Let $\mathbf{X} = (X_1, \dots, X_p)^\mathsf{T}$ be a $p$-dimensional random variable following a multivariate normal distribution with mean vector $\boldsymbol{\mu}$ and covariance matrix $\boldsymbol{\Sigma}$, i.e. $\mathbf{X} \sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$.
Let $X_{(1)}$ the minimum of the components of $\mathbf{X}$, that is to say, $X_{(1)} = \min\{X_1,\dots,X_p\}$.
What are the cumulative distribution function (cdf) and the probability distribution function (pdf) of $X_{(1)}$?
I've read about the distribution of the maximum here and also here. I guess that both concepts must be tightly connected, since $\min\{x_1,\dots,x_p\} = - \max\{-x_1,\dots,-x_p\}$.
However, I still have doubts. I read this expression somewhere, but I think it is not totally right, or at least I am not able to deduce it by my own:
The cdf of $X_{(1)}$ is
$F(x) = G\left( x \mathbf{1}_p \;|\; -\boldsymbol{\mu}, \boldsymbol{\Sigma} \right)$,
where
$\mathbf{1}_p = (1,\dots,1)^\mathsf{T} \in \mathbb{R}^p$
and
$G$ is the cdf of a multivariate normal distribution $\mathbf{X}$, that is to say,
$G(\mathbf{x}) = \mathrm{Pr}(\mathbf{X} \le \mathbf{x}) = \mathrm{Pr}(X_1 \le x_1 \;\;\text{and}\;\; \cdots \;\;\text{and}\;\; X_p \le x_p)$ .
I have tried to reproduce this result, but all I get is this:
$F(x) = \mathrm{Pr}(X_{(1)} \le x) = \mathrm{Pr}(X_1 \le x \;\;\text{or}\;\; \cdots \;\;\text{or}\;\; X_p \le x) = $
$1 - \mathrm{Pr}(X_1 > x \;\;\text{and}\;\; \cdots \;\;\text{and}\;\; X_p > x) =$
$1-\mathrm{Pr}(- X_1 < -x \;\;\text{and}\;\; \cdots \;\;\text{and}\;\; -X_p < -x)$,
and I do not know how to continue.
I know that, if $\mathbf{X} \sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$, then $-\mathbf{X} \sim \mathcal{N}(-\boldsymbol{\mu}, \boldsymbol{\Sigma})$. And this is all I am able to figure out.
EDIT:
I've been told that my question is already answered here, but I admit that I am not able to deduce the answer to my question from this other post, and that is why I would like to reopen my question.
