advanced34 min studyLesson 200 in Mathematics

Central limit theorems and tightness

Explain Gaussian fluctuations, the role of normalization and characteristic functions, and why tightness alone does not imply a central limit theorem.

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What you’ll understand

  • Normalize sums correctly and distinguish a central limit theorem from a law of large numbers.
  • Connect the quadratic expansion of a characteristic function to Gaussian limits.
  • Use tightness and Lindeberg’s condition to diagnose why a proposed Gaussian limit may fail.

A law of large numbers says that an average approaches its mean. A central limit theorem asks what remains visible when we magnify the shrinking fluctuations. For independent identically distributed observations with a finite positive variance, centering the sum and dividing by its standard deviation produces a limiting standard normal law. The Gaussian shape describes the distribution of normalized errors, not the raw data distribution, and does not mean that one sequence of normalized errors converges pointwise to a particular normal random variable.

Keep the centering and the scale visible

Let the observations have mean μ and variance σ² strictly between zero and infinity. Their sum has mean nμ and variance nσ². Subtracting the mean removes the deterministic drift, and dividing by σ√n keeps the variance equal to one. The iid central limit theorem states convergence in distribution to a standard normal variable. The equivalent statement for the sample average multiplies its error by √n/σ. Omitting the square root or forgetting to center changes the proposed limit completely.

This theorem concerns fixed continuity thresholds of the limiting normal law. It does not, from finite variance alone, supply a universal finite-sample error rate, accurate extreme-tail probabilities, or a density approximation for every distribution. Quantitative results such as Berry–Esseen require stronger moment assumptions. At lattice thresholds, a continuity correction can improve a normal approximation, but it remains an approximation whose accuracy must be assessed for the actual sample size and distribution.

Zn=∑j=1nXj−nμσn=n(X‾n−μ)σ⇒N(0,1).Z_n=\frac{\sum_{j=1}^nX_j-n\mu}{\sigma\sqrt n}=\frac{\sqrt n(\overline X_n-\mu)}{\sigma}\quad\Rightarrow\quad N(0,1).
This classical form assumes iid observations and a finite, strictly positive variance.↔ Scroll wide equations horizontally.

A local quadratic expansion controls the global limit

After standardizing one observation to mean zero and variance one, its characteristic function near zero is one minus t²/2 plus a remainder smaller than t². Independence makes the transform of the normalized sum the nth power of the single-observation transform evaluated at t/√n. The quadratic term accumulates to a finite exponent, while the smaller remainder disappears. The resulting transform exp(−t²/2) identifies the standard normal law through the continuity theorem.

Tightness explains another necessary piece of limit reasoning: normalized mass must not escape arbitrarily far. Here Chebyshev gives a uniform tail bound of one over M² because every Zn has variance one. On the real line, tightness ensures that every subsequence has a further weakly convergent subsequence. Identification then matters: the transform argument forces all such limits to be the same Gaussian law. Tightness supplies subsequences, not their Gaussian identity.

φY(u)=1−u22+o(u2),φZn(t)=[φY ⁣(tn)]n⟶e−t2/2,sup⁡nP(∣Zn∣>M)≤M−2.\varphi_Y(u)=1-\frac{u^2}{2}+o(u^2),\qquad\varphi_{Z_n}(t)=\left[\varphi_Y\!\left(\frac t{\sqrt n}\right)\right]^n\longrightarrow e^{-t^2/2},\qquad\sup_n\mathbb P(|Z_n|>M)\leq M^{-2}.
Y is a standardized observation; the second-moment assumption justifies the expansion and the tail bound.↔ Scroll wide equations horizontally.

Many small contributions is a hypothesis, not a picture

A triangular array allows each row to contain different independent centered variables. Normalize each row so its total variance is one. Lindeberg’s sufficient condition asks that, for every positive ε, the total second-moment contribution of terms larger than ε tends to zero. It prevents a few macroscopic jumps from controlling the row sum. Under this condition the row sums converge to a standard normal law, extending the iid theorem to many unequal small contributions.

