NAVIGATION

Home

Research

Bookshelf

Garden

FIND ME ON

GitHub

LinkedIn

Email

Home

Research

Bookshelf

Garden

Existence of Sequences of Independent rvs

Theorem (7.1.1)

Let (μn)n≥1(\mu_{n})_{n\ge 1} be a sequence of Borel probability measures on (R,B(R))(\mathbb{R},\mathcal{B}(\mathbb{R})). Then ∃(Ω,F,P)\exists(\Omega,\mathcal{F},\mathbb{P}) and ∃(Xn)n≥1\exists(X_{n})_{n\ge 1} defined on (Ω,F,P)(\Omega,\mathcal{F},\mathbb{P}) such that (Xn)n≥1(X_{n})_{n\ge 1} are Independent random variables and Xn∼μnX_{n}\sim\mu _n,∀n≥1\forall n\ge 1.

Remark

What this is saying is that for any sequence of distributions we can find a representative sequence of random variable that is distributed by the represented distribution AND they’re all independent.

In order to prove this we first begin with a lemma:

>[!lem|7.1.2] >Let FF be the CDF of some rv XX. Then for U∼Unif([0,1])U\sim \text{Unif}([0,1]), if ϕ(u)=inf⁡{x∈R:F(x)≥u}=inf⁡F(x)≥u{x∈R},for u∈(0,1)\phi(u)=\inf\{x\in \mathbb{R}: F(x)\ge u \}=\inf_{F(x)\ge u}\{ x\in \mathbb{R} \},\quad\text{for }u\in(0,1)then we have X∼ϕ(U)X\sim\phi(U) i.e. P(ϕ(U)≤x)=F(x),∀x∈R\mathbb{P}(\phi(U)\le x)=F(x),\quad\forall x\in \mathbb{R}or the CDF of ϕ(U)\phi(U) is FF.
\begin{proof} Since FF is Right Continuous then ϕ(u)=inf⁡F(t)≥ut=arg⁡min⁡t∈R+{F(t)≥u}\phi(u)=\inf_{F(t)\ge u}t=\underset{t\in \mathbb{R}^{+}}{\arg\min}\{ F(t)\ge u \}(i.e. ϕ(u)=F−1(u)\phi(u)=F^{-1}(u) ) and this is well-defined ∀u∈(0,1)\forall u\in(0,1) by other properties of FF. Thus
ϕ(u)≤t  ⟺  u≤F(t)\phi(u)\le t\iff u\le F(t) arg⁡min⁡t∈R+{F(t)≥u}≤t  ⟺  u≤F(t)\underset{t\in \mathbb{R}^{+}}{\arg\min}\{ F(t)\ge u \}\le t\iff u\le F(t) Hence, P(ϕ(u)≤t)=P(u≤F(t))=F(t)\mathbb{P}(\phi(u)\le t)=\mathbb{P}(u\le F(t))=F(t)for U∼Unif([0,1])U\sim \text{Unif}([0,1]).
\end{proof}

Now we prove the theorem: \begin{proof} Got lazy… \end{proof}

Linked from

Skohorod's Theorem

Summary of MATH 895