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# 14 Central limit theorems

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## The case of i.i.d. r.v.'s

We are now ready to prove:

Theorem 14.1: The central limit theorem

Let $$X_1, X_2, \ldots$$ be an i.i.d.\ sequence of r.v.'s with finite variance. Denote $$\mu = \expec X_1$$, $$\sigma = \sigma(X_1)$$ and $$S_n = \sum_{k=0}^n X_k$$. Then as $$n\to\infty \( we have the convergence in distribution $\frac{S_n-n\mu}{\sqrt{n}\sigma} \implies N(0,1).$ Proof For convenience, denote \(\hat{X}_k = (X_k - \mu)/\sigma$$ and $$\hat{S}_n = \sum_{k=0}^n \hat{X}_k$$.
Then

$\varphi_{\hat{S}_n/\sqrt{n}}(t) = \varphi_{\hat{S}_n}(t/\sqrt{n}) = \prod_{k=1}^n \varphi_{\hat{X}_k}(t/\sqrt{n}) = \left(\varphi_{\hat{X}_1}(t/\sqrt{n})\right)^n.$

Note that $$\expec \hat{X}_1 = 0$$ and $$\var(\hat{X}_1)=\expec \hat{X}_1^2 = 1$$. Therefore by Theorem~\ref{thm-charfun-secondmoment}, $$\varphi_{\hat{X}_1}$$ satisfies

$\varphi_{\hat{X}_1}(u) = 1 - \frac{u^2}{2} + o(u^2)$

as $$u\to0$$. It follows that

$\varphi_{\hat{S}_n/\sqrt{n}}(t) = \left(1 - \frac{t^2}{2n} + o\left(\frac{t^2}{n}\right) \right)^n \xrightarrow[n\to\infty]{} e^{-t^2/2}$

for any $$t\in\R$$. Using the continuity theorem (Theorem~\ref{thm-continuity}) and our previous computations, it follows that $$\hat{S}_n \implies N(0,1)$$, as claimed.

## Generalizations

The CLT can be generalized in many ways. None of the assumptions (independence, identical distributions, even finite variance) are entirely necessary. A central paradigm of probability theory is that any random quantity that arises as a sum of many small contributions that are either independent or not too strongly dependent, will converge to the normal distribution in some asymptotic limit. Thousands of examples exist, but there is no single all-encompassing theorem that includes all of them as a special case. Rather, probabilists have a toolbox of tricks and techniques that they try to apply in order to prove normal convergence in any given situation. characteristic functions are among the more useful techniques. Another important technique, the so-called \textbf{moment method}, involves the direct use of moments: If we can show that $$\expec(W_n^k)\to \expec(Z^k)$$, where $$(W_n)_{n=1}^\infty$$ is the (normalized) sequence being studied, and $$Z\sim N(0,1)$$, then by Theorem 3.3.12 in [Dur2010] (p.\ 105) or Theorem (3.12) in [Dur2004] (p.\ 109), that implies that $$W_n\implies N(0,1)$$.

We now discuss several examples of interesting generalizations of CLT.

## Triangular arrays

Theorem 14.2: Lindeberg-Feller CLT for triangular arrays

Let $$(X_{n,k})_{1\le k\le n<\infty}$$ be a triangular array of r.v.'s. Denote $$S_n = \sum_{k=1}^n X_{n,k}$$ (the sum of the $$n$$-th row). Assume that:

1. For each $$n$$, the r.v.'s $$(X_{n,k})_{k=1}^n$$ are independent.
2. $$\expec X_{n,k}=0$$ for all $$n,k$$.
3. $$\var(S_n) = \sigma_n^2 \to \sigma^2<\infty$$ as $$n\to\infty$$.
4. For all $$\epsilon>0$$, $$\lim_{n\to\infty}} \sum_{k=1}^n \expec\left( X_{n,k}^2 \ind_{\{|X_{n,k}|>\epsilon\}}\right)=$$.

Then $$S_n \implies N(0,\sigma^2)$$ as $$n\to\infty$$.

Proof
See [Dur2010], p. 110--111 or [Dur2004], p. 115--116. The proof uses the characteristic function technique and is a straightforward extension of the proof for the i.i.d.\ case.
Example 14.3: Record times and cycles in permutations

Let $$X_1, X_2, \ldots$$ be i.i.d.\ $$U(0,1)$$ r.v.'s. Let $$A_n$$ be the event that $$\{X_n = \max(X_1,\ldots,X_n)\}$$ (in this case, we say that $$n$$ is a \textbf{record time}). Let $$S_n = \sum_{k=1}^n \ind_{A_k}$$ be the number of record times up to time $$n$$. We saw in a homework exercise that the $$A_k$$'s are independent events and $$\prob(A_k) = 1/k$$. This implies that $$\expec(S_n) = \sum_{k=1}^n \frac{1}{k} = H_n$$ (the $$n$$-th \textbf{harmonic number}) and $$\var(S_n) = \sum_{k=1}^n \frac{k-1}{k^2}$$. Note that both $$\expec(S_n)$$ and $$\var(S_n)$$ are approximately equal to $$\log n$$, with an error term that is $$O(1)$$. Now taking $$X_{n,k} = (\ind_{A_k}-k^{-1})/\sqrt{\var(S_n)}$$ in Theorem~\ref{thm-lindeberg-feller}, it is easy to check that the assumptions of the theorem hold. It follows that

