# Components of an n-dimensional random variable from a uniform distribution are independent

For a random variable $$X \in \mathcal{R}^n$$ from a uniform distribution, are the components independent? i.e,

$$Pr(X_1,X_2,...,X_n) = Pr(X_1)Pr(X_2)....Pr(X_n)$$

If so, does it depend on the uniform distribution? e.g if the uniform distribution is a p-dimensional hypercube.

• Depends on what you consider to be a multivariate uniform distribution. If $X_1=\dots=X_n$ then each variable is uniform but the variables are not independent. – Galton Nov 13 '19 at 18:35

Suppose $$(X_1, ... , X_n)$$ is uniformly distributed on $$M_1\times...\times M_n \subset \mathbb{R}^n$$, where all $$M_i$$ are measurable subsets of $$\mathbb{R}$$ with finite positive measure. Then $$\forall k < n$$ we have $$P(X_k \in M_i)=1$$ and $$\forall K_1\times...\times K_n \subset M_1\times...\times M_n$$, such that all $$K_i$$ are measurable subsets of $$M_i$$, we have $$P((X_1, ... , X_n) \in K_1\times...\times K_n) = \frac{\mu(K_1\times...\times K_n)}{\mu(M_1\times...\times M_n)}=\frac{\Pi_{i=1}^n \mu(K_i)}{\Pi_{i=1}^n \mu(M_i)}=\Pi_{i=1}^n\frac{\mu(K_i)}{\mu(M_i)}=\Pi_{i=1}^nP(X_i \in K_i)$$
where $$\mu$$ stands for Lebesgue measure. From that we can conclude, that $$X_1, ... , X_n$$ are independent.
However, for arbitrary measurable $$A \subset \mathbb{R}^n$$ with $$0<\mu(A)<+\infty$$, the components of a random variable, uniformly distributed on it are not always independent. For, example, if $$n = 2$$ and $$A = \{(x, y)\in [0;2]^2|x + y \geq 2\}$$, then $$P((X_1, X_2)\in[0;1]^2) = 0$$, despite both $$P(X_1\in[0;1])$$ and $$P(X_2 \in [0;1])$$ being non-zero.