# Distribution of the second minimum of a set of random variables

Given $$n$$ i.i.d. random variables $$X_1,...X_n$$, what is the distribution of the second smallest value ?

I know from this question that CDF of the minimum value is $$1 - (1-F(x))^n$$ where $$F(x)$$ is the CDF of $$X$$.

Moreover, how is distributed the difference between the minimum and the second minimum value of this set ?

First, suppose that $$X_{1},\ldots,X_{n}$$ are $$n$$ independent random variables, each with cdf $$F(x)$$. Let $$F_{(r)}(x)$$ with $$(r=1,\ldots, n)$$ denote the cdf of the $$r$$th order statistic $$X_{(r)}$$. The cdf of the $$r$$th order statistic is:

\begin{align} F_{(r)}(x) &=\mathrm{P}(X_{(r)}\leq x)\\ &=\mathrm{P}(\text{at least r of the X_{i} are less than or equal to x})\\ &=\sum_{i=r}^{n}{{n\choose i}}F(x)^{i}\left[1-F(x)\right]^{n-i} \end{align}

An alternative form of the cdf is $$F_{(r)}(x) = F(x)^{r}\sum_{j=0}^{n-r}{{r+j-1\choose r-1}}\left[1-F(x)\right]^{j}$$

So for $$r=2$$ this gives: $$F_{(2)}(x)= 1 + (1 - F(x))^{n - 1}(F(x)-F(x)n - 1)$$

The distribution of the difference between two order statistics is given in this answer.

Reference

David HA, Nagaraja HN (2003): Order Statistics. 3rd ed. Wiley.

• Very good answer, and thank you for giving a reference for the proof ! May 16 '20 at 15:49