advanced30 min studyLesson 662 in Mathematics

Filtrations, adaptedness, and predictability

Track when information becomes available and distinguish a valid pre-increment choice from a rule that sees the next outcome.

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What you’ll understand

  • Distinguish adapted from predictable sequences with explicit indexing.
  • Recognize stopping times through information available by each time.
  • Explain why martingale increments can be weighted by predictable choices but not arbitrary current observations.

A stochastic process cannot be called fair without specifying what its observer already knows. The same numerical sequence can be a martingale for one information flow and fail for a larger one that reveals future outcomes. Filtrations record that flow, adaptedness prohibits hidden current values, and predictability requires a coefficient to be chosen before the increment it multiplies. These distinctions are easiest to audit on a finite coin tree, where every event has an explicit list of compatible histories.

An increasing information sequence

A discrete filtration is an increasing sequence of sigma-algebras indexed by nonnegative integers. At time n, its events are those decidable using information available by that time. A process is adapted when its time-n variable is measurable for the time-n sigma-algebra. Integrability is an additional condition, not part of adaptedness. A martingale is adapted and integrable at each time, and its next conditional expectation equals its present value. A submartingale replaces equality with a conditional lower bound on the next value.

For independent fair signs, the natural filtration reveals the first n signs and their partial sum is a martingale. If the initial information instead reveals the first sign, that first increment has nonzero conditional value on each revealed branch. The underlying outcome probabilities have not changed, but the fairness assertion has. Adding information can invalidate a martingale property, whereas replacing a filtration by a smaller one preserves it only when the process remains adapted; the tower property then supplies the required conditional equality.

F0⊂F1⊂⋯ ,Mn∈L1(Fn),E[Mn∣Fn−1]=Mn−1(n≥1)\mathcal F_0\subset\mathcal F_1\subset\cdots,\qquad M_n\in L^1(\mathcal F_n),\qquad \mathbb E[M_n\mid\mathcal F_{n-1}]=M_{n-1}\quad(n\ge1)
All equalities and measurability statements are understood up to the relevant null sets. A filtration is information, while a process is a collection of random variables.↔ Scroll wide equations horizontally.

Predictable means known before the increment

We index a weight H at time k by the increment from k minus one to k that it multiplies. Predictability means that this weight is measurable for the information at k minus one. Its product with a martingale difference therefore has conditional mean zero, provided the product is integrable. A bounded predictable weight always preserves this integrability for an integrable difference. The weight may depend on the whole past, including previous gains, and need not be deterministic or independent of earlier increments.

An adapted coefficient can already use the current sign, so adaptedness alone is too weak for a pre-increment rule. Choosing the weight equal to that same fair sign makes its product with the sign identically one. The positive mean comes from using unavailable information, not from exploiting a flaw in the conditional expectation identity. This simple check is a useful way to detect an indexing error in a stochastic sum before invoking any general theorem about transforms or integration.

Hk is Fk−1-measurable,ΔMk=Mk−Mk−1⟹E[HkΔMk∣Fk−1]=0H_k\text{ is }\mathcal F_{k-1}\text{-measurable},\quad \Delta M_k=M_k-M_{k-1}\Longrightarrow \mathbb E[H_k\Delta M_k\mid\mathcal F_{k-1}]=0
For this displayed implication assume Hk is bounded, or impose integrability conditions that justify taking out the known factor. Later L2 integration uses an explicit energy condition.↔ Scroll wide equations horizontally.

A stopping rule uses no future information

A nonnegative-integer-valued random time, possibly infinite, is a stopping time when the event that it has occurred by n belongs to the information at n. The first occurrence of heads is a stopping time for coin observations. The last heads time before a fixed future horizon usually is not: deciding whether it is already past may require inspecting later tosses. A rule can be a well-defined random variable and still fail this information requirement. Its name or operational interpretation does not establish measurability.

For a stopping time T, the indicator that T is at least k is known at time k minus one, because it is the complement of the event that stopping occurred by that earlier time. This makes the stopped sum a predictable transform of the increments. By contrast, the indicator that T equals k can depend on the newly observed increment. Keeping these two indicators distinct explains why stopping a martingale at a bounded stopping time preserves fairness even though the stopping decision itself can use the current observation.

Audit a two-step predictable rule

Let ξ1,ξ2\xi_1,\xi_2 be independent fair signs. Use H1=2H_1=2 and H2=1+ξ1H_2=1+\xi_1 to weight the two signs. List the total gains and compare with the anticipative choice H~2=ξ2\widetilde H_2=\xi_2.

  1. The first weight is deterministic and the second is measurable after the first sign, taking values two or zero. Thus both satisfy the pre-increment indexing convention.

  2. In the sign order (+,+),(+,−),(−,+),(−,−)(+,+),(+,-),(-,+),(-,-), the total 2ξ1+(1+ξ1)ξ22\xi_1+(1+\xi_1)\xi_2 has values 4,0,−2,−24,0,-2,-2. Their equal-weight mean is zero. On each first-sign branch, the second weighted increment also has mean zero.

  3. For the anticipative coefficient, H~2ξ2=ξ22=1\widetilde H_2\xi_2=\xi_2^2=1 on every path. It is adapted at time two but is not measurable at time one, so the predictable-transform conclusion does not apply.

The result

Past-dependent weights preserve conditional fairness here, while using the next sign as its own weight introduces a deterministic positive increment.

Explore the model

Interactive lab

Interactive lab · probability

A terminal count forgets the observation order

Pause and Reset, set both p=0.5,n=2p=0.5,n=2, and predict count masses 1/4,1/2,1/41/4,1/2,1/4. The middle bar merges histories HT and TH even though their time-one information differs. Reset and set both p=0.5,n=1p=0.5,n=1 to inspect the information available after only one toss. The instrument shows exact binomial laws and highlights outcomes, not a sampled filtration or an ordered coin tree. Control edits retain time; Reset restores both controls and time.

A terminal count forgets the observation orderThese 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 forProcesses and predictable rules depend on observation order, which a final count distribution alone does not retain.

Check your understanding

With Hk multiplying Mk − Mk−1, which measurability condition is predictability?

Try it yourself

For two fair coin tosses, let L be the last heads time, using zero when no heads occurs. Show L is not a stopping time for the natural filtration, and compare the first-heads time truncated at two.

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

Give me a hint

Determine the event L ≤ 1 and ask whether the first observation can distinguish its outcomes.

Compare with the solution

The event {L≤1}\{L\le1\} is exactly that the second toss is tails. It contains HT and TT but excludes HH and TH, so it cannot be decided from the first toss and is not in the time-one sigma-algebra. The truncated first-heads time equals one after a first heads and two otherwise; its event of occurring by one is the first-heads event, and it always occurs by two. Hence the truncated first-heads rule is a stopping time.

Take it with you

  • Martingale fairness is relative to a filtration.
  • A predictable weight is known before its indexed increment.
  • Stopping-time survival indicators are predictable even when the stopping decision uses current data.

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