Steven R. Dunbar
Department of Mathematics
203 Avery Hall
University of Nebraska-Lincoln
Lincoln, NE 68588-0130
http://www.math.unl.edu
Voice: 402-472-3731
Fax: 402-472-8466

Math 489/Math 889
Stochastic Processes and
Advanced Mathematical Finance
Dunbar, Fall 2009

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Laws of Large Numbers

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Rating

Mathematically Mature: may contain mathematics beyond calculus with proofs.

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QuestionofDay

Question of the Day

Consider a fair ( $p = 1 ∕ 2 = q$ ) coin tossing game carried out for 1000 tosses. Explain in a sentence what the “law of averages” says about the outcomes of this game.

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Key Concepts

The precise statement, meaning and proof of the Weak Law of Large Numbers.
The precise statement and meaning of the Strong Law of Large Numbers.

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Vocabulary

The Weak Law of Large Numbers is a precise mathematical statement of what is usually loosely referred to as the “law of averages”. Precisely, let $X_{1}, \dots, X_{n}$ be independent, identically distributed random variables each with mean $μ$ and variance $σ^{2}$ . Let $S_{n} = X_{1} + \dots + X_{n}$ and consider the sample mean or more loosely, the “average” $S_{n} ∕ n$ . Then the Weak Law of Large Numbers says that the sample mean $S_{n} ∕ n$ converges in probability to the population mean $μ$ . That is: $lim_{n \to \infty} ℙ n [| S_{n} ∕ n - μ | > ϵ] = 0$
In words, the proportion of those samples whose sample mean differs significantly from the population mean diminishes to zero as the sample size increases.
The Strong Law of Large Numbers says that $S_{n} ∕ n$ converges to $μ$ with probability $1$ . That is: $ℙ [lim_{n \to \infty} S_{n} ∕ n = μ] = 1$
In words, the Strong Law of Large Numbers “almost every” sample mean approaches the population mean as the sample size increases.

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Mathematical Ideas

The Weak Law of Large Numbers

Lemma 1 (Markov’s Inequality). If $X$ is a random variable that takes only nonnegative values, then for any $a > 0$ :

ℙ [X \geq a] \leq E [X] ∕ a

Proof. Here is a proof for the case where $X$ is a continuous random variable with probability density $f$ :

\begin{array}{l} E [X] & = \int_{0}^{\infty} x f (x) d x \\ = \int_{0}^{a} x f (x) d x + \int_{a}^{\infty} x f (x) d x \\ \geq \int_{a}^{\infty} x f (x) d x \\ \geq \int_{a}^{\infty} a f (x) d x \\ = a \int_{a}^{\infty} f (x) d x \\ = a ℙ [X \geq a] . \end{array}

(The proof for the case where $X$ is a purely discrete random variable is similar with summations replacing integrals. The proof for the general case is exactly as given with $d F (x)$ replacing $f (x) d x$ and interpreting the integrals as Riemann-Stieltjes integrals.) □

Lemma 2 (Chebyshev’s Inequality). If $X$ is a random variable with finite mean $μ$ and variance $σ^{2}$ , then for any value $k > 0$ :

ℙ [| X - μ | \geq k] \leq σ^{2} ∕ k^{2} .

Proof. Since ${(X - μ)}^{2}$ is a nonnegative random variable, we can apply Markov’s inequality (with $a = k^{2}$ ) to obtain

ℙ [{(X - μ)}^{2} \geq k^{2}] \leq E [{(X - μ)}^{2}] ∕ k^{2} .

But since ${(X - μ)}^{2} \geq k^{2}$ if and only if $| X - μ | \geq k$ , the inequality above is equivalent to:

ℙ [| X - μ | \geq k] \leq σ^{2} ∕ k^{2}

and the proof is complete. □

Theorem 3 (Weak Law of Large Numbers). Let $X_{1}, X_{2}, X_{3}, \dots,$ be independent, identically distributed random variables each with mean $μ$ and variance $σ^{2}$ . Let $S_{n} = X_{1} + \dots + X_{n}$ . Then $S_{n} ∕ n$ converges in probability to $μ$ . That is:

lim_{n \to \infty} ℙ [| S_{n} ∕ n - μ | > ϵ] = 0 .

Proof. Since the mean of a sum of random variables is the sum of the means, and scalars factor out of expectations:

E [S_{n} ∕ n] = (1 ∕ n) \sum_{i = 1}^{n} E [X_{i}] = (1 ∕ n) (n μ) = μ .

Since the variance of a sum of independent random variables is the sum of the variances, and scalars factor out of variances as squares:

Var [S_{n} ∕ n] = (1 ∕ n^{2}) \sum_{i}^{n} Var [X_{i}] = (1 ∕ n^{2}) (n σ^{2}) = σ^{2} ∕ n .

Fix a value $ϵ > 0$ . Then using elementary definitions for probability measure and Chebyshev’s Inequality:

0 \leq ℙ n [| S_{n} ∕ n - μ | > ϵ] \leq ℙ n [| S_{n} ∕ n - μ | \geq ϵ] \leq σ^{2} ∕ (n ϵ^{2}) .

