Discrete-Time Martingale Decomposition Theorems

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 X=(X_{n})_{n=1}^{\infty} adapted to a filtration (\mathcal{F}_{n})_{n=1}^{\infty}, such that X_{n}\in L^{1}, can be (a.s.) uniquely written

\displaystyle X_{n}=Y_{n}+Z_{n},

where (Y_{n})_{n=1}^{\infty} is a martingale and (Z_{n})_{n=1}^{\infty} is a predictable process with Z_{n}\in L^{1} for all n. . If (X_{n}) is a submartingale (supermartingale), then Z_{n} \leq (\geq) Z_{n+1} a.s.

Proof. Suppose (X_{n}) has such a Doob decomposition. Then, for n\geq 1,

\begin{array}{lcl}\displaystyle\mathbb{E}[X_{n+1}-X_{n}\mid\mathcal{F}_{n}]&=&\mathbb{E}[(Y_{n+1}+Z_{n+1})-(Y_{n}+Z_{n})\mid\mathcal{F}_{n}]\\&=&\displaystyle\mathbb{E}[Y_{n+1}-Y_{n}\mid\mathcal{F}_{n}]+\mathbb{E}[Z_{n+1}-Z_{n}\mid\mathcal{F}_{n}]\\&=&\displaystyle Z_{n+1}-Z_{n},\end{array}

where we use the martingale property of (Y_{n}) and the predictability of (Z_{n}). Note that if (X_{n}) is a submartingale, then it is obvious from \mathbb{E}[X_{n+1}-X_{n}\mid\mathcal{F}_{n}]\geq 0 that Z_{n+1}\geq Z_{n}, and analogously for the supermartingale case. We use this identity to define Z_{n} by setting Z_{1}:=0 and

\displaystyle Z_{n}:=\mathbb{E}[X_{n}-X_{n-1}\mid\mathcal{F}_{n-1}]+Z_{n-1}=\sum_{j=2}^{n}\mathbb{E}[X_{j}-X_{j-1}\mid\mathcal{F}_{j-1}], \indent \forall n > 1

It is tautological that Z_{n} is \mathcal{F}_{n-1}-measurable for all n \geq 1, where \mathcal{F}_{0}:=\left\{\emptyset,\Omega\right\} is the trivial \sigma-algebra. That Z_{n}\in L^{1} follows from the hypothesis that X_{n}\in L^{1} and the triangle inequality.

We define Y_{n}:=X_{n}-Z_{n}. Clearly, (Y_{n}) is adapted to the filtration. To see that it defines a martingale, observe that for n\geq 2,

\begin{array}{lcl}\displaystyle\mathbb{E}[Y_{n+1}\mid\mathcal{F}_{n}]=\mathbb{E}[X_{n+1}-Z_{n+1}\mid\mathcal{F}_{n}]&=&\mathbb{E}[X_{n+1}\mid\mathcal{F}_{n}]-\mathbb{E}[Z_{n+1}\mid\mathcal{F}_{n}]\\&=&\displaystyle\mathbb{E}[X_{n+1}\mid\mathcal{F}_{n}]-Z_{n+1}\\&=&\displaystyle\mathbb{E}[X_{n+1}\mid\mathcal{F}_{n}]-\sum_{j=2}^{n+1}\mathbb{E}[X_{j}-X_{j-1}\mid\mathcal{F}_{j-1}]\\&=&\displaystyle\mathbb{E}[X_{n}\mid\mathcal{F}_{n}]-\sum_{j=2}^{n}\mathbb{E}[X_{j}-X_{j-1}\mid\mathcal{F}_{j-1}]\\&=&\displaystyle X_{n}-Z_{n}\\&=&\displaystyle Y_{n}\end{array}

We now prove the uniqueness claim. Suppose X_{n}=\tilde{Y}_{n}+\tilde{Z}_{n} is another Doob decomposition. Our first identity shows that the predictable process is uniquely determined, hence \tilde{Z}_{n}=Z_{n} a.s. for all n\geq 1. Hence,

\displaystyle Y_{n}=X_{n}-Z_{n}=X_{n}-\tilde{Z}_{n}=\tilde{Y}_{n},\indent\forall n\geq 1

\Box

Recall that a stochastic process (X_{n})_{n=1}^{\infty} adapted to a filtration (\mathcal{F}_{n})_{n=0}^{\infty} is a semi-martingale if X_{n}=Y_{n}+Z_{n}, where (Y_{n})_{n=0}^{\infty} is martingale and Z_{n}=U_{n}-V_{n} such that U_{n} \leq U_{n+1}, V_{n}\leq V_{n+1}, and U_{n},V_{n}\in L^{1}. 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 \left\{X_{n}\right\}_{n=1}^{\infty} be a submartingale martingale. Then there is a representation of X_{n} := U_{n}-V_{n}, where \left\{U_{n}\right\}_{n=1}^{\infty} and \left\{V_{n}\right\}_{n=1}^{\infty} are nonnegative martingales, if and only if \lim_{n \rightarrow \infty}\mathbb{E}\left[\left|X_{n}\right|\right] < \infty.

