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Density of an Order Statistic

Theorem

Let X1,…,XnX_1,\dots,X_n be i.i.d. continuous random variables with PDF ff and CDF FF, let X(1)≤⋯≤X(n)X_{(1)}\le\cdots\le X_{(n)} be their order statistics, and write Fˉ=1−F\bar F=1-F. For 1≤k≤n1\le k\le n, the kkth order statistic has density

fX(k)(x)=nf(x)(n−1k−1)×F(x)k−1Fˉ(x)n−k.f_{X_{(k)}}(x)= n f(x) {\binom{n-1}{k -1}} \times F(x)^{k-1} \bar F(x)^{n-k}.

The formula holds for every xx except possibly on a set of zero length, for example at finitely many points.

The i.i.d. assumption makes the probabilities of falling below a threshold equal and independent. It produces the binomial count underlying the order-statistic CDF and density.


Interpretation

An order statistic is a sorted observation. Its density combines the chance of an observation near a value with the chances that the required numbers of observations fall below and above it.

Densities of the minimum, median and maximum of three uniform observations on 0 to 1, peaking at 0, in the middle and at 1, with the flat density of one observation for comparison.

For three i.i.d. Unif⁡(0,1)\operatorname{Unif}(0,1) observations, the kkth order statistic has density fX(k)(x)=3(2k−1)xk−1(1−x)3−kf_{X_{(k)}}(x)=3\binom{2}{k-1}x^{k-1}(1-x)^{3-k} for 0<x<10<x<1 and k=1,2,3k=1,2,3: the minimum 3(1−x)23(1-x)^2 peaks at 00, the median 6x(1−x)6x(1-x) peaks at 1/21/2 and the maximum 3x23x^2 peaks at 11. The dashed line is the density of a single observation.

For X(k)X_{(k)} to lie in a short interval of length Δx\Delta x at xx, one of the nn observations must lie there (factor nn, probability about f(x)Δxf(x)\Delta x), and of the remaining n−1n-1, exactly k−1k-1 must fall below xx. That count is binomial with n−1n-1 trials and success probability F(x)F(x), which gives (n−1k−1)F(x)k−1Fˉ(x)n−k\binom{n-1}{k-1}F(x)^{k-1}\bar F(x)^{n-k}. The i.i.d. assumption gives every observation the same probability F(x)F(x) and makes the trials independent. Continuity makes exact ties have probability zero, but several observations can still fall in the same short interval. At a point where ff is continuous and finite, the probability of two or more observations in that interval is of order (Δx)2(\Delta x)^2, so the one-observation calculation gives the first-order term used for the density. For discrete observations ties occur and the formula does not apply.

For k=nk=n and k=1k=1 the formula is the derivative of the CDFs F(x)nF(x)^n and 1−Fˉ(x)n1-\bar F(x)^n from Order Statistic, with Fˉ\bar F the survivor function.

The factor nn is easy to forget. Without it the density of the maximum of three uniforms would be x2x^2, which integrates to 1/31/3 instead of 11.


Key Results

  • For the maximum, FX(n)(x)=F(x)nF_{X_{(n)}}(x)=F(x)^n and fX(n)(x)=nf(x)F(x)n−1f_{X_{(n)}}(x)=nf(x)F(x)^{n-1}.
  • For the minimum, FX(1)(x)=1−(1−F(x))nF_{X_{(1)}}(x)=1-(1-F(x))^n and fX(1)(x)=nf(x)(1−F(x))n−1f_{X_{(1)}}(x)=nf(x)(1-F(x))^{n-1}.

Examples

  • For the maximum of three i.i.d. Unif⁡(0,1)\operatorname{Unif}(0,1) variables, the density is 3x23x^2 on (0,1)(0,1) and P(X(3)≤1/2)=1/8\mathbb{P}(X_{(3)}\le1/2)=1/8.
  • The minimum of four independent Exp⁡(2)\operatorname{Exp}(2) variables has tail e−8x\mathrm{e}^{-8x} for x≥0x\ge0, hence distribution Exp⁡(8)\operatorname{Exp}(8).

Prerequisites

Sources

Symbols: delta

Last updated: 27 October 2026 11:52