Conditional Laws Given an Order Statistic

Why Conditioning Splits the Sample

In an iid continuous sample, knowing the value of the kth order statistic gives more information than simply knowing that one observation equals that value. It tells us that exactly \(k-1\) observations lie below it and \(n-k\) lie above it. The values on each side still have a tractable distribution: they behave like samples from the parent distribution restricted to the corresponding side of the observed order statistic.

The precise statement concerns the ordered observations below and above the kth value. Conditional on that value, the lower group has the distribution of \(k-1\) draws from the parent distribution truncated below the value, and the upper group has the distribution of \(n-k\) draws truncated above it. These two ordered groups are conditionally independent.

Truncated Parent Distributions

Let \(X_1,\ldots,X_n\) be iid with a continuous distribution function \(F\) and density \(f\). Write \(X_{(1)}\le\cdots\le X_{(n)}\) for their order statistics. Fix \(k\) with \(1\le k\le n\). At a value \(x\) with \(0<F(x)<1\), the parent distribution restricted to values below \(x\) has conditional CDF \(F(t)/F(x)\) for \(t<x\). Restricted to values above \(x\), it has conditional CDF \((F(t)-F(x))/(1-F(x))\) for \(t>x\).

The conditioning event \(X_{(k)}=x\) has probability zero for a continuous sample. Conditional distributions are therefore understood through conditional densities, which specify a regular conditional law for almost every \(x\) at which the density of \(X_{(k)}\) is positive. The formulas below give a natural version at any such \(x\) with \(0<F(x)<1\) and \(f(x)>0\).

Conditional Joint Law

The central result identifies the full conditional law of both ordered groups, not just their individual marginal distributions.

on the region \(u_1<\cdots<u_{k-1}<x<v_1<\cdots<v_{n-k}\), and is zero elsewhere. Empty products are one. Consequently, the ordered lower group and ordered upper group are conditionally independent. Each group has the order-statistic distribution from an iid sample of the relevant size from its corresponding truncated parent law.

Proof. The joint density of all \(n\) order statistics is \(n!\prod_{i=1}^n f(x_i)\) on \(x_1<\cdots<x_n\). Integrating over the \(k-1\) ordered coordinates below \(x\) and the \(n-k\) ordered coordinates above \(x\) gives the marginal density

Here the integral of \(\prod f(u_i)\) over \(u_1<\cdots<u_{k-1}<x\) is \(F(x)^{k-1}/(k-1)!\): the integral over the full \((k-1)\)-dimensional region with every coordinate below \(x\) is \(F(x)^{k-1}\), and its \((k-1)!\) orderings have equal integrals. The corresponding upper integral is \([1-F(x)]^{n-k}/(n-k)!\). Dividing the full joint density \(n!f(x)\prod f(u_i)\prod f(v_j)\) by this marginal density cancels \(n!f(x)\) and yields the displayed conditional density. Its factorization into a function of the \(u_i\)'s and a function of the \(v_j\)'s proves conditional independence. Each factor is the usual ordered-sample density formed from the appropriate truncated density. \(\square\)

In particular, for \(a<x\), one observation drawn from the lower-truncated law satisfies \(P(X_1\le a\mid X_1<x)=F(a)/F(x)\). For \(b>x\), one observation drawn from the upper-truncated law satisfies \(P(X_1>b\mid X_1>x)=[1-F(b)]/[1-F(x)]\). Conditional probabilities involving several observations use these probabilities together with the appropriate group size and order.

Conditional Interval Counts

An immediate consequence is a binomial law for counts within either truncated group. For example, let \(a<b<x\), and let \(N_-\) count how many of the \(k-1\) observations below \(X_{(k)}\) fall in \((a,b]\). Conditional on \(X_{(k)}=x\), each lower-group draw has probability \([F(b)-F(a)]/F(x)\) of falling in that interval.

For an interval \(x<a<b\) in the upper group, the corresponding count has law \(\operatorname{Bin}(n-k,[F(b)-F(a)]/[1-F(x)])\).

Derivation. The theorem says that the lower group has the distribution of the order statistics of \(k-1\) iid draws from the lower-truncated law. Counting observations in \((a,b]\) is unaffected by sorting. Each draw has probability \(p=[F(b)-F(a)]/F(x)\) of falling in the interval. Independence of the draws therefore gives \(P(N_-=r\mid X_{(k)}=x)=\binom{k-1}{r}p^r(1-p)^{k-1-r}\) for \(r=0,\ldots,k-1\), which is the stated binomial law. The upper-group result follows by the same argument using the upper-truncated law. Since the two ordered groups are conditionally independent, counts formed separately from the lower and upper groups are conditionally independent as well. \(\square\)

Worked Conditional Probabilities

Common Mistake and Why the Result Matters

A common error is to treat all \(n\) observations as independent draws from the parent distribution truncated at \(x\). That cannot be right: given \(X_{(k)}=x\), exactly \(k-1\) observations are below \(x\), exactly \(n-k\) are above it, and the kth observation is fixed at \(x\). The lower and upper truncated laws apply to their respective groups, with these different group sizes.

A second pitfall is to use an ordinary conditional-probability ratio with \(P(X_{(k)}=x)\) in the denominator. That probability is zero in a continuous model. The valid calculation conditions through the density of \(X_{(k)}\), as in the derivation above. The resulting formulas make conditional questions about extremes, quantiles, and observations near a sample rank into ordinary calculations under truncated distributions.