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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?

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

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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 Var, the probability $P(Y\leq y) = P(\min(X_1\dots X_n)\leq y)$ implies that at least one $X_i$ be 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 $y$.

If the $X_i$'s are iid, then the probability that all $X_i$ is greater than $y = (1-F(y))^n$. Therefore, the original prob is $1-(1-F(y))^n$.

Say $X_i \sim \text{Uniform} (0,1)$, then intuitively the probability $\min(X_1\dots X_n)\leq 1$ should be equivalent 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 Apr 28 '11 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 Mar 10 '13 at 19:56
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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 $\endgroup$ – robin girard Jul 23 '10 at 21:56
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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.

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  • $\begingroup$ @gung I understand why you would conclude that, but this answer doesn't apply to the IID setting of the question: it therefore comes across as a (correct and potentially interesting) comment to the question itself. $\endgroup$ – whuber Dec 8 '17 at 17:37
  • $\begingroup$ It's up to you, @whuber, if you what to convert this to a comment, it's your call. $\endgroup$ – gung Dec 8 '17 at 18:29

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