advanced32 min studyLesson 663 in Mathematics

Martingale differences and Doob decomposition

Separate an integrable adapted process into its predictable conditional drift and an unpredictable martingale component.

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What you’ll understand

  • Center increments by their previous-information conditional means.
  • Construct and prove uniqueness of the discrete Doob decomposition.
  • Identify when the predictable component is increasing and when it can change direction.

An adapted process can contain both systematic conditional drift and genuinely mean-zero new information. Subtracting an unconditional mean does not generally separate them, because the expected next change can depend on the history. The discrete Doob decomposition subtracts that conditional expectation one increment at a time. We use the convention that the predictable part starts at zero and the martingale part starts at the original initial value. Under this convention the decomposition is unique up to almost-sure equality.

Differences isolate new information

A martingale difference sequence consists of integrable time-k measurable variables whose conditional means given time k minus one are zero. Summing these differences from an integrable initial value gives a martingale. Conversely, the increments of every martingale form such a sequence. The differences need not be independent or identically distributed; their conditional distributions may change with the past. Zero unconditional mean follows by taking expectations, but zero unconditional mean alone does not establish the stronger conditional property.

When differences are square integrable, distinct increments are orthogonal in L2. For an earlier increment and a later one, condition their product on the information immediately before the later increment. The earlier factor is measurable there and the later factor has zero conditional mean. Square integrability justifies the product's integrability by Cauchy–Schwarz. Hence the variance of a sum of martingale differences is the sum of their variances, even though independence may fail. Conditional variances can still be random and will later form the predictable bracket.

dk=ΔXk−E[ΔXk∣Fk−1],E[dk∣Fk−1]=0,E[djdk]=0 (j<k, dj,dk∈L2)d_k=\Delta X_k-\mathbb E[\Delta X_k\mid\mathcal F_{k-1}],\qquad \mathbb E[d_k\mid\mathcal F_{k-1}]=0,\qquad \mathbb E[d_jd_k]=0\ (j<k,\ d_j,d_k\in L^2)
X is integrable and adapted. Conditional centering produces martingale differences without assuming independent increments.↔ Scroll wide equations horizontally.

Accumulate the drift before each increment

For an integrable adapted process X, define the predictable component A by summing the conditional means of its increments. At time n this sum is measurable for the information at n minus one, because each summand uses no later information. Set the martingale component to X minus A. Its next increment is precisely the conditionally centered difference, so it is a martingale. All terms are integrable at each finite time. No square-moment hypothesis is needed for this construction.

To see uniqueness, compare two decompositions with predictable components starting at zero. The difference of their martingale components is also predictable and begins at zero. Its time-k increment is known at time k minus one, while the martingale property says its conditional mean is zero. A known integrable variable equals its own conditional expectation, so every such increment is zero. This argument identifies exactly why predictability, rather than merely adaptedness, is needed for a canonical decomposition.

A0=0,An=∑k=1nE[Xk−Xk−1∣Fk−1],Mn=Xn−An,M0=X0A_0=0,\qquad A_n=\sum_{k=1}^n\mathbb E[X_k-X_{k-1}\mid\mathcal F_{k-1}],\qquad M_n=X_n-A_n,\qquad M_0=X_0
The convention here is X = M + A with A0 = 0. Some texts instead write X = X0 + a zero-start martingale + A; the two conventions differ only in the initial term.↔ Scroll wide equations horizontally.

Submartingale drift has a sign

If X is a submartingale, each conditional drift increment is nonnegative, so A is increasing. If X is a supermartingale, the same convention makes A decreasing. For a general integrable adapted process, A may change direction, and describing it as increasing would be false. The decomposition is algebraically available in all three cases; the submartingale inequality is the extra condition that makes its drift monotone. This sign distinction is useful when deciding whether a proposed compensation has the correct orientation.

For a fair random walk, its square has one unit of expected increase per step, so subtracting time produces a martingale. A cubic function has a drift proportional to the present position, whose sign changes. Such examples show that a nonlinear function of a martingale need not be a martingale itself. Convex functions give submartingales when integrability permits conditional Jensen, while other functions can create drift without a fixed sign. Compensation must be calculated from the actual conditional increment rather than guessed from an ordinary derivative.

Compensate a shifted square

Let SnS_n be a fair sign random walk with S0=0S_0=0, and set Xn=(Sn+2)2X_n=(S_n+2)^2. Find its Doob decomposition and check the martingale through two steps.

  1. Expand the next increment: Xk−Xk−1=2(Sk−1+2)ξk+1X_k-X_{k-1}=2(S_{k-1}+2)\xi_k+1. Its previous-information conditional mean is one, so An=nA_n=n and Mn=(Sn+2)2−nM_n=(S_n+2)^2-n.

  2. Initially M0=4M_0=4. At time one the two possible values are eight and zero. At time two, in path order (+,+),(+,−),(−,+),(−,−)(+,+),(+,-),(-,+),(-,-), the values are 14,2,2,−214,2,2,-2.

  3. Conditional averages of the two time-two children are (14+2)/2=8(14+2)/2=8 and (2−2)/2=0(2-2)/2=0, matching time one. The unconditional mean remains (14+2+2−2)/4=4(14+2+2-2)/4=4.

The result

The shifted square is a submartingale with predictable increasing drift n. The compensated process retains initial mean four and satisfies the conditional, not merely unconditional, equality.

Explore the model

Interactive lab

Interactive lab · probability

Recover the compensated mean from exact endpoints

Pause and Reset; set both p=0.5,n=2p=0.5,n=2, predicting count masses 1/4,1/2,1/41/4,1/2,1/4. Convert the count K to S2=2K−2S_2=2K-2 and then to (S2+2)2−2(S_2+2)^2-2; the transformed values are −2,2,14-2,2,14 and their weighted mean is four. Reset and set both p=0.5,n=1p=0.5,n=1 to compare transformed values 0,80,8, again averaging four. The lab shows endpoint laws rather than conditional tree branches. Reset clears both controls and clock; edits alone retain time.

Recover the compensated mean from exact endpointsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forMatching endpoint means is a useful check, while the martingale claim still requires conditional averages along every branch.

Check your understanding

Under X = M + A with A0 = 0, which formula gives the predictable increment of A?

Try it yourself

For the same fair random walk, find the Doob decomposition of Xn=Sn3X_n=S_n^3. Is its predictable component necessarily increasing?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Expand the cube of S plus the next sign and use the first three moments of a fair sign.

Compare with the solution

The conditional increment mean is E[3Sk−12ξk+3Sk−1ξk2+ξk3∣Fk−1]=3Sk−1\mathbb E[3S_{k-1}^2\xi_k+3S_{k-1}\xi_k^2+\xi_k^3\mid\mathcal F_{k-1}]=3S_{k-1}. Therefore An=3∑k=1nSk−1A_n=3\sum_{k=1}^nS_{k-1} and Mn=Sn3−AnM_n=S_n^3-A_n, with both initial values zero. The drift can increase after a positive position and decrease after a negative one; the cube is not a globally convex function on the real line and this process is not generally a submartingale.

Take it with you

  • Conditional centering produces differences with zero predictable drift.
  • The zero-start predictable component fixes the decomposition convention.
  • Only a submartingale assumption guarantees increasing drift under X = M + A.

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