothesis that . Hence, is a.s. finite for all . By the conditional monotone convergence theorem,

which shows that is martingale.

Define . By the martingale property of and the submartingale property of , we have that

which shows that is a submartingale.

In fact, . Indeed,

so it suffices to show that . Since , we have that

We can combine the Doob and Krickeberg decompositions to make the following statements about martingales and submartingales. Any submartingale is bounded from below by martingale. If, in addition, the submartingale is bounded in , then it is bounded from above by an -bounded martingale.

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Does this actually work for theorem 2? To get the indicator outside the expectation in showing the martingale property, would you need $$1_{\{X_{n+1} \geq 0\}} \in \mathcal{F}_n$$?

Hi Peter, thank you for your comment. It appears that I made a few mistakes with regards to Theorem 2. One is that the statement of the theorem should read “martingale” instead of “submartingale.” I post a corrected version soon.

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Im thankful for the article post. Will read on…

Theorem 3 use the monotone convergence theorem, which works for nonnegative R.V.s, but E(X_m | F_n) is not. So if this proof still legal?

The same trick using in proof of Riesz decomposition in next blog. But here exists upper and lower bound: X_n> E(X_m | F_n)> E(Y_m | F_n)> Y_n, dominated convergence works.

Theorem 3 use the monotone convergence theorem, which works for nonnegative R.V.s, but E(X_m | F_n) is not. So if this proof still legal?

The same trick using in proof of Riesz decomposition in next blog. But here exists upper and lower bound: X_n> E(X_m | F_n)> E(Y_m | F_n)> Y_n, dominated convergence works.

The sequence $\mathbb{E}[X_{m}|\mathcal{F}_{n}] – X_{n}$ is nonnegative by the submartingale hypothesis.

You are right, thank you.

No problem; thank you for taking the time to comment. I will edit the post to make the use of the monotone convergence theorem more clear.

By the way, if sup $ E(|X_n|)<\infty $ can be extended to $ sup E(X_n)<\infty $?

since finiteness of E(Y_n) can be gotten by $ E(X_n)<\infty $ as well.