Bernoulli sample frequencies

IID finite-variance weak law: Let $(X_k{)}_{k\ge 1}$ be IID square-integrable real random variables, with $\mu =\mathbb{E}{X}_1$ and ${\sigma }^2=\mathrm{Var}(X_1)$. For $S_n={\sum }_{k=1}^n{X}_k$, $\mathbb{E}\mathrm{\mid }{S}_n\mathrm{/}n-\mu {\mathrm{\mid }}^2={\sigma }^2\mathrm{/}n,$ and $S_n\mathrm{/}n\to \mu$ in $L^2$ and in probability. Also $\mathbb{P}(\mathrm{\mid }{S}_n\mathrm{/}n-\mu \mathrm{\mid }\ge \epsilon )\le {\sigma }^2\mathrm{/}(n{\epsilon }^2)$ for $\epsilon >0$.

Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials: For $p\in [0,1]$, a Bernoulli$(p)$ random variable takes the value $1$ with probability $p$ and $0$ with probability $1-p$. For $n\in \mathbb{N}$, a binomial$(n,p)$ random variable is a sum $X={\sum }_{i<n}{X}_i$ of $n$ mutually independent Bernoulli$(p)$ variables. When $n=0$, this is the constant zero random variable.

A Bernoulli$(p)$ variable has mean $p$ and variance $p(1-p)$; a binomial$(n,p)$ variable has mean $np$ and variance $np(1-p)$: If $X$ is Bernoulli$(p)$, then $\mathbb{E}[X]=p$ and $\mathrm{Var}(X)=p(1-p)$. If $S$ is binomial$(n,p)$, then $\mathbb{E}[S]=np,\mathrm{Var}(S)=np(1-p)\mathrm{.}$ These formulas include $p=0$, $p=1$, and $n=0$.

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure $\nu$ on $(S,\mathrm{\Sigma })$ is the common law of a countable independent family of $S$-valued random elements.