advanced36 min studyLesson 669 in Mathematics

Martingale central limit theorems

Replace independence by conditional centering while retaining variance stabilization and a conditional small-jump hypothesis.

Before you begin

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The question for today

What you’ll understand

State a scalar martingale-array CLT with conditional Lindeberg and variance conditions.
Verify those conditions for bounded history-dependent differences.
Distinguish Gaussian limits from large-jump and random-variance failures.

Many asymptotic sums contain successive errors whose distributions depend on earlier observations. Independence is too restrictive, but zero conditional mean still cancels first-order terms in a characteristic-function or smooth-test expansion. A martingale central limit theorem adds two safeguards: the accumulated conditional variance must settle to a deterministic value, and no large jump may carry a nonvanishing share of the squared fluctuations. Those hypotheses are substantive conditions, not conclusions implied by the word martingale.

Normalize each row of differences

Consider a triangular array of real square-integrable martingale differences. Each row has its own finite sequence of sigma-algebras and a finite number of increments. The increments are measurable at their current row time and have conditional mean zero given the preceding row time. Their sum is the row's terminal fluctuation. Allowing the row length and normalization to change accommodates averages, adaptive measurements, and other sequences whose asymptotic scale is not a single fixed finite tree.

The predictable variance of a row is the sum of its conditional second moments. Assume this random quantity converges in probability to a finite positive deterministic variance. Also require that, for every fixed positive jump threshold, the sum of conditional second moments restricted to increments exceeding that threshold converges in probability to zero. Under these hypotheses the row sum converges in distribution to the centered normal law with the stated variance. We invoke this scalar martingale-array CLT; a full proof requires truncation and a conditional replacement argument.

V_n=\sum_{k=1}^{m_n}\mathbb E[d_{n,k}^2\mid\mathcal F_{n,k-1}]\xrightarrow{\mathbb P}\sigma^2>0,\qquad \sum_{k=1}^{m_n}\mathbb E[d_{n,k}^2\mathbf1_{\{|d_{n,k}|>\varepsilon\}}\mid\mathcal F_{n,k-1}]\xrightarrow{\mathbb P}0\quad(\varepsilon>0)

Each row is a square-integrable martingale difference sequence. The variance limit is deterministic, and the second displayed convergence is the conditional Lindeberg condition.↔ Scroll wide equations horizontally.

Why conditional quantities enter

A Taylor expansion of a smooth test function has a linear term whose conditional expectation vanishes for a martingale difference. Its quadratic term depends on the conditional second moment, which can vary with the past. Lindeberg's condition controls the Taylor remainder by excluding a significant total contribution from large increments. Thus the mechanism resembles the independent CLT but uses conditional rather than unconditional local moments. A zero unconditional mean cannot replace the first step, and matching one unconditional terminal variance cannot replace stabilization of the conditional variance.

For instance, choose an initial random amplitude, equal to one or two with equal probability, and multiply every subsequent independent fair sign by that same amplitude. Dividing the sum by the square root of its length gives a mixture of centered Gaussian limits with variances one and four, rather than a single Gaussian of variance five halves. The predictable row variance retains the initial random amplitude squared. Its unconditional average stabilizes, but it does not converge in probability to the required deterministic constant.

\sum_{k=1}^{m_n}d_{n,k}\ \Rightarrow\ N(0,\sigma^2),\qquad |D_k|\le C\Longrightarrow \frac1n\sum_{k=1}^n\mathbb E[D_k^2\mathbf1_{\{|D_k|>\varepsilon\sqrt n\}}\mid\mathcal F_{k-1}]=0\ \text{for sufficiently large }n

The normal conclusion uses both array hypotheses. The bounded-difference observation verifies conditional Lindeberg for normalized increments Dk divided by square root n; it does not by itself verify variance convergence.↔ Scroll wide equations horizontally.

