Conditional Order Statistics Between Fixed Ranks

Why Conditioning on Two Ranks Changes the Distribution

Suppose a sample is known to have its \(r\)th order statistic at \(a\) and its \(s\)th order statistic at \(b\), where \(r<s\). The observations in ranks \(r+1,\ldots,s-1\) must lie between \(a\) and \(b\). Their conditional distribution is therefore not the original parent distribution: it is the parent distribution restricted to that interval. Once this restriction is accounted for, the intervening ranks can be analyzed as order statistics from a smaller sample.

This result is useful when a problem gives information about two sample ranks and then asks about a rank between them. The main tasks are to count the intervening observations correctly and to replace the parent cdf with its conditional, truncated version.

Definition: Let \(X_1,\ldots,X_n\) be iid observations from an absolutely continuous distribution with cdf \(F\) and density \(f\). Write their order statistics as \(X_{(1)}<\cdots<X_{(n)}\), which are distinct with probability 1. Fix ranks \(r<s\) and feasible endpoint values \(a<b\), with \(F(b)>F(a)\). Conditional on \(X_{(r)}=a\) and \(X_{(s)}=b\), there are \(m=s-r-1\) intervening observations. Their ordered values have the distribution of the order statistics of \(m\) iid observations from the parent distribution truncated to \((a,b)\).

Because the endpoints are continuous random variables, the event that they equal particular values has probability zero. Conditioning on these exact values is understood through the conditional density (a regular conditional distribution). The formulas below apply at feasible endpoints where the relevant densities and conditional distributions are defined.

The Truncated Parent Distribution

For one parent observation, conditioning on its value lying in \((a,b)\) gives the truncated cdf \(G\) and density \(g\) below. The denominator \(F(b)-F(a)\) is the parent probability of falling in that interval.

G(x)= \begin{cases} 0, & x\le a,\\[4pt] \dfrac{F(x)-F(a)}{F(b)-F(a)}, & a<x<b,\\[8pt] 1, & x\ge b, \end{cases} \qquad g(x)=\frac{f(x)}{F(b)-F(a)},\quad a<x<b.

The conditional density of the entire ordered interior vector is

f_{\mathrm{int}\mid a,b}(x_{r+1},\ldots,x_{s-1}) = \frac{m!}{[F(b)-F(a)]^m} \prod_{j=r+1}^{s-1} f(x_j), \quad a<x_{r+1}<\cdots<x_{s-1}<b.

The factor \(m!\) reflects the ordering of the \(m\) observations. Equivalently, each interior observation has truncated density \(g\), and the interior order statistics are the order statistics of \(m\) iid draws from that truncated distribution.

Why the Conditional Density Has This Form

The joint density of all \(n\) iid order statistics is \(n!\prod_{j=1}^n f(x_j)\) on the region \(x_1<\cdots<x_n\). To obtain the joint density of the two endpoint ranks, integrate out the observations below \(a\), between \(a\) and \(b\), and above \(b\). For \(q\) ordered observations in a region with parent probability \(p\), the integral of their product densities over that ordered region is \(p^q/q!\): the unordered region has total integral \(p^q\), and the \(q!\) orderings have equal integrals.

Applying that fact to the three groups gives the endpoint density. Here \(m=s-r-1\):

f_{X_{(r)},X_{(s)}}(a,b) = \frac{n!}{(r-1)!\,m!\,(n-s)!} F(a)^{r-1} [F(b)-F(a)]^m [1-F(b)]^{n-s} f(a)f(b),\quad a<b.

Now divide the joint density of all order statistics by this endpoint density. The factors for observations below \(a\), above \(b\), and the endpoint densities cancel, leaving the conditional density of the intervening ordered values:

\frac{n!\prod_{j=1}^n f(x_j)} {f_{X_{(r)},X_{(s)}}(a,b)} \quad\longrightarrow\quad \frac{m!}{[F(b)-F(a)]^m} \prod_{j=r+1}^{s-1}f(x_j).

This derivation explains both parts of the result: the interior sample size is \(m\), not \(n\), and its density is normalized by the parent probability of the interval \((a,b)\).