Unit variance alone does not establish this condition. A row consisting of one symmetric sign and any number of zero variables has total variance one and is tight, but its sum always has a two-point law. The sign remains a full-size contribution as the row grows. Heavy tails, dependence, or a dominant coordinate can similarly defeat a casual Gaussian claim. Diagnose the assumptions before treating a smooth-looking finite histogram as proof of a central limit theorem.

∑kEYn,k2=1,∀ε>0: ∑kE ⁣[Yn,k21{∣Yn,k∣>ε}]⟶0.\sum_k\mathbb EY_{n,k}^2=1,\qquad\forall\varepsilon>0:\ \sum_k\mathbb E\!\left[Y_{n,k}^2\mathbf1_{\{|Y_{n,k}|>\varepsilon\}}\right]\longrightarrow0.
Lindeberg’s condition is stated for variables independent within each row, centered, with the displayed unit total variance normalization.↔ Scroll wide equations horizontally.

Approximate a symmetric binomial window

For 100 independent fair Bernoulli trials, approximate the probability of between 40 and 60 successes inclusive using a normal approximation with continuity correction.

  1. The success count K has mean 50 and variance 25, so its standard deviation is 5. Centering and scaling gives Z = (K − 50)/5.

  2. The inclusive integer window 40 through 60 is approximated by the continuous interval from 39.5 to 60.5. Its standardized endpoints are −2.1 and 2.1.

  3. For a standard normal distribution, the probability between these endpoints is Φ(2.1) − Φ(−2.1) = 2Φ(2.1) − 1, approximately 0.9643. The exact binomial probability is a separate finite sum.

The result

The continuity-corrected normal approximation is about 96.43%. The calculation illustrates a finite-sample approximation rather than an exact equality or a general tail-error guarantee.

Explore the model

Interactive lab

Interactive lab · probability

Compare raw spread with standardized fluctuations

Fix p = 0.50 and compare n = 10, 20, and 30. Predict mean n/2 and count variance n/4. The raw distribution becomes wider; to discuss a central limit theorem, mentally recenter at n/2 and divide count offsets by √n/2.

Compare raw spread with standardized fluctuationsThese 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 forThe bars and moments are exact binomial quantities. The lab does not draw a Gaussian overlay, standardize the displayed axis, estimate approximation error, or simulate convergence. Small finite-n shapes support an intuition whose proof needs the stated hypotheses.

Check your understanding

A sequence of centered real variables has variance one at every index. What follows from this alone?

Try it yourself

For n ≥ 2, let R and R2,…,Rn be independent symmetric signs. Form Tn = R/√2 + (R2 + ⋯ + Rn)/√(2(n−1)). Show that Tn has variance one and is tight, but its limit is not standard normal. Check Lindeberg’s condition at ε = 0.5.

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

Give me a hint

Apply the iid central limit theorem only to the second term. Keep the first sign term intact, and multiply the two limiting characteristic functions.

Compare with the solution

The two independent terms have variances one half and one half, so Tn has variance one; Chebyshev proves tightness. The second term converges to N(0,1/2), independently of the unchanged first sign. Thus the limiting transform is cos(t/√2) exp(−t²/4), describing an equal mixture of normals with means ±1/√2 and variance 1/2. At t = π/√2 this transform is zero, whereas the standard normal transform is positive, proving the laws differ. For sufficiently large n, only the first term exceeds 0.5 in absolute value. Its contribution to the Lindeberg sum is one half, which does not vanish. The dominant sign prevents the sufficient small-contribution condition from holding.

Take it with you

  • A central limit theorem describes normalized fluctuations, while a law of large numbers describes the unscaled average.
  • Tightness prevents escape of probability mass and does not identify a Gaussian limit.
  • Lindeberg’s condition rules out persistent large summands in independent triangular arrays.

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