advanced28 min studyLesson 198 in Mathematics

Laws of large numbers

Explain why averages stabilize, distinguish weak and strong laws, and identify when dependence or nonintegrability defeats averaging.

On this page

What you’ll understand

  • Derive a finite-variance weak law from the variance of an average.
  • Distinguish probability convergence from the pathwise conclusion of the strong law.
  • Check independence and integrability before treating a long average as an expectation.

A long average is often used as an estimate of an expectation, but length alone does not make the estimate reliable. Repeating the same underlying random outcome a million times adds no new information. The laws of large numbers explain stabilization under explicit assumptions about the variables being averaged. They also distinguish two claims: the probability of a bad average becomes small at a large fixed sample size, or almost every infinite sequence of observations eventually stays accurate. These statements answer different questions about repeated measurement.

A weak law from one variance calculation

Suppose the observations are independent and identically distributed, with mean μ and finite variance σ². Linearity of expectation makes their average unbiased. Independence removes the cross-covariances, leaving variance σ² divided by the sample size. Chebyshev’s inequality then bounds the probability of an error exceeding any fixed tolerance. The bound tends to zero, proving convergence in probability. Pairwise uncorrelated observations with a common mean μ and a common finite variance would already suffice for this particular calculation.

The bound is a guarantee rather than an exact tail probability, and can be conservative. It shows the characteristic square-root relation: reducing the root-mean-square error by a factor of two requires four times as many independent observations. The numerator contains variance, so this proof cannot be used unchanged when the variance is infinite. That limitation belongs to the proof; a weak law can still hold for independent identically distributed integrable variables with infinite variance.

X‾n=1n∑j=1nXj,Var⁡(X‾n)=σ2n,P(∣X‾n−μ∣≥ε)≤σ2nε2.\overline X_n=\frac1n\sum_{j=1}^nX_j,\qquad \operatorname{Var}(\overline X_n)=\frac{\sigma^2}{n},\qquad \mathbb P(|\overline X_n-\mu|\geq\varepsilon)\leq\frac{\sigma^2}{n\varepsilon^2}.
The displayed variance and probability calculation assumes finite variance and vanishing cross-covariances.↔ Scroll wide equations horizontally.

Why a pathwise law needs a stronger argument

The strong law for independent identically distributed real variables assumes finite expected absolute value and concludes almost sure convergence of the averages to the mean. A finite variance is sufficient but unnecessary. Almost sure convergence means that nearly every complete observation sequence eventually stays within every fixed tolerance. There is no single deterministic sample size after which all possible sequences become accurate, and no assertion that every finite prefix improves on the one before it.

Even in the finite-variance case, summing the preceding Chebyshev bounds over every sample size gives a divergent harmonic series, so the first Borel–Cantelli lemma cannot finish the proof directly. One route uses Kolmogorov’s maximal inequality at dyadic sample sizes. For centered partial sums, the probability that any of the first 2^m sums exceeds ε2^m is at most σ²/(ε²2^m), which is summable in m. Borel–Cantelli then controls whole dyadic blocks. Extending the theorem to merely integrable observations requires a truncation argument.

E∣X1∣<∞,Xj i.i.d.⟹X‾n→a.s.EX1.\mathbb E|X_1|<\infty,\quad X_j\ \text{i.i.d.}\quad\Longrightarrow\quad \overline X_n\xrightarrow{\mathrm{a.s.}}\mathbb EX_1.
This is the strong law. The maximal-inequality sketch explains a finite-variance case, while the stated integrable case is more general.↔ Scroll wide equations horizontally.

Audit the data-generating assumptions

If every observation equals one shared Bernoulli variable Y, the average equals Y at every sample size. Its variance never decreases. Each marginal distribution is still identical, but the observations are perfectly dependent. Some dependent sequences obey other laws of large numbers, for example under suitable stationarity and ergodicity assumptions, but those are additional theorems rather than consequences of the independent-observation calculation.

An expectation must also exist in the relevant sense. A symmetric Cauchy law has no finite expected absolute value and no ordinary mean; a symmetry argument does not create one. Averages of independent standard Cauchy variables remain standard Cauchy, so the usual stabilization claim fails. In applications, inspect what constitutes an independent observation, whether measurements share a common shock, and whether extreme values are controlled. A precise theorem cannot repair a misspecified sampling model.

Xj=Y for every j⟹X‾n=Y,Var⁡(X‾n)=Var⁡(Y).X_j=Y\ \text{for every }j\quad\Longrightarrow\quad \overline X_n=Y,\qquad \operatorname{Var}(\overline X_n)=\operatorname{Var}(Y).
Identical marginal laws do not replace the independence assumption needed for the elementary variance reduction.↔ Scroll wide equations horizontally.

Choose a conservative sample size

Independent Bernoulli observations have success probability 0.3. Use Chebyshev’s inequality to guarantee that the sample proportion differs from 0.3 by at least 0.1 with probability at most 0.01.

  1. The Bernoulli mean is 0.3 and its variance is 0.3 × 0.7 = 0.21. For n independent observations, the sample proportion has variance 0.21/n.

  2. Chebyshev bounds the failure probability by 0.21/(n × 0.1²) = 21/n. Requiring 21/n ≤ 0.01 yields n ≥ 2100.

  3. At n = 2100 the success count has mean 630 and variance 441. Dividing its standard deviation 21 by 2100 gives a proportion standard deviation of 0.01, consistent with the variance calculation.

The result

A sample size of 2100 is sufficient under the stated model. This is a conservative bound; it does not claim the exact binomial failure probability is one percent or that this is the smallest adequate sample size.

Explore the model

Interactive lab

Interactive lab · probability

Watch a finite family of average laws concentrate

Hold p = 0.30 and compare n = 5, 15, and 30. Predict the mean success count np and then divide each count variance by n². The corresponding proportion variances are 0.042, 0.014, and 0.007.

Watch a finite family of average laws concentrateThese 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 count bars are exact finite-n binomial probabilities and retain an integer count axis. Interpreting k/n connects them to sample proportions; the lab neither draws an infinite observation sequence nor proves a strong law.

Check your understanding

For iid real observations, which assumption suffices for the usual strong law of large numbers?

Try it yourself

A sensor reports Xi = Y + εi, where Y is one shared random offset with mean zero and positive finite variance. The εi are iid, integrable, independent of Y, and have mean μ. What is the almost sure limit of the average, and why is it generally not μ?

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

Give me a hint

Separate the average into the unchanged common offset and the average of the independent noise terms.

Compare with the solution

The sample average equals Y + (ε1 + ⋯ + εn)/n. The strong law applies to the iid noise, so its average tends almost surely to μ. Adding the same finite random variable Y at every index gives an almost sure limit Y + μ. Because Y has positive variance, it is not almost surely zero. Collecting more readings removes the independent noise variation but does not remove the shared offset. Treating these readings as independent observations would therefore give the wrong limiting target.

Take it with you

  • The weak law controls error probabilities at a fixed large sample size; the strong law controls almost every entire sequence.
  • Variance reduction proves a useful weak law, while absolute integrability supports the full iid strong law.
  • Repeated measurements only average away uncertainty that the sampling assumptions actually allow to vary.

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