Limit theorems · Definition

Law of large numbers

probabilitylimits

Before you start you need to know

It explains why frequencies and averages stabilize, and so justifies estimating an unknown expectation from data.

Definition

The law of large numbers states that the average \(\bar X_n\) of \(n\) independent observations of a variable \(X\) approaches its expectation \(\mu\) as \(n\) grows.

\[\bar X_n=\dfrac1n\sum_{i=1}^{n}X_i\ \xrightarrow[n\to\infty]{}\ \mu\]

Let \(X_1,X_2,\dots\) be independent, with the same expectation \(\mu\) and finite variance \(\sigma^2\). The (weak) law says that, for every \(\varepsilon>0\), \(P(|\bar X_n-\mu|\ge\varepsilon)\to0\) as \(n\to\infty\). The strong form says more: with probability 1, the sequence of averages converges to \(\mu\).

With finite variance the proof is a direct application of Chebyshev: since \(E[\bar X_n]=\mu\) and \(\operatorname{Var}(\bar X_n)=\sigma^2/n\), we get \(P(|\bar X_n-\mu|\ge\varepsilon)\le\sigma^2/(n\varepsilon^2)\), which tends to 0. It also gives an explicit bound: to guarantee an error below \(\varepsilon\) with failure probability \(\delta\), it suffices that \(n\ge\sigma^2/(\delta\varepsilon^2)\).

A guiding example is the relative frequency of an event: if \(X_i=1\) when the event occurs in repetition \(i\), then \(\bar X_n\) is the proportion of times it occurs and tends to its probability. This justifies reading a probability as a long-run frequency, and also simulation and estimation methods.

Two typical errors: believing the law «compensates» past streaks (trials are independent) and thinking the absolute difference \(S_n-n\mu\) vanishes, when what stabilizes is the average. The speed and shape of the remaining fluctuation are described by the central limit theorem.

Think about it

Why does the law not contradict the fact that, after 1000 tosses, the number of heads is almost never exactly 500?

Because it only ensures the proportion approaches \(\tfrac12\); the absolute difference from 500 can be several units while the proportion is still very close to 0.5.

With \(\sigma^2=4\), how many observations does Chebyshev guarantee for an error below 0.2 with failure at most 0.1?

\(n\ge\sigma^2/(\delta\varepsilon^2)=4/(0.1\cdot0.04)=1000\).

Exercises 10

What comes next

References

  • Joseph K. Blitzstein, Jessica Hwang, Introduction to Probability, 2nd edition, CRC Press, 2019. §10.1
  • Sheldon Ross, A First Course in Probability, 10th edition, Pearson, 2019. §8.2
  • Ronald E. Walpole, Raymond H. Myers, Sharon L. Myers, Keying Ye, Probabilidad y estadística para ingeniería y ciencias, 9.ª edición, Pearson, 2012. §8.5