Doob decomposition theorem

Proof

Existence

Using conditional expectations, define the processes A and M, for every n ∈ I, explicitly by

and

where the sums for n = 0 are empty and defined as zero. Here A adds up the expected increments of X, and M adds up the surprises, i.e., the part of every X_k that is not known one time step before. Due to these definitions, A_n+1 (if n + 1 ∈ I) and M_n are F_n-measurable because the process X is adapted, E[|A_n|] < ∞ and E[|M_n|] < ∞ because the process X is integrable, and the decomposition X_n = M_n + A_n is valid for every n ∈ I. The martingale property

${\displaystyle \mathbb{E}[M_n-M_{n-1}|{\mathcal{F}}_{n-1}]=0}$ a.s.

also follows from the above definition (2), for every n ∈ I \ {0}.

Uniqueness

To prove uniqueness, let X = M' + A' be an additional decomposition. Then the process Y := M − M' = A' − A is a martingale, implying that

${\displaystyle \mathbb{E}[Y_n|{\mathcal{F}}_{n-1}]=Y_{n-1}}$ a.s.,

and also predictable, implying that

${\displaystyle \mathbb{E}[Y_n|{\mathcal{F}}_{n-1}]=Y_n}$ a.s.

for any n ∈ I \ {0}. Since Y₀ = A'₀ − A₀ = 0 by the convention about the starting point of the predictable processes, this implies iteratively that Y_n = 0 almost surely for all n ∈ I, hence the decomposition is almost surely unique.

Corollary

A real-valued stochastic process X is a submartingale if and only if it has a Doob decomposition into a martingale M and an integrable predictable process A that is almost surely increasing. It is a supermartingale, if and only if A is almost surely decreasing.

Proof

If X is a submartingale, then

${\displaystyle \mathbb{E}[X_k|{\mathcal{F}}_{k-1}]\ge X_{k-1}}$ a.s.

for all k ∈ I \ {0}, which is equivalent to saying that every term in definition (1) of A is almost surely positive, hence A is almost surely increasing. The equivalence for supermartingales is proved similarly.

Example

Let X = (X_n)_{n∈${\displaystyle {\mathbb{N}}_0}$} be a sequence in independent, integrable, real-valued random variables. They are adapted to the filtration generated by the sequence, i.e. F_n = σ(X₀, . . . , X_n) for all n ∈ ${\displaystyle {\mathbb{N}}_0}$. By (1) and (2), the Doob decomposition is given by

${\displaystyle A_n=\sum _{k=1}^n(\mathbb{E}[X_k]-X_{k-1}),n\in {\mathbb{N}}_0,}$

and

${\displaystyle M_n=X_0+\sum _{k=1}^n(X_k-\mathbb{E}[X_k]),n\in {\mathbb{N}}_0.}$

If the random variables of the original sequence X have mean zero, this simplifies to

${\displaystyle A_n=-\sum _{k=0}^{n-1}{X}_k}$ and ${\displaystyle M_n=\sum _{k=0}^n{X}_k,n\in {\mathbb{N}}_0,}$

hence both processes are (possibly time-inhomogeneous) random walks. If the sequence X = (X_n)_{n∈${\displaystyle {\mathbb{N}}_0}$} consists of symmetric random variables taking the values +1 and −1, then X is bounded, but the martingale M and the predictable process A are unbounded simple random walks (and not uniformly integrable), and Doob's optional stopping theorem might not be applicable to the martingale M unless the stopping time has a finite expectation.

Application

In mathematical finance, the Doob decomposition theorem can be used to determine the largest optimal exercise time of an American option. Let X = (X₀, X₁, . . . , X_N) denote the non-negative, discounted payoffs of an American option in a N-period financial market model, adapted to a filtration (F₀, F₁, . . . , F_N), and let ${\displaystyle \mathbb{Q}}$ denote an equivalent martingale measure. Let U = (U₀, U₁, . . . , U_N) denote the Snell envelope of X with respect to ${\displaystyle \mathbb{Q}}$. The Snell envelope is the smallest ${\displaystyle \mathbb{Q}}$-supermartingale dominating X and in a complete financial market it represents the minimal amount of capital necessary to hedge the American option up to maturity. Let U = M + A denote the Doob decomposition with respect to ${\displaystyle \mathbb{Q}}$ of the Snell envelope U into a martingale M = (M₀, M₁, . . . , M_N) and a decreasing predictable process A = (A₀, A₁, . . . , A_N) with A₀ = 0. Then the largest stopping time to exercise the American option in an optimal way is

${\displaystyle {\tau }_{\text{max}}:=\{\begin{array}{ll}N& \text{if }{A}_N=0,\\ min\{n\in \{0,\dots ,N-1\}\mid A_{n+1}<0\}& \text{if }{A}_N<0.\end{array}}$

Since A is predictable, the event {τ_max = n} = {A_n = 0, A_n+1 < 0} is in F_n for every n ∈ {0, 1, . . . , N − 1}, hence τ_max is indeed a stopping time. It gives the last moment before the discounted value of the American option will drop in expectation; up to time τ_max the discounted value process U is a martingale with respect to ${\displaystyle \mathbb{Q}}$.

Generalization

The Doob decomposition theorem can be generalized from probability spaces to σ-finite measure spaces.