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\begin{thm}[Central limit theorem (CLT)] Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of iid r.v.'s with values in [math]\R[/math]. We assume that [math]\E[X_k^2] \lt \infty[/math] (i.e. [math]X_k\in L^2(\Omega,\A,\p)[/math]) and let [math]\sigma^2=Var(X_k)[/math] for all [math]k\in\{1,...,n\}[/math]. Then for all [math]k\in\{1,...,n\}[/math] we get

[[math]] \sqrt{n}\left(\frac{1}{n}\sum_{i=0}^nX_i-\E[X_k]\right)\xrightarrow{n\to\infty\atop law}\mathcal{N}(0,\sigma^2). [[/math]]

Equivalently, for all [math]a,b\in\bar\R[/math] with [math]a \lt b[/math] and for all [math]k\in\{1,...,n\}[/math] we get

[[math]] \lim_{n\to\infty}\p\left[\sum_{i=0}^nX_i\in \left[n\E[X_k]+a\sqrt{n},n\E[X_k]+b\sqrt{n}\right]\right]=\frac{1}{\sigma\sqrt{2\pi}}\int_a^be^{-\frac{x^2}{2\sigma^2}}dx. [[/math]]

\end{thm}

Example

If [math]\E[X_k]=0[/math] for all [math]k\in\{1,...,n\}[/math], then

[[math]] \lim_{n\to\infty}\p\left[\frac{\sum_{i=1}^nX_i-n\E[X_k]}{\sqrt{n}}\in[a,b]\right]=\frac{1}{\sigma\sqrt{2\pi}}\int_a^be^{-\frac{x^2}{2\sigma^2}}dx. [[/math]]

\begin{proof} Without loss of generality we can assume that [math]\E[X_k]=0[/math], for all [math]k\in\{1,...,n\}[/math]. Define now a sequence [math]Z_n=\frac{\sum_{i=1}^nX_i}{\sqrt{n}}[/math]. Then we can obtain

[[math]] \Phi_{Z_n}(\xi)=\E\left[e^{i\xi Z_n}\right]=\E\left[e^{i\xi \left(\frac{\sum_{i=1}^nX_i}{\sqrt{n}}\right)}\right]=\prod_{j=1}^n\E\left[e^{i\xi \frac{X_j}{\sqrt{n}}}\right]=\E\left[e^{i\xi\frac{X_k}{\sqrt{n}}}\right]^n=\Phi_{X_k}^n(\xi). [[/math]]

We have already seen that

[[math]] \Phi_{X_k}(\xi)=1+n\xi\E[X_k]-\frac{1}{2}\xi^2\E[X_k^2]+O(\xi^2)=1-\frac{\sigma^2}{2}\xi^2+O(\xi^2). [[/math]]

Finally, for fixed [math]\xi[/math], we have

[[math]] \Phi_{X_k}\left(\frac{\xi}{\sqrt{n}}\right)=1-\frac{\sigma^2\xi^2}{2n}+O\left(\frac{1}{n}\right) [[/math]]

[[math]] \lim_{n\to\infty}\Phi_{X_k}^n(\xi)=\lim_{n\to\infty}\left(1-\frac{\sigma^2\xi^2}{2}+O\left(\frac{1}{n}\right)\right)^n=\E\left[e^{i\xi \mathcal{N}(0,\sigma^2)}\right]. [[/math]]

\end{proof}

For the illustration of the CLT, the distribution of a Gaussian distributed r.v. [math]X[/math] with [math]\mu=100[/math] and [math]\sigma=10[/math] ([math]X\sim \mathcal{N}(\mu,\sigma^2)[/math]).

The distribution of the mean of [math]n=5[/math] i.i.d. Gaussian r.v.'s [math](X_n)[/math] with [math]\mu=100[/math] and [math]\sigma^2=10[/math] ([math]X_k\sim \mathcal{N}(\mu,\sigma^2)[/math]). The figure illustrates the behavior of the distribution of the mean, which is given by [math]\bar X_n=\frac{1}{n}\sum_{k=1}^nX_k[/math].

The distribution of the mean of [math]n=20[/math] i.i.d. Gaussian r.v.'s [math](X_n)[/math] with [math]\mu=100[/math] and [math]\sigma^2=10[/math] ([math]X_k\sim \mathcal{N}(\mu,\sigma^2)[/math]). The figure illustrates the behavior of the distribution of the mean, which is given by [math]\bar X_n=\frac{1}{n}\sum_{k=1}^nX_k[/math].

The distribution of the mean of [math]n=100[/math] i.i.d. Gaussian r.v.'s [math](X_n)[/math] with [math]\mu=100[/math] and [math]\sigma^2=10[/math] ([math]X_k\sim \mathcal{N}(\mu,\sigma^2)[/math]). The figure illustrates the behavior of the distribution of the mean, which is given by [math]\bar X_n=\frac{1}{n}\sum_{k=1}^nX_k[/math].