Proof. First, note that since \left\{X_{n}\right\} is a martingale, we have by Jensen’s inequality that

\displaystyle\mathbb{E}\left[\left|X_{n+1}\right|\mid X_{n}\right]\geq\left|\mathbb{E}\left[X_{n+1}\mid X_{n}\right]\right|=\left|X_{n}\right|,

so that

\displaystyle\mathbb{E}\left[\left|X_{n+1}\right|\right]=\mathbb{E}\left[\mathbb{E}\left[\left|X_{n+1}\right|\mid X_{n}\right]\right]\geq\mathbb{E}\left[\left|X_{n}\right|\right]

Thus, (\mathbb{E}[\left|X_{n}\right|])_{n=1}^{\infty} is an increasing sequence, and therefore it suffices to show that \sup_{n}\mathbb{E}[\left|X_{n}\right|] < \infty by the monotone convergence theorem.

If such a representation X_{n} = U_{n}-V_{n} exists, then since \mathbb{E}[U_{n}] = \mathbb{E}[U_{1}] and \mathbb{E}[V_{n}]=\mathbb{E}[V_{1}] for all n \geq 1, we have that

\begin{array}{lcl}\displaystyle\mathbb{E}\left[\left|X_{n}\right|\right]=\mathbb{E}\left[\left|U_{n}-V_{n}\right|\right]\leq\mathbb{E}\left[\left|U_{n}\right|+\left|V_{n}\right|\right]&=&\displaystyle\mathbb{E}\left[U_{n}+V_{n}\right]\\&=&\displaystyle\mathbb{E}\left[U_{n}\right]+\mathbb{E}\left[V_{n}\right]\\&=&\displaystyle\mathbb{E}\left[U_{1}\right]+\mathbb{E}\left[V_{1}\right]\end{array}

for all n \geq 1.

Now suppose that \lim_{n \rightarrow \infty}\mathbb{E}[\left|X_{n}\right|] < \infty. By definition of the Lebesgue integral, there exist nonnegative L^{1} functions X_{n}^{+}, X_{n}^{-} such that X_{n}=X_{n}^{+}-X_{n}^{-}. Let (\mathcal{F}_{n})_{n=0}^{\infty} denote the filtration with respect to which X_{n} is adapted.

Since the mappings x\mapsto x^{+} and x\mapsto x^{-} define convex functions, it follows from the conditional version of Jensen’s inequality that \left\{X_{n}^{+}\right\} and \left\{X_{n}^{-}\right\} are nonnegative submartingales. Hence, for all n\geq 1

\displaystyle\mathbb{E}[X_{n+1}^{+}\mid\mathcal{F}]=\mathbb{E}\left[\mathbb{E}[X_{n+1}^{+}\mid\mathcal{F}]\mid\mathcal{F}\right]\geq\mathbb{E}\left[X_{n}^{+}\mid\mathcal{F}\right]

By induction, it follows that the sequence \mathbb{E}[X_{n+m}^{+}\mid\mathcal{F}] is a nondecreasing sequence of functions. Since \max\left\{X_{n}^{+},X_{n}^{-}\right\}\leq\left|X_{n}\right| and \lim_{n\rightarrow\infty}\mathbb{E}[\left|X_{n}\right|] exists, we see from the monotone convergence theorem that

\displaystyle U_{n}:=\lim_{m\rightarrow\infty}\mathbb{E}[X_{n+m}^{+}\mid\mathcal{F}_{n}]\text{ and }V_{n}:=\lim_{m\rightarrow\infty}\mathbb{E}[X_{n+m}^{-}\mid\mathcal{F}_{n}]

exist for all n\geq 1. We complete by observing that

\displaystyle U_{n}-V_{n}=\lim_{m\rightarrow\infty}\mathbb{E}\left[X_{n+m}^{+}-X_{n+m}^{-}\mid\mathcal{F}_{n}\right]=\lim_{m\rightarrow\infty}X_{n}=X_{n}

\Box

We can generalize Theorem 2 to the case where (X_{n}) is a submartingale.

