How is the minimum of a set of IID random variables distributed?

Question

If $X_1, ..., X_n$ are independent identically-distributed random variables, what can be said about the distribution of $\min(X_1, ..., X_n)$ in general?

Answer 1

If the CDF of $X_i$ is denoted by $F(x)$, then the CDF of the minimum is given by $1-[1-F(x)]^n$.

Reasoning: given $n$ random variables, the probability $P(Y\leq y) = P(\min(X_1\dots X_n)\leq y)$ implies that at least one $X_i$ is smaller than $y$.

The probability that at least one $X_i$ is smaller than $y$ is equivalent to one minus the probability that all $X_i$ are greater than $y$, i.e. $P(Y\leq y) = 1 - P(X_1 \gt y,\dots, X_n \gt y)$.

If the $X_i$'s are independent identically-distributed, then the probability that all $X_i$ are greater than $y$ is $[1-F(y)]^n$. Therefore, the original probability is $P(Y \leq y) = 1-[1-F(y)]^n$.

Example: say $X_i \sim \text{Uniform} (0,1)$, then intuitively the probability $\min(X_1\dots X_n)\leq 1$ should be equal to 1 (as the minimum value would always be less than 1 since $0\leq X_i\leq 1$ for all $i$). In this case $F(1)=1$ thus the probability is always 1.

Answer 2

If the cdf of $X_i$ is denoted by $F(x)$, then the cdf of the minimum is given by $1-[1-F(x)]^n$.

Answer 3

Rob Hyndman gave the easy exact answer for a fixed n. If you're interested in asymptotic behavior for large n, this is handled in the field of extreme value theory. There is a small family of possible limiting distributions; see for example the first chapters of this book.