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?

  • $\begingroup$ Mathematical answer or numerical? Your question is very general. For a fast algorithm see the article here: medium.com/@mike.roweprediger/… but I'm not sure if that is the question. $\endgroup$ Commented Mar 4, 2023 at 15:30

3 Answers 3


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


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.

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    $\begingroup$ This site supports Markdown syntax for editing, and also LATEX for mathematical expressions. Further information can be found here: stats.stackexchange.com/editing-help. $\endgroup$
    – chl
    Commented Apr 28, 2011 at 9:47
  • $\begingroup$ Thanks for providing the reasoning. I had a problem with non-identically-distributed variables, but the minimum logic still applied well :) $\endgroup$
    – Matchu
    Commented Mar 10, 2013 at 19:56
  • $\begingroup$ I think that answer 1-(1-F(x))^n is correct in special cases. Special cases is condition that pmf of r.v. is based on a formula for domain of r.v. If it be different in various parts of domain above mentioned formula deviates a little from actual simulation results. $\endgroup$ Commented Nov 13, 2016 at 14:46
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    $\begingroup$ Can anything be said about the mean and the standard deviation? $\endgroup$
    – vinntec
    Commented Apr 27, 2020 at 13:11

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.

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    $\begingroup$ My opinion is that this book is THE book about extrem value theory $\endgroup$ Commented Jul 23, 2010 at 21:56

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