Conditional Distribution of One Interior Rank

Let \(X_{(r+k)}\) be the \(k\)th order statistic among the \(m\) interior observations, where \(1\le k\le m\). At a threshold \(x\) between the endpoints, each truncated draw is at or below \(x\) with probability \(G(x)\). The event \(X_{(r+k)}\le x\) occurs exactly when at least \(k\) of the \(m\) draws are at or below \(x\). Thus the conditional cdf is a binomial tail:

P\bigl(X_{(r+k)}\le x\mid X_{(r)}=a,\ X_{(s)}=b\bigr) = \sum_{j=k}^{m} \binom{m}{j}G(x)^j[1-G(x)]^{m-j}, \quad a<x<b.

Differentiating this cdf, or using the order-statistic density for \(m\) iid draws from \(g\), gives

f_{X_{(r+k)}\mid a,b}(x) = \frac{m!}{(k-1)!(m-k)!} G(x)^{k-1}[1-G(x)]^{m-k}g(x), \quad a<x<b.

These formulas concern only ranks strictly between \(r\) and \(s\). If \(s=r+1\), then \(m=0\) and there is no interior order statistic to analyze.

Worked Example: A Middle Rank in a Uniform Sample

Let \(X_1,\ldots,X_7\) be iid uniform on \((0,10)\). Given \(X_{(2)}=2\) and \(X_{(6)}=8\), find \(P(X_{(4)}\le 5\mid X_{(2)}=2,X_{(6)}=8)\).

Method and interior count. The endpoint ranks are 2 and 6, so \(m=6-2-1=3\) observations lie between them. Within these three interior observations, \(X_{(4)}\) is rank \(k=4-2=2\).

For a uniform parent distribution on \((0,10)\), \(F(x)=x/10\) on that interval. The truncated cdf at 5 is

G(5)=\frac{F(5)-F(2)}{F(8)-F(2)} =\frac{0.5-0.2}{0.8-0.2} =\frac{0.3}{0.6} =\frac12.

The second of three truncated observations is at or below 5 if at least two of the three are at or below 5. Therefore,

P(X_{(4)}\le5\mid X_{(2)}=2,X_{(6)}=8) = \binom32\left(\frac12\right)^2\left(\frac12\right) + \binom33\left(\frac12\right)^3 = \frac38+\frac18 = \frac12.

The conditional probability is \(0.5000\) to four decimal places. The result is plausible by symmetry: 5 is the midpoint of the interval \((2,8)\), and the second order statistic among three draws has probability one-half of being at or below that midpoint. The method avoids treating \(X_{(4)}\) as a rank in the original seven-observation sample after the two endpoints have been fixed.

Worked Example: Truncation for an Exponential Parent

Suppose \(X_1,\ldots,X_5\) are iid exponential with rate \(\lambda=\ln 2\) per time unit, so \(F(x)=1-e^{-(\ln 2)x}\) for \(x\ge0\). Given \(X_{(1)}=1\) and \(X_{(5)}=3\), find the conditional probability that \(X_{(3)}\le2\).

Method and rank. There are \(m=5-1-1=3\) observations between the fixed minimum and maximum. The requested \(X_{(3)}\) is the second interior order statistic, so \(k=2\).

Since \(e^{-(\ln 2)x}=2^{-x}\), the needed parent cdf values are

F(1)=1-\frac12=\frac12,\qquad F(2)=1-\frac14=\frac34,\qquad F(3)=1-\frac18=\frac78.

The conditional probability that one interior draw is at or below 2 is

G(2)=\frac{F(2)-F(1)}{F(3)-F(1)} =\frac{\frac34-\frac12}{\frac78-\frac12} =\frac{\frac14}{\frac38} =\frac23.

At least two of the three interior draws must be at or below 2. Hence

\begin{aligned} P(X_{(3)}\le2\mid X_{(1)}=1,X_{(5)}=3) &=\binom32\left(\frac23\right)^2\left(\frac13\right) +\binom33\left(\frac23\right)^3\\ &=\frac49+\frac8{27} =\frac{20}{27} \approx 0.7407. \end{aligned}

The answer is \(0.7407\) to four decimal places. This probability is conditional on the two endpoint ranks; it is not the probability that at least two of all five original observations are at or below 2.

Worked Example: A Joint Event for Two Interior Ranks

Let \(X_1,\ldots,X_8\) be iid uniform on \((0,1)\). Given \(X_{(2)}=0.2\) and \(X_{(7)}=0.8\), find the probability that \(X_{(4)}>0.5\) and \(X_{(6)}\le0.65\).

Method and rank interpretation. There are \(m=7-2-1=4\) interior observations. Thus \(X_{(4)}\) is the second interior order statistic, and \(X_{(6)}\) is the fourth, or largest, interior order statistic.

