Minimum of a set of random variables that are normally distributed I have random variables $X_0,X_1,…,X_n$ that are IID, and are normally distributed with mean $\mu$ and variance $\sigma$. I want to find the distribution of the minimum value, in the set $X_i$, let's call it Y.
I'm aware of the answer as listed here : 
How is the minimum of a set of random variables distributed?
My problem is that for a normally distributed variable $X$, its cdf is given by:
$\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^{2}/2}dt $ 
which has no analytical solution. Furthermore, even if I did have the answer to an approximation to the integral, I'm not sure how to calculate  the distribution of $Y$.
 A: I'll present the standard normal case. Let $Z_1, ..., Z_n \sim \ \text{iid} \  \mathcal N(0,1)$. Then $P(Z_{(1)} \leq z) = 1 - P(Z_{(1)} \geq z) = 1 - P(Z_1 \geq z, ..., Z_n \geq z) = 1 - (1 - \Phi(z))^n$.
We can differentiate with respect to $z$ to find the pdf:
$$
f_{(1)}(z) = \frac{d}{dz} \left(1 - (1 - \Phi(z))^n\right) = n(1-\Phi(z))^{n-1}\varphi(z)
$$
This is the distribution of the minimum value of your set of $n$ iid standard normal random variables. No integration required (unless you want to actually evaluate it).
Update: limiting case
Suppose we let $n \rightarrow \infty$. 
$$
\lim_{n \rightarrow \infty} P(Z_{(1)} \leq z) = \lim_{n \rightarrow \infty} 1 - (1 - \Phi(z))^n.
$$
Because $\forall z \in \mathbb R$ $\Phi(z) \in (0,1)$, we have that
$$
\lim_{n \rightarrow \infty} P(Z_{(1)} \leq z) = 1 - 0 = 1
$$
for an arbitrary $z \in \mathbb R$. This means that if we have a large enough collection of iid standard normal RVs, eventually we'll get one that's less than any particular real number $z$. This makes sense since the support of $\mathcal N(0,1)$ is all of $\mathbb R$.
