Discrete-Time Martingale Decomposition Theorems II

A few days ago, I wrote a post on some decomposition theorems for discrete-time martingales. Today, I want to talk a little bit more about decomposition theorems. In particular, I want to show that the Krickeberg decomposition theorem is not (almost surely) unique, unlike the Doob decomposition. I also want to cover another result known as the Riesz decomposition, which makes use of the celebrated martingale convergence theorem.

In what follows, all random variables are taken to defined on a complete probability space (\Omega,\mathcal{F},\mathbb{P}) and all filtrations are assumed to be complete.

As we will need the definition in the statement of the Riesz decomposition theorem, we first introduce the stochastic idea of potential. I have not studied enough probability theory and PDE to say for sure, but I imagine that the term “potential” is motivated by the connections between (sub-,super-)martingales and (sub-,super-)harmonic functions.

Definition 1. We say that a nonnegative supermartingale \left\{(X_{n},\mathcal{F}_{n})\right\}_{n=0}^{\infty} is a potential if \mathbb{E}[X_{n}]\rightarrow 0 as n\rightarrow\infty.

Recall that a supermartingale has the defining property \mathbb{E}[X_{n+1}\mid\mathcal{F}_{n}]\leq X_{n}, so by the tower property of conditional expectation, the sequence (\mathbb{E}[X_{n}])_{n=0}^{\infty} is decreasing. In the case of a potential, we have that \mathbb{E}[X_{n}]\downarrow 0.

To see that the Krickeberg decomposition is not (a.s.) unique, we need the martingale convergence theorem, which tells us that any nonnegative martingale \left\{(X_{n},\mathcal{F}_{n})\right\}_{n=0}^{\infty} converges almost surely to an L^{1} random variable X_{\infty} as n\rightarrow\infty. I claim that \left\{(\mathbb{E}[X_{\infty}\mid\mathcal{F}_{n}],\mathcal{F}_{n})\right\}_{n=0}^{\infty} is a nonnegative martingale. Nonnegativity is immediate from X_{\infty} \geq 0 a.s., and for any A\in\mathcal{F}_{n},

\displaystyle\int_{A}\mathbb{E}\left[\mathbb{E}[X_{\infty}\mid\mathcal{F}_{n+1}]\mid\mathcal{F}_{n}\right]d\mathbb{P}=\int_{A}\mathbb{E}[X_{\infty}\mid\mathcal{F}_{n+1}]d\mathbb{P}=\int_{A}X_{\infty}d\mathbb{P}=\int_{A}\mathbb{E}\left[X_{\infty}\mid\mathcal{F}_{n}\right]d\mathbb{P},

since \mathcal{F}_{n}\subset\mathcal{F}_{n+1}.

Applying the Krickeberg decomposition, we can write

\displaystyle X_{n}=M_{n}^{(1)}-M_{n}^{(2)},

where \left\{(M_{n}^{(1)},\mathcal{F}_{n})\right\}_{n=0}^{\infty},\left\{(M_{n}^{(2)},\mathcal{F}_{n})\right\}_{n=0}^{\infty} are nonnegative martingales. Since the sum of nonnegative martingales is again a nonnegative martingale, we see that

\displaystyle X_{n}=\left(M_{n}^{(1)}+\mathbb{E}[X_{\infty}\mid\mathcal{F}_{n}]\right)-\left(M_{n}^{(2)}+\mathbb{E}[X_{\infty}\mid\mathcal{F}_{n}]\right)

is another Krickeberg decomposition for X_{n}.

We’re now ready to give the proof of the Riesz decomposition, which is similar to that of the Krickeberg decomposition.

Theorem 2. Any supermartingale \left\{(X_{n},\mathcal{F}_{n})\right\}_{n=0}^{\infty} satisfying \inf_{n}\mathbb{E}[X_{n}] > -\infty can be uniquely written

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

where \left\{(M_{n},\mathcal{F}_{n})\right\}_{n=0}^{\infty}is a martingale and \left\{(Z_{n},\mathcal{F}_{n})\right\}_{n=0}^{\infty} is a potential.