Theorem 3. Suppose (X_{n})_{n=1}^{\infty} is a submartingale which is bounded in L^{1} (i.e. \sup_{n}\left\|X_{n}\right\|_{L^{1}}<\infty). Then we can write X_{n}=Y_{n}-Z_{n}, where (Y_{n}) is a martingale and (Z_{n}) is a nonnegative supermartingale.

Proof. Let n,m\in\mathbb{Z}^{\geq 1}. If m \geq n, then by repeated conditioning

\displaystyle\mathbb{E}[X_{m+1}\mid\mathcal{F}_{n}]\geq\mathbb{E}[X_{n+1}\mid\mathcal{F}_{n}]\geq X_{n}

by definition of submartingale. Thus Y_{n}:=\lim_{m\rightarrow\infty}\mathbb{E}[X_{m}\mid\mathcal{F}_{n}] exists a.s. as increasing (possibly infinite) limit and Y_{n}-X_{n}\geq 0. Since \mathbb{E}[X_{m}\mid\mathcal{F}_{n}] is \mathcal{F}_{n}-measurable for all m and pointwise limits of measurable functions are again measurable, we have that Y_{n} is \mathcal{F}_{n}-measurable.By the monotone convergence theorem,

\displaystyle\mathbb{E}[Y_{n}]=\lim_{m\rightarrow\infty}\mathbb{E}\left[\mathbb{E}[X_{m}\mid\mathcal{F}_{n}]\right]=\lim_{m\rightarrow\infty}\mathbb{E}[X_{m}]<\infty

by hypothesis that \sup_{m}\left\|X_{m}\right\|_{L^{1}}<\infty. Hence, Y_{n} is a.s. finite for all n. By the conditional monotone convergence theorem,

\displaystyle\mathbb{E}[Y_{n+1}\mid\mathcal{F}_{n}]=\lim_{m\rightarrow\infty}\mathbb{E}\left[\mathbb{E}[X_{m}\mid\mathcal{F}_{n}]\mid\mathcal{F}_{n}\right]=\lim_{m\rightarrow\infty}\mathbb{E}\left[X_{m}\mid\mathcal{F}_{n}\right]=Y_{n},

which shows that (Y_{n})_{n=1}^{\infty} is martingale.

Define Z_{n}:=Y_{n}-X_{n}\geq 0. By the martingale property of (Y_{n}) and the submartingale property of (X_{n}), we have that

\displaystyle\mathbb{E}[Z_{n+1}\mid\mathcal{F}_{n}]=\mathbb{E}\left[Y_{n+1}-X_{n+1}\mid\mathcal{F}_{n}\right]=Y_{n}-\mathbb{E}[X_{n+1}\mid\mathcal{F}_{n}]\leq Y_{n}-X_{n}=Z_{n},

which shows that (Z_{n})_{n=1}^{\infty} is a submartingale. \Box

In fact, \max\left\{\sup_{n}\left\|Y_{n}\right\|_{L^{1}},\sup_{n}\left\|Z_{n}\right\|_{L^{1}}\right\}<\infty. Indeed,

\displaystyle\left\|Y_{n}\right\|_{L^{1}}\leq\sup_{n}\left\|X_{n}\right\|_{L^{1}}+\sup_{n}\left\|Z_{n}\right\|_{L^{1}},

so it suffices to show that \sup_{n}\left\|Z_{n}\right\|_{L^{1}}<\infty. Since Z_{n}\geq 0, we have that

\displaystyle\left\|Z_{n}\right\|_{L^{1}}\leq\left\|Z_{1}\right\|_{L^{1}}\leq\left\|X_{1}\right\|_{L^{1}}+\left\|Y_{1}\right\|_{L^{1}}

We can combine the Doob and Krickeberg decompositions to make the following statements about martingales and submartingales. Any submartingale (X_{n})_{n=1}^{\infty} is bounded from below by martingale. If, in addition, the submartingale (X_{n}) is bounded in L^{1}, then it is bounded from above by an L^{1}-bounded martingale.

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11 Responses to Discrete-Time Martingale Decomposition Theorems

  1. Pingback: Discrete-Time Martingale Decomposition Theorems II | Math by Matt

  2. Peter says:

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

    • matthewhr says:

      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.

  3. Pingback: Correction | Math by Matt

  4. Im thankful for the article post. Will read on…

  5. Anonymous says:

    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.

  6. Xia Panqiu says:

    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.

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