$\frac{S_n-H_n}{\sigma(S_n)} \implies N(0,1).$

Equivalently, because of the asymptotic behavior of $$\expec(S_n)$$ and $$\var(S_n)$$ it is also true that

$\frac{S_n-\log n}{\sqrt{\log n}} \implies N(0,1).$

Note: $$S_n$$ describes the distribution of another interesting statistic on random permutations. It is not too difficult to show by induction (using an amusing construction often referred to as the Chinese restaurant process) that if $$\sigma \in S_n$$ is a uniformly random permutation on $$n$$ elements, then the number of cycles in $$\sigma$$ is a random variable which is equal in distribution to $$S_n$$.

## Erdös-Kac Theorem

Theorem 14.4: Erdös-Kac theorem (1940)

Let $$g(m)$$ denote the number of prime divisors of an integer $$k$$ (for example, $$g(28)=2)$$. For each $$n\ge 1$$, let $$X_n$$ be a uniformly random integer chosen in $$\{1,2,\ldots,n\}$$, and let $$Y_n=g(X_n)$$ be the number of prime divisors of $$X_n$$. Then we have

$\frac{Y_n - \log\log n}{\sqrt{\log\log n}} \implies N(0,1).$

In other words, for any $$x\in\R$$ we have

$\frac{1}{n}\# \left\{ 1\le k\le n: g(k) \le \log\log n + k \sqrt{\log\log n} \right\} \xrightarrow[n\to\infty]{} \Phi(x).$

Proof 14.2
See [Dur2010], p.\ 114--117 or [Dur2004], p.\ 119--124. The proof uses the moment method.

Note that $$Y_n$$ can be written in the form $$\sum_{p\le n} \ind_{\{p | X_n\}}$$, namely the sum over all primes $$p\le n$$ of the indicator of the event that $$X_n$$ is divisible by $$p$$. The probability that $$X_n$$ is divisible by $$p$$ is roughly $$1/p$$, at least if $$p$$ is significantly smaller than $$n$$. Therefore we can expect $$Y_n$$ to be on the average around

$\sum_{\textrm{prime }p\le n} \frac{1}{p},$

a sum that is known (thanks to Euler) to behave roughly like $$\log\log n$$. The Erdös-Kac theorem is intuitively related to the observation that these indicators $$\ind_{\{p | X_n\}}$$ for different $$p$$s are also close to being independent (a fact which follows from the Chinese remainder theorem). Of course, they are only approximately independent, and making these observations precise is the challenge to proving the theorem. In fact, many famous open problems in number theory (even the Riemann Hypothesis, widely considered to be the most important open problem in mathematics) can be formulated in terms of a statement about approximate independence (in some loose sense) of some arithmetic sequence relating to the prime numbers.

## The Euclidean algorithm

As a final example from number theory, consider the following problem: For some $$n\ge 1$$, choose $$X_n$$ and $$Y_n$$ independently and uniformly at random in $$\{1,2,\ldots,n\}$$, and compute their greatest common divisor (g.c.d.) using the Euclidean algorithm. Let $$N_n$$ be the number of division (with remainder) steps that were required. For example, if $$X = 58$$ and $$Y=24$$ then the application of the Euclidean algorithm would result in the sequence of steps

$(58,24) \to (24,10) \to (10,4) \to (4,2) \to (2,0),$

so 4 division operations were required (and the g.c.d.\ is 2).

Theorem: CLT for the number of steps in the Euclidean algorithm;

There exists a constant $$\sigma_\infty$$ (which has a very complicated definition) such that

$\frac{N_n - \frac{12 \log 2}{\pi^2} \log n}{\sigma_\infty \sqrt{\log n}} \implies N(0,1).$

Hensley's theorem was in recent years significantly generalized and the techniques extended by Brigitte Vall\'ee, a French mathematician. The fact that the average value of $$N_n$$ is approximately $$(12\log 2/\pi^2)\log n$$ was previously known from work of Heilbronn and Dixon in 1969--1970, using ideas dating back to Gauss, who discovered the probability distribution now called the Gauss measure''. This is the probability distribution on $$(0,1)$$ with density $$\frac{1}{\log 2(1+x)}$$, which Gauss found (but did not prove!) describes the limiting distribution of the ratio of a pair of independent $$U(0,1)$$ random variables after many iterations of the division-with-remainder step in the Euclidean algorithm.