advanced34 min studyLesson 666 in Mathematics

Martingale convergence theorems

Distinguish pathwise convergence, preservation of expectations, uniform integrability, and stronger Lp limits.

Before you begin

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The question for today

What you’ll understand

State separate sufficient hypotheses for almost-sure and L1 convergence.
Relate uniformly integrable martingales to integrable terminal closure.
Calculate a martingale-series limit and an explicit failure of expectation preservation.

A martingale can settle down on almost every path while losing expectation in the limit. This is possible because almost-sure convergence does not control the contribution of rare large values. Martingale convergence theory separates three questions: whether a finite limit exists, whether convergence holds in an integrable norm, and whether the limiting variable closes the earlier forecasts. Uniform integrability connects the latter two questions, while boundedness in a stronger Lp space gives a convenient sufficient condition.

Upcrossings force pathwise limits

If a real sequence has distinct lower and upper limits, it repeatedly crosses some interval with rational endpoints. The upcrossing inequality bounds the expected number of such crossings for a submartingale in terms of an integrability bound. For a discrete martingale bounded in L1, this implies almost-sure convergence to a finite integrable limit. A nonnegative supermartingale also converges almost surely to a finite limit because its expectations are bounded by its initial expectation and it has the required one-sided control.

These conclusions concern the existence of a limit along paths. They do not claim that the expectations of the limit equal the limiting expectations of the process. Fatou gives integrability and, for nonnegative variables, an upper bound on the limit's mean. Equality can fail when probability mass moves into increasingly rare events with increasingly large values. A uniform bound on first moments is therefore enough for an almost-sure martingale limit but is not by itself enough to recover the original means from that limit.

\sup_n\mathbb E|M_n|<\infty\Longrightarrow M_n\to M_\infty\text{ a.s.},\quad M_\infty\in L^1;\qquad \{M_n\}\text{ UI}\Longrightarrow M_n\to M_\infty\text{ in }L^1

M is an integrable discrete-time martingale. Uniform integrability strengthens the bounded-first-moment condition and makes expectations pass to the limit.↔ Scroll wide equations horizontally.

Uniform integrability is equivalent to closure

For a discrete martingale, uniform integrability is equivalent to convergence almost surely and in L1, and is also equivalent to having an integrable closing variable Z with each time-n value equal to its conditional expectation given the time-n information. If L1 convergence holds, conditioning the terminal limit and using contraction recovers the earlier values. Conversely, the family of conditional expectations of a single integrable variable over arbitrary smaller sigma-algebras is uniformly integrable, so closure supplies exactly the needed tail control.

The closing variable need not be measurable for the information eventually observed. In that case the limit is its conditional expectation given the sigma-algebra generated by all finite-time information, not necessarily Z itself. If Z is measurable for that eventual information, the limit equals Z. This distinction prevents treating every forecast process as revealing all hidden randomness. A martingale summarizes an information flow, and even infinitely many observations can leave some independent randomness unobserved.

Stronger moments give stronger convergence

If the martingale has uniformly bounded pth moments for some finite p greater than one, it converges both almost surely and in Lp. One proof combines pathwise convergence with the strong Doob bound: the infinite-time maximum has a finite pth moment, furnishing an integrable dominating variable for the pth power of the error. In particular an L2-bounded martingale has an L2 terminal variable, and its squared increments have summable expected energy by orthogonality.

A series of independent centered increments with square-summable variances provides a useful class of examples. Its partial sums are an L2-bounded martingale, so the theorem gives almost-sure and L2 convergence without requiring absolute convergence. When the coefficients are also absolutely summable and increments bounded, one can check convergence directly and compute a deterministic tail bound as a second verification. These two arguments answer different questions: the martingale theorem applies to more series, while an absolute bound supplies explicit pathwise numerical accuracy.

M_n=\sum_{k=1}^n3^{-k}\xi_k,\qquad \mathbb E M_n^2=\sum_{k=1}^n9^{-k}\le\frac18,\qquad \mathbb E|M_\infty-M_n|^2=\frac{9^{-n}}8

The signs are independent and fair. The last identity uses orthogonality and the exact geometric tail; it is a mean-square error, not its square-root L2 norm.↔ Scroll wide equations horizontally.

Work it through

Two certificates for a series tail

For the displayed sign series, determine the mean, variance, and deterministic absolute bound of its limit. Give both error bounds after the first three increments.

The partial sums have mean zero and bounded second moments, so the L2 convergence theorem gives limit mean zero and variance $\sum_{k\ge1}9^{-k}=1/8$ . The absolute coefficient sum is $\sum_{k\ge1}3^{-k}=1/2$ , bounding every limit by one half in magnitude.
After three increments, the deterministic absolute tail is at most $\sum_{k\ge4}3^{-k}=1/54$ . This holds on every sign sequence, not only almost surely.
The tail mean square is $9^{-3}/8=1/5832$ . Its L2 norm is $1/\sqrt{5832}=1/(54\sqrt2)$ , distinct from the mean-square quantity and smaller than the deterministic bound.

The result

The same limit has a pathwise error certificate and a sharper averaged error scale. Its expectation survives because convergence is in L2 and hence also in L1.

The living notebook

Explore the model

Interactive lab

Interactive lab · probability

Rare large values can retain a mean

Pause and Reset, set both $p=0.25,n=1$ , and assign value four to success and zero to failure on paper; predict weighted mean one. Reset and set both $p=0.01,n=1$ , now assigning value one hundred to success; the mean remains one despite the rarer nonzero bar. The model displays the Bernoulli masses, not the rescaled values or a coupled martingale. This illustrates the tail-control issue without proving convergence. Control changes retain time; Reset clears both controls and clock.

Exact probabilityHighlighted outcomeMean np

Mean · np: 5
Variance · np(1−p): 2.5
Total probability: 1

Success probability p0.5Trials n10

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 forRare values need not contribute little to expectation when their sizes increase at the same time.

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Check your understanding

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Read the checkpoint explanation

A bounded first-moment martingale converges almost surely but may lose mean. Uniform integrability upgrades that convergence to L1. A common deterministic bound is sufficient for UI but is far stronger than its definition.

Your turn

Try it yourself

Observe independent fair coin pairs and let $Z_n=4^n$ if the first n pairs were all HH, and zero otherwise, with $Z_0=1$ . Verify the martingale property, find its almost-sure limit, and test uniform integrability directly.

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

Give me a hint

Given survival through n pairs, the next pair continues survival with probability one quarter. For any fixed cutoff choose n so that the surviving value exceeds it.

Compare with the solution

Conditioned on survival, the next expected value is $(1/4)4^{n+1}=4^n$ ; after failure all future values remain zero. Thus it is a nonnegative mean-one martingale. Survival forever has probability zero, so the limit is zero almost surely. For any cutoff K and n with $4^n>K$ , the tail expectation is $4^n\cdot4^{-n}=1$ . Hence the family is not UI, does not converge in L1, and its almost-sure zero limit has smaller expectation.

Keep the idea

Take it with you

Bounded first moments can give a pathwise limit without preserving means.
Uniform integrability, L1 convergence, and integrable closure are equivalent for martingales.
A finite p-moment bound with p greater than one yields Lp convergence.

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