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
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
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
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
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
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
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
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
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