If $U_1,...,U_n$ are unif(0,1) , then is $(U_1,...,U_n)$ unif$[0,1]^n$?

Question

Let $U_1,...,U_n$ be identicaly distributed random variable unif(0,1).

Consider random vector $U=(U_1,...,U_n)$.

How can I show that $U$ is distributed as unif$(0,1)^n$?

Intuitively, It is obvious.

Answer 1

Let $F_i$ denote the distribution of $U_i$ (for each $i$) and let $F_*$ denote the distribution of $(U_1,\ldots,U_n)$. By definition, we have that $F_i(u_i)=u_i$ for every $u_i\in(0,1)$. Due to the independence property, we then have that $F_*(u_1,...,u_n)=\prod_{i=1}^n F_i(u_i)=\prod_{i=1}^n u_i$. This is all we need to show since the right-hand side is the distribution of the $n$-dimensional Uniform distribution.