For one interior draw, the truncated probabilities are

G(0.5)=\frac{0.5-0.2}{0.8-0.2}=\frac12, \qquad G(0.65)=\frac{0.65-0.2}{0.8-0.2}=\frac{0.45}{0.6}=\frac34.

The event \(X_{(6)}\le0.65\) requires all four interior draws to be at or below 0.65. The event \(X_{(4)}>0.5\) requires the second interior order statistic to exceed 0.5, so at most one draw can be at or below 0.5. Among draws already required to be at or below 0.65, each draw has probability \(1/2\) of falling at or below 0.5, and probability \(3/4-1/2=1/4\) of falling in \((0.5,0.65]\), relative to the truncated distribution. The joint event consists of either zero draws in the lower part or exactly one:

\begin{aligned} P(X_{(4)}>0.5,\ X_{(6)}\le0.65\mid X_{(2)}=0.2,X_{(7)}=0.8) &=\left(\frac14\right)^4 +\binom41\left(\frac12\right)\left(\frac14\right)^3\\ &=\frac1{256}+\frac8{256} =\frac9{256} \approx0.0352. \end{aligned}

The answer is \(0.0352\) to four decimal places. As a check, the event is a subset of \(X_{(6)}\le0.65\), whose conditional probability is \((3/4)^4=81/256\); \(9/256\) is smaller, as required.

Common Exam Trap: Using the Parent Cdf Without Truncation

In the exponential example, a tempting but incorrect step is to use \(F(2)=3/4\) as the probability for each interior observation. That is the probability an unconditioned parent observation is at or below 2. The interior observations are already known to lie between 1 and 3, so the correct probability is \(G(2)=2/3\).

The incorrect calculation would treat the second of three observations as at or below 2 with probability

\binom32\left(\frac34\right)^2\left(\frac14\right) +\left(\frac34\right)^3 = \frac{27}{64}+\frac{27}{64} = \frac{27}{32} = 0.84375.

The correct probability is \(20/27\approx0.7407\), not \(0.84375\). The error ignores the conditioning that excludes values below 1 and above 3. A second frequent error is to count \(n\) observations rather than \(s-r-1\): once the endpoint ranks are fixed, only the observations strictly between those ranks enter the conditional order-statistic calculation.

Key takeaway: Given \(X_{(r)}=a\) and \(X_{(s)}=b\), analyze the \(m=s-r-1\) interior observations using the truncated cdf \(G(x)=[F(x)-F(a)]/[F(b)-F(a)]\). For an interior rank, the conditional cdf is the probability that a binomial count with \(m\) trials and success probability \(G(x)\) is at least the corresponding interior rank.

Check Your Understanding

For each problem, identify the number of interior observations before applying the truncated distribution.

Let \(X_1,\ldots,X_9\) be iid uniform on \((0,10)\). Given \(X_{(2)}=1\) and \(X_{(7)}=7\), find \(P(X_{(4)}\le4\mid X_{(2)}=1,X_{(7)}=7)\). (Answer: \(0.6875\))
Let \(X_1,\ldots,X_5\) be iid exponential with rate \(\ln 2\) per time unit. Given \(X_{(1)}=2\) and \(X_{(5)}=4\), find \(P(X_{(3)}\le3\mid X_{(1)}=2,X_{(5)}=4)\). (Answer: \(20/27\approx0.7407\))
Let \(X_1,\ldots,X_8\) be iid uniform on \((0,1)\). Given \(X_{(2)}=0.2\) and \(X_{(7)}=0.8\), find the probability that \(X_{(4)}>0.5\) and \(X_{(6)}\le0.65\). (Answer: \(9/256\approx0.0352\))
For an iid continuous sample, suppose \(F(a)=0.2\), \(F(b)=0.8\), and \(F(c)=0.5\), where \(a<c<b\). Given \(X_{(4)}=a\) and \(X_{(10)}=b\) in a sample of size 12, find \(P(X_{(7)}\le c\mid X_{(4)}=a,X_{(10)}=b)\). (Answer: \(0.5000\))
Let \(X_1,\ldots,X_8\) be iid uniform on \((0,12)\). Given \(X_{(2)}=3\) and \(X_{(7)}=9\), find the conditional density of \(X_{(4)}\) at \(x=6\). (Answer: \(0.25\) per unit)

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