# Given a random variable $X_1$, does there always exsist an i.i.d. sequence $\{X_n\}$?

Below is from Tao's lecture note and he says there exists an i.i.d. sequence of random variables $$\{X_n\}_n$$ such that each $$X_i$$ is uniformly distributed. For any given random variable $$X_1$$, does there exist an i.i.d. sequence$$\{X_n\}_n$$ ?

https://terrytao.wordpress.com/2015/10/23/275a-notes-3-the-weak-and-strong-law-of-large-numbers/#sme

Proposition 6 Let $${\varepsilon > 0}$$. Then, for sufficiently large $${n}$$, a proportion of at least $$1-\varepsilon$$ of the cube $${[-1,1]^n}$$ (by $${n}$$-dimensional Lebesgue measure) is contained in the annulus $${\{ x \in {\bf R}^n: (1-\varepsilon) \sqrt{n/3} \leq |x| \leq (1+\varepsilon) \sqrt{n/3} \}}$$.

Proof: Let $${X_1,X_2,\dots}$$ be iid random variables drawn uniformly from $${[-1,1]}.$$

## 1 Answer

Let $$P$$ be the distribution of $$X_1$$, $$\Omega=\mathbb{R}^{\mathbb{N}}$$, $$\mathcal{F}=\mathcal{B}(\mathbb{R})^{\otimes\mathbb{N}}$$, where $$\mathcal{B}(\mathbb{R})$$ is the Borel $$\sigma$$-algebra on $$\mathbb{R}$$ and $$\mathbb{P}=P^{\otimes\mathbb{N}}$$. For each $$n\in\mathbb{N}$$, let $$\pi_n:\Omega\to\mathbb{R}$$ be defined as $$\pi_n(\omega)=\omega_n$$, where $$\omega=\{\omega_n\}_{n\in\mathbb{N}}$$. We have $$\{\pi_n\}_{n\in\mathbb{N}}$$ is a sequence of $$\mathbb{P}$$-independent random variables such that for each $$n\in\mathbb{N}$$, $$\pi_n$$ has distribution $$P$$, i.e., $$\pi_n$$ and $$X_1$$ have the same distribution.