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 ?


1 Answer 1


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.


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

  • 2
    $\begingroup$ Very good answer, and thank you for giving a reference for the proof ! $\endgroup$
    – Toool
    Commented May 16, 2020 at 15:49

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