Proof. We first prove uniqueness. Suppose M_{n}^{(1)}+Z_{n}^{(1)}=X_{n}=M_{n}^{(2)}+Z_{n}^{(2)} are two Riesz decompositions. Then

\begin{array}{lcl} \displaystyle\mathbb{E}[Z_{n+k}^{(2)}\mid\mathcal{F}_{n}]-\mathbb{E}[Z_{n+k}^{(1)}\mid\mathcal{F}_{n}]&=&\displaystyle\mathbb{E}\left[X_{n+k}-M_{n+k}^{(2)}\mid\mathcal{F}_{n}\right]-\mathbb{E}\left[X_{n+k}-M_{n+k}^{(1)}\mid\mathcal{F}_{n}\right]\\[.7 em]&=&\displaystyle M_{n}^{(1)}-M_{n}^{(2)}\\[.7 em]&=&\displaystyle Z_{n}^{(2)}-Z_{n}^{(1)}\end{array}

The monotone convergence theorem implies that

\displaystyle 0 \leq\mathbb{E}\left[\lim_{k\rightarrow\infty} \mathbb{E}\left[Z_{n+k}^{(i)}\mid\mathcal{F}_{n}\right]\right]\leq\lim_{k\rightarrow\infty}\mathbb{E}[Z_{n+k}^{(i)}]=0

since (Z_{n+k}^{(1)},\mathcal{F}_{n}),(Z_{n+k}^{(2)},\mathcal{F}_{n}) are potentials. Hence, M_{n}^{(2)}-M_{n}^{(1)}=0=Z_{n}^{(1)}-Z_{n}^{(2)} for all n.

We now prove the existence of the Riesz decomposition. Since (X_{n},\mathcal{F}_{n}) is a submartingale, we have for k\geq 0,

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

By induction, we see that

\displaystyle\mathbb{E}\left[X_{n+k+1}\mid\mathcal{F}_{n}\right]\leq\mathbb{E}\left[X_{n+k}\mid\mathcal{F}_{n}\right]\leq X_{n}

By the monotone convergence theorem, M_{n}:=\lim_{k\rightarrow\infty}\mathbb{E}[X_{n+k}\mid\mathcal{F}_{n}] is a \mathcal{F}_{n}-measurable random variable (in the extended sense). Since \inf_{n}\mathbb{E}[X_{n}] > -\infty by hypothesis, we by the monotone convergence theorem that M_{n} is a.s. finite and

\begin{array}{lcl}\displaystyle\mathbb{E}[M_{n}]=\mathbb{E}\left[\lim_{k\rightarrow\infty}\mathbb{E}\left[X_{n+k}\mid\mathcal{F}_{n}\right]\right]=\lim_{k\rightarrow\infty}\mathbb{E}\left[\mathbb{E}[X_{n+k}\mid\mathcal{F}_{n}]\right]&=&\displaystyle\lim_{k\rightarrow\infty}\mathbb{E}[X_{n+k}]\\[.7 em]&=&\displaystyle\inf_{m}\mathbb{E}[X_{m}]\end{array}

To see that (M_{n},\mathcal{F}_{n}) is a martingale, observe that by the conditional monotone convergence theorem,

\begin{array}{lcl}\displaystyle\mathbb{E}[M_{n+1}\mid\mathcal{F}_{n}]=\mathbb{E}\left[\lim_{k\rightarrow\infty}\mathbb{E}\left[X_{n+k+1}\mid\mathcal{F}_{n+1}\right]\mid\mathcal{F}_{n}\right]&=&\displaystyle\lim_{k\rightarrow\infty}\mathbb{E}\left[\mathbb{E}\left[X_{n+k+1}\mid\mathcal{F}_{n+1}\right]\mid\mathcal{F}_{n}\right]\\[.7 em]&=&\displaystyle\lim_{k\rightarrow\infty}\mathbb{E}\left[X_{n+k+1}\mid\mathcal{F}_{n}\right]\\[.7 em]&=&\displaystyle M_{n}\end{array}

We complete the proof by showing that Z_{n}:=X_{n}-M_{n} is a potential. It is evident from above that Z_{n}\geq 0 and

\displaystyle\lim_{n\rightarrow\infty}\mathbb{E}[Z_{n}]=\lim_{n\rightarrow\infty}\mathbb{E}[X_{n}]-\mathbb{E}[M_{n}]=\lim_{n\rightarrow\infty}\mathbb{E}[X_{n}]-\inf_{m}\mathbb{E}[X_{m}]=0

for all n. That (Z_{n},\mathcal{F}_{n}) is a supermartingale is immediate from the martingale property of (M_{n},\mathcal{F}_{n}) and supermartingale property of (X_{n},\mathcal{F}_{n}). \Box

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