Standard Bivariate/ Multivariate Gaussians and Independence

Question

Suppose you have $Z=(Z_1,Z_2,\ldots,Z_n)^T$, where $Z$ has the standard Gaussian distribution on $\Bbb R^n$. Is it true that the $Z_i$, $i=1,2,\ldots,n$ are all independent, since they have the Standard Gaussian distribution on $\Bbb R^n$?

Answer 1

It's true that $Z_i$ are independent, and it follows from the equality $$Ee^{i(\vec{t}, \vec{Z})} = Ee^{i \sum_{k=1}^n t_i Z_k} = \prod_{k=1}^n Ee^{i t_k Z_k}$$ This equality may be checked easily since we know characteristic funcitons of $\vec{Z}$ and $Z_k$.

But it's not true that $Z_i$ are independent since "they have $N(0,1)$ distribution". They are independent for another reason.

Answer 2

Another approach: If the random vector $\mathbf Z\in{\mathbb R}^n$ has standard Gaussian distribution, then the joint density of $\mathbf Z:=(Z_1,\ldots,Z_n)^\top$ is $$ f(x_1,\ldots,x_n)=\frac1{(2\pi)^{n/2}}\exp[-\frac12 {\mathbf x}^\top{\mathbf x}]=\frac1{(2\pi)^{n/2}}\exp[-\frac12(x_1^2+\cdots+x_n^2)] $$ which can be factored into the form $$ \prod_{i=1}^n\left(\frac1{\sqrt{2\pi}}\exp(-\frac12 x_i^2)\right)=\prod_{i=1}^n \varphi(x_i) $$ where $\varphi$ is the standard Gaussian density. Thus we've decomposed the joint density into the product of the marginal densities, implying that $Z_1,\ldots,Z_n$ are independent and identically distributed standard Gaussian random variables.