The distribution of the mean of [math]n=1000[/math] i.i.d. Gaussian r.v.'s [math](X_n)[/math] with [math]\mu=100[/math] and [math]\sigma^2=10[/math] ([math]X_k\sim \mathcal{N}(\mu,\sigma^2)[/math]). The figure illustrates the behavior of the distribution of the mean, which is given by [math]\bar X_n=\frac{1}{n}\sum_{k=1}^nX_k[/math]. Clearly we see that it's converging to [math]\mu[/math] for [math]n\to\infty[/math].

Theorem

Let [math](X_n)_{n\geq 1}[/math] be independent r.v.'s but not necessarily i.i.d. We assume that [math]\E[X_j]=0[/math] and that [math]\E[X_j^2]=\sigma_j^2 \lt \infty[/math], for all [math]j\in\{1,...,n\}[/math]. Assume further that [math]\sup_n\E[\vert X_n\vert^{2+\delta}] \lt \infty[/math] for some [math]\delta \gt 0[/math], and that [math]\sum_{j=1}^\infty \sigma_j^2 \lt \infty.[/math] Then

[[math]] \frac{\sum_{j=1}^nX_j}{\sqrt{\sum_{j=1}^n \sigma_j^2}}\xrightarrow{n\to\infty\atop law}\mathcal{N}(0,1). [[/math]]

Example

We got the following examples:

• Let [math](X_n)_{n\geq 1}[/math] be i.i.d. r.v.'s with [math]\p[X_n=1]=p[/math] and [math]\p[X_n=0]=1-p[/math]. Then [math]S_n=\sum_{i=1}^nX_i[/math] is a binomial r.v. [math]\B(p,n)[/math]. We have [math]\E[S_n]=np[/math] and [math]Var(S_n)=np(1-p)[/math]. Now with the strong law of large numbers we get [math]\frac{S_n}{n}\xrightarrow{n\to\infty\atop a.s.}p[/math] and with the central limit theorem we get
  [[math]] \frac{S_n-np}{\sqrt{np(1-p)}}\xrightarrow{n\to\infty\atop law}\mathcal{N}(0,1). [[/math]]
• Let [math]\mathcal{P}[/math] be the set of prime numbers. For [math]p\in\mathcal{P}[/math], define [math]\B_p[/math] as [math]\p[\B_p=1]=\frac{1}{p}[/math] and [math]\p[\B_p=0]=1-\frac{1}{p}[/math]. We take the [math](\B_p)_{p\in\mathcal{P}}[/math] to be independent and
  [[math]] W_n=\sum_{\mathcal{P}\subset\N\atop p\in\mathcal{P}}\B_p, [[/math]]
  the probabilistic model for the total numbers of distinct prime divisors of [math]n:=W(0)[/math]. It's a simple exercise to check that [math](W_n)_{n\geq 1}[/math] satisfies the assumption of theorem 13.2 and using the fact that [math]\sum_{p\leq n\atop p\in\mathcal{P}}\frac{1}{p}\sim \log\log n[/math] and we obtain
  [[math]] \frac{W_n-\log\log n}{\sqrt{\log\log n}}\xrightarrow{n\to\infty\atop law}\mathcal{N}(0,1). [[/math]]

  

  An illustration of the CLT, where the r.v.'s [math](X_n)[/math] are i.i.d. exponentially distributed. Also with the cumulative distribution functions. Here we have [math]n=2[/math] exponentially distributed r.v's [math]\left(X_k\sim \lambda e^{-\lambda}\right)[/math]. On the left side, the black curve represents the density function of the r.v.'s and the red curve represents the density of a Gaussian r.v. [math]Y[/math] with [math]\mu=0[/math] ([math]Y\sim\mathcal{N}(0,\sigma^2)[/math]). On the right side, the black curve represents the cumulative distribution function of the r.v.'s [math]X_k[/math] and the red curve represents the cumulative distribution function of a Gaussian r.v. [math]Y[/math].

  

  An illustration of the CLT, where the r.v.'s [math](X_n)[/math] are i.i.d. exponentially distributed. Also with the cumulative distribution functions. Here we have [math]n=4[/math] exponentially distributed r.v's [math]\left(X_k\sim \lambda e^{-\lambda}\right)[/math]. On the left side, the black curve represents the density function of the r.v.'s and the red curve represents the density of a Gaussian r.v. [math]Y[/math] with [math]\mu=0[/math] ([math]Y\sim\mathcal{N}(0,\sigma^2)[/math]). On the right side, the black curve represents the cumulative distribution function of the r.v.'s [math]X_k[/math] and the red curve represents the cumulative distribution function of a Gaussian r.v. [math]Y[/math].

