How do extreme values scale with sample size? Assume I have a random vector $X = \{x_1, x_2, ..., x_N\}$, composed of i.i.d. binomially distributed values.  If it would simplify the problem substantially, we can approximate them as normally distributed.  Given that all other parameters are fixed, I want to know how $E[min(X)]$ (the expected value of the smallest number in the vector $X$) scales with $N$.  
I don't care about a precise answer.  I just want to know how it scales, i.e. linearly (obviously not), exponentially, power law, etc.
 A: The table in this page of this book might help you. The explicit formulas for the expectation of minimum of sample of binomial distributions is given in the page before.
A: Assume that the random variables $x_k$ are i.i.d., nonnegative, integer valued, bounded by $n$, and such that $P(x_k=0)$ and $P(x_k=1)$ are both positive. For every $N\ge1$, let
$$
X_N= \min\{x_1,\ldots,x_N\}.
$$ 
Then, when $N\to+\infty$,
$$
E(X_N)=c^N(1+o(1)),
$$ 
where $c<1$ is independent of $N$ and given by 
$$
c=P(x_k\ge1).
$$
Hence $E(X_N)$ is exponentially small. When each $x_k$ is Binomial $(n,p)$ with $n\ge1$ and $p$ in $(0,1)$ fixed, the result holds with $c=1-(1-p)^n$. 

To see this, note that $[X_N\ge i]=[x_1\ge i]\cap\cdots\cap[x_N\ge i]$ for every $i$ and that, since $X_N$ is nonnegative and integer valued, $E(X_N)$ is the sum over $i\ge1$ of $P(X_N\ge i)$, hence
$$
E(X_N)=\sum_{i\ge 1}P(x_1\ge i)^N.
$$
For every $i\ge n+1$, $P(x_1\ge i)=0$. For every $2\le i\le n$, $0\le P(x_1\ge i)\le P(x_1\ge 2)$. Hence
$$
c^N\le E(X_N)\le c^N+(n-1)d^N,
$$ 
with
$$
c=P(x_1\ge1),\quad d=P(x_1\ge 2).
$$
Because $P(x_k=1)$ is positive, one knows that $d<c$, hence $E(X_N)\sim c^N$ when $N\to+\infty$.
A: The distribution of the minimum of any set of N iid random variables is:
$$f_{min}(x)=Nf(x)[1-F(x)]^{N-1}$$
Where $f(x)$ is the pdf and $F(x)$ is the cdf (this is sometime called a $Beta-F$ distribution, because it is a compound of a Beta distribution and an arbitrary distribution).  Hence the expectation (in this particular case) is given by:
$$E[min(X)] = N\sum_{x=0}^{x=n} xf(x)[1-F(x)]^{N-1}$$
Which means that $E[min(X)]=NE(x_1[1-F(x_1)]^{N-1})$.  Using the "delta method" to approximation this expectation $E[g(x)]\approx g(E[X])$ gives
$$E[min(X)]=NE(x_1[1-F(x_1)]^{N-1})\approx N(E(x_1)[1-F(E(x_1))]^{N-1})$$
Substituting $np=E[x_1]$ then gives the approximation:
$$E[min(X)]\approx Nnp[1-F(np)]^{N-1}$$
Note that $F(np)\approx \frac{1}{2}$ (via normal approx.) to give
$$E[min(X)]\approx \frac{Nnp}{2^{N-1}}$$