Small jumps and deterministic variance are separate

A row containing one fair sign of size one and otherwise zeros has predictable variance one for every row, but its sum remains a two-point law. The conditional Lindeberg quantity is one for any threshold below one, so that hypothesis detects the failure. Conversely a uniform bound shrinking with row length makes Lindeberg straightforward, but a random variance limit can still produce mixtures. Both checks must be performed on the proposed normalization rather than inferred from a finite variance or a large number of summands.

The theorem gives weak convergence of the terminal sum, not almost-sure convergence and not a finite-sample error rate. A functional CLT for a whole time-indexed path adds hypotheses about accumulated variance at intermediate times and tightness. Quantitative normal-approximation bounds require additional information as well. These distinctions matter when interpreting a numerical histogram: a short finite example can verify algebraic variance calculations, while an asymptotic theorem is justified by checking its hypotheses across the entire sequence of problems.

Work it through

A bounded adaptive-variance sequence

Let fair signs be independent, set $D_1=\xi_1$ , and for $k\ge2$ set $D_k=H_k\xi_k$ , with $H_k=1$ if $\xi_{k-1}=1$ and $H_k=2$ otherwise. Find the limit of $\sum_{k=1}^nD_k/\sqrt n$ .

The coefficient is known before step k, so every Dk is conditionally centered. Its conditional variance is $H_k^2$ , equal to one or four according to the preceding sign.
The law of large numbers for the independent preceding signs gives $(1+\sum_{k=2}^nH_k^2)/n\to(1+4)/2=5/2$ almost surely. Also $|D_k|\le2$ , so every conditional Lindeberg term is zero once $\varepsilon\sqrt n>2$ .
The CLT therefore gives $N(0,5/2)$ . For a finite check at n equal to two, the unnormalized sum has values $2,0,1,-3$ , giving mean zero and second moment $7/2$ ; the normalized second moment is $7/4$ , not yet its asymptotic value.

The result

History-dependent magnitudes are compatible with a Gaussian limit because the conditional variance averages to a constant and the bounded jumps become negligible under the square-root normalization.

The living notebook

Explore the model

Interactive lab

Interactive lab · sampling

An independent special case of the CLT

Pause and Reset, set both $p=0.5,n=25$ , and predict sample-mean variance $0.01$ . Reset and set both $p=0.5,n=100$ , predicting variance $0.0025$ . Standardize on paper as $2\sqrt n(K/n-1/2)$ to obtain variance one. The lab compares exact independent-binomial masses with normal cell probabilities and highlights exact outcomes; it neither samples an adaptive martingale nor verifies its conditional variance hypotheses. Edits retain time; Reset restores both controls and clock.

Exact binomial massNormal interval massHighlighted outcome / mean

Mean sample proportion: 0.5
Sample-mean variance: 0.0125
Uniform CDF error bound: ≤ 0.1062

Success probability p0.5Independent samples n20

t = 0 s

Exact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). 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 forNormal approximations for independent averages illustrate the scale, while dependent martingale limits require their own conditional checks.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Conditional centering cancels first-order drift, variance convergence fixes the Gaussian scale, and Lindeberg prevents a few large jumps from retaining non-Gaussian structure. None of the listed weaker conditions supplies all three roles.

Your turn

Try it yourself

Let $D_k=a_k\xi_k$ with independent fair signs and deterministic amplitudes alternating one and three. Find the limiting law of the sum divided by $\sqrt n$ , and compute the normalized variance at n equal to four.

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

Give me a hint

Average the squared amplitudes and use their uniform bound to check the large-jump condition.

Compare with the solution

The predictable variance divided by n tends to $(1^2+3^2)/2=5$ . The differences are bounded by three, so normalized Lindeberg terms vanish for large n at each fixed threshold. Therefore the limit is $N(0,5)$ . At n equal to four the sum's variance is $1+9+1+9=20$ , giving normalized variance five already. This finite equality checks the scale but does not make the four-step distribution exactly Gaussian.

Keep the idea

Take it with you

Martingale CLTs use conditional centering and conditional variance.
Lindeberg excludes significant large jumps after normalization.
A weak Gaussian limit requires separate hypotheses from path convergence or an error rate.

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