Then by the squeeze theorem for limits

lim_{n \to \infty} ℙ [| S_{n} ∕ n - μ | > ϵ] = 0 .

□

Jacob Bernoulli originally proved the Weak Law of Large Numbers in 1713 for the special case when the $X_{i}$ are binomial random variables. Bernoulli had to create an ingenious proof to establish the result, since Chebyshev’s inequality was not known at the time. The theorem then became known as Bernoulli’s Theorem. Simeon Poisson proved a generalization of Bernoulli’s binomial Weak Law and first called it the Law of Large Numbers. In 1929 the Russian mathematician Aleksandr Khinchin proved the general form of the Weak Law of Large Numbers presented here. Many other versions of the Weak Law are known, with hypotheses that do not require such stringent requirements as being identically distributed, and having finite variance.

The Strong Law of Large Numbers

Theorem 4 (Strong Law of Large Numbers). Let $X_{1}, X_{2}, X_{3}, \dots,$ be independent, identically distributed random variables each with mean $μ$ and variance $E [X_{j}^{2}] < \infty$ . Let $S_{n} = X_{1} + \dots + X_{n}$ . Then $S_{n} ∕ n$ converges with probability $1$ to $μ$ ,

ℙ [lim_{n \to \infty} \frac{S_{n}}{n} = μ] = 1 .

The proof of this theorem is beautiful and deep, but would take us too far afield to prove it. The Russian mathematician Andrey Kolmogorov proved the Strong Law in the generality stated here, culminating a long series of investigations through the first half of the 20th century.

Discussion of the Weak and Strong Laws of Large Numbers

In probability theory a theorem that tells us how a sequence of probabilities converges is called a weak law. For coin tossing, the sequence of probabilities is the sequence of binomial probabilities associated with the first $n$ tosses. The Weak Law of Large Numbers says that if we take $n$ large enough, then the binomial probability of the mean over the first $n$ tosses differing “much” from the theoretical mean should be small. This is what is usually popularly referred to as the law of averages. However, this is a limit statement and the Weak law of Large Numbers above does not indicate the rate of convergence, nor the dependence of the rate of convergence on the difference $ϵ$ . Note furthermore that the Weak Law of Large Numbers in no way justifies the false notion called the “Gambler’s Fallacy”, namely that a long string of successive Heads indicates a Tail “is due to occur soon”. The independence of the random variables completely eliminates that sort of prescience.

A strong law tells how the sequence of random variables as a sample path behaves in the limit. That is, among the infinitely many sequences (or paths) of coin tosses we select one “at random” and then evaluate the sequence of means along that path. The Strong Law of Large Numbers says that with probability $1$ that sequence of means along that path will converge to the theoretical mean. The formulation of the notion of probability on an infinite (in fact an uncountably infinite) sample space requires mathematics beyond the scope of the course, partially accounting for the lack of a proof for the Strong Law here.

Note carefully the difference between the Weak Law of Large Numbers and the Strong Law. We do not simply move the limit inside the probability. These two results express different limits. The Weak Law is a statement that the group of finite-length experiments whose sample mean is close to the population mean approaches all of the possible experiments as the length increases. The Strong Law is an experiment-by-experiment statement, it says (almost every) sequence has a sample mean that approaches the population mean. Weak laws are usually much easier to prove than strong laws.

Mathematical Ideas

This section is adapted from Chapter 8, “Limit Theorems”, A First Course in Probability, by Sheldon Ross, Macmillan, 1976.

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Problems to Solve

Problems to Work for Understanding

Suppose $X$ is a continuous random variable with mean and variance both equal to $20$ . What can be said about $ℙ [0 \leq X \leq 40]$ ?
Suppose X is an exponentially distributed random variable with mean $E [X] = 1$ . For $x = 0.5$ , $1$ , and $2$ , compare $ℙ [X \geq x]$ with the Markov inequality bound.
Suppose $X$ is a Bernoulli random variable with $ℙ [X = 1] = p$ and $ℙ [X = 0] = 1 - p = q$ . Compare $ℙ [X \geq 1]$ with the Markov inequality bound.
Let $X_{1}, X_{2}, \dots, X_{10}$ be independent Poisson random variables with mean $1$ . First use the Markov Inequality to get a bound on $ℙ [X_{1} + \dots + X_{10} > 15]$ . Next find the exact probability that $ℙ [X_{1} + \dots + X_{10} > 15]$ using that the fact that the sum of independent Poisson random variables with parameters $λ_{1}$ , $λ_{2}$ is again Poisson with parameter $λ_{1} + λ_{2}$ .

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Reading Suggestion

Reading Suggestion:

References

[1] Emmanuel Lesigne. Heads or Tails: An Introduction to Limit Theorems in Probability, volume 28 of Student Mathematical Library. American Mathematical Society, 2005.

[2] Sheldon Ross. A First Course in Probability. Macmillan, 1976.

[3] Sheldon M. Ross. Introduction to Probability Models. Elsevier, 8th edition edition, 2003.

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Links

Outside Readings and Links:

Virtual Laboratories in Probability and Statistics.. Search the page for Weak Law and then run the Binomial Coin Experiment and the Matching Experiment.

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