  

  An illustration of the CLT, where the r.v.'s [math](X_n)[/math] are i.i.d. exponentially distributed. Also with the cumulative distribution functions. Here we have [math]n=10[/math] exponentially distributed r.v's [math]\left(X_k\sim \lambda e^{-\lambda}\right)[/math]. On the left side, the black curve represents the density function of the r.v.'s and the red curve represents the density of a Gaussian r.v. [math]Y[/math] with [math]\mu=0[/math] ([math]Y\sim\mathcal{N}(0,\sigma^2)[/math]). On the right side, the black curve represents the cumulative distribution function of the r.v.'s [math]X_k[/math] and the red curve represents the cumulative distribution function of a Gaussian r.v. [math]Y[/math].

  

  An illustration of the CLT, where the r.v.'s [math](X_n)[/math] are i.i.d. exponentially distributed. Also with the cumulative distribution functions. Here we have [math]n=20[/math] exponentially distributed r.v's [math]\left(X_k\sim \lambda e^{-\lambda}\right)[/math]. On the left side, the black curve represents the density function of the r.v.'s and the red curve represents the density of a Gaussian r.v. [math]Y[/math] with [math]\mu=0[/math] ([math]Y\sim\mathcal{N}(0,\sigma^2)[/math]). On the right side, the black curve represents the cumulative distribution function of the r.v.'s [math]X_k[/math] and the red curve represents the cumulative distribution function of a Gaussian r.v. [math]Y[/math]. Now we can see that both, the density and the cumulative distribution function, are converging to a Gaussian density and a Gaussian cumulative distribution function for [math]n\to\infty[/math].

  Theorem (Erdös-Kac)

  Let [math]N_n[/math] be a r.v. with uniformly distribution in [math]\{1,...,n\}[/math], then

  [[math]] \frac{W(N_n)-\log\log n}{\sqrt{\log\log n}}\xrightarrow{n\to\infty\atop law}\mathcal{N}(0,1), [[/math]]

  where [math]W(n)=\sum_{p\leq n\atop p\in\mathcal{P}}\one_{p|n}[/math].

• Suppose that [math](X_n)_{n\geq 1}[/math] are i.i.d. r.v.'s with distribution function [math]F(x)=\p[X_1\leq x][/math]. Let [math]Y_n(x)=\one_{X_n\leq x}[/math], where [math](Y_n)_{n\geq 1}[/math] are i.i.d. Define [math]F_n(x)=\frac{1}{n}\sum_{k=1}^nY_k(x)=\frac{1}{n}\sum_{k=1}^n\one_{X_k\leq x}[/math]. [math]F_n[/math] is called the empirical distribution function. With the strong law of large numbers we get [math]\lim_{n\to\infty\atop a.s.}F_n(x)=\E[Y_1(x)][/math] and
  [[math]] \E[Y_1(x)]=\E[\one_{X_1\leq x}]=\p[X_1\leq x]=F(x). [[/math]]
  In fact, it is a theorem (Gliwenko-Cantelli) which says that
  [[math]] \sup_{x\in\R}\vert F_n(x)-F(x)\vert\xrightarrow{n\to\infty\atop a.s.}0. [[/math]]
  Next we note that
  [[math]] \sqrt{n}(F_n(x)-F(x))=\sqrt{n}\left(\frac{1}{n}\sum_{k=1}^nY_k(x)-\E[Y_1(x)]\right)=\frac{1}{\sqrt{n}}\left(\sum_{k=1}^nY_k(x)-n\E[Y_1(x)]\right) [[/math]]
  Now with the central limit theorem we get
  [[math]] \sqrt{n}(F_n(x)-F(x))\xrightarrow{n\to\infty\atop law}\mathcal{N}(0,\sigma^2(x)), [[/math]]
  where [math]\sigma^2(x)=Var(Y_1)=\E[Y_1^2(x)]=\E[Y_1(x)]=F(x)-F^2(x)=F(x)(1-F(x))[/math]. Hence
  [[math]] \sqrt{n}(F_n(x)-F(x))\xrightarrow{n\to\infty\atop law}\mathcal{N}(0,F(x)(1-F(x))). [[/math]]

Theorem (Berry-Esseen)

Let [math](X_n)_{n\geq 1}[/math] be i.i.d. r.v's and suppose that [math]\E[\vert X_k\vert^3] \lt \infty[/math], [math]\forall k\in\{1,...,n\}[/math]. Let

[[math]] G_n(x)=\p\left[\frac{\sum_{i=1}^nX_i-n\E[X_k]}{\sigma\sqrt{n}}\leq x\right],\forall k\in\{1,...,n\}, [[/math]]

where [math]\sigma^2=\E[X_k^2][/math] [math]\forall k\in\{1,...,n\}[/math] and [math]\Phi(x)=\p\left[\mathcal{N}(0,1)\leq x\right]=\int_{-\infty}^xe^{-\frac{u^2}{2}}\frac{1}{\sqrt{2\pi}}du[/math]. Then

[[math]] \sup_{x\in\R}\vert G_n(x)-\Phi(x)\vert\leq C\frac{\E[\vert X_k\vert^3]}{\sigma^3\sqrt{n}},\forall k\in\{1,...,n\} [[/math]]

where [math]C[/math] is a universal constant.

General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].