Order Statistics For Threshold And Reliability Questions

Why Thresholds Turn Order Statistics Into Counts

In reliability problems, a system may operate if enough components survive a specified time. In a sample, a question may ask whether the second-smallest observation is below a threshold. These sound like different problems, but each can be expressed by counting how many observations lie on one side of the threshold. For an independent, identically distributed sample, that count has a binomial distribution, which gives a direct route to the desired probability.

The key is to interpret the order statistic carefully. The minimum is at most a threshold if at least one observation is at most that threshold. The maximum is at most the threshold only if every observation is. For a specified rank between these extremes, the event depends on how many observations meet the threshold condition.

Definition: Given a sample \(X_1,\ldots,X_n\), arrange its values in ascending order. The \(k\)th order statistic, written \(X_{(k)}\), is the value in position \(k\), where \(1\le k\le n\). Thus \(X_{(1)}\) is the minimum and \(X_{(n)}\) is the maximum. Unless stated otherwise, the formulas below assume the \(X_i\) are independent and identically distributed, each with cumulative distribution function \(F_X(x)=P(X_i\le x)\).

The General Threshold Formula

Fix a threshold \(x\), and let \(N_x\) be the number of sample observations at most \(x\). Each observation meets this condition with probability \(F_X(x)\). Under the iid assumption, the indicators of these events are independent, so \(N_x\) has a binomial distribution with parameters \(n\) and \(F_X(x)\). The \(k\)th observation in ascending order is at most \(x\) exactly when at least \(k\) observations are at most \(x\).

P(X_{(k)}\le x) = \sum_{j=k}^{n}\binom{n}{j} [F_X(x)]^j[1-F_X(x)]^{n-j}.

This formula applies to discrete as well as continuous distributions. In particular, \(F_X(x)\) includes any probability mass at \(x\). A continuous distribution has no probability mass at a single point, so strict and non-strict threshold events at a fixed point have the same probability. For a discrete distribution, that need not be true.

The extreme order statistics have especially simple formulas. The minimum is at most \(x\) if at least one observation is at most \(x\); the maximum is at most \(x\) if all \(n\) observations are at most \(x\).

P(X_{(1)}\le x)=1-[1-F_X(x)]^n, \qquad P(X_{(n)}\le x)=[F_X(x)]^n.

Taking complements gives useful forms for strict exceedance events:

P(X_{(1)}>x)=[1-F_X(x)]^n, \qquad P(X_{(n)}>x)=1-[F_X(x)]^n.

Why the Formula Holds

The general formula follows by partitioning the event according to the exact number of observations at most \(x\). The event \(X_{(k)}\le x\) occurs for \(N_x=k,k+1,\ldots,n\). For any particular \(j\), there are \(\binom{n}{j}\) ways to choose which \(j\) observations are at most \(x\). Each such choice has probability \([F_X(x)]^j[1-F_X(x)]^{n-j}\), by independence. These disjoint possibilities add to the stated sum.

For the minimum, it is often shorter to count the complement: the minimum exceeds \(x\) precisely when all observations exceed \(x\). Independence makes that probability \([1-F_X(x)]^n\), so subtracting it from one gives \(P(X_{(1)}\le x)\). For the maximum, the event that it is at most \(x\) requires all observations to be at most \(x\), giving \([F_X(x)]^n\).

Identify the rank.
Write the event using \(X_{(k)}\), and remember the observations are ordered from smallest to largest.

Translate the event into a count.
For \(X_{(k)}\le x\), at least \(k\) sample values must be at most \(x\). For \(X_{(k)}>x\), fewer than \(k\) can be at most \(x\).

Find the one-observation probability.
Use \(F_X(x)\) for “at most \(x\)” and \(1-F_X(x)\) for “greater than \(x\).”

Apply the binomial count.
Use a power formula for the minimum or maximum, or sum the relevant binomial probabilities for an interior rank.

Worked Threshold Problems

Worked Example: The Second-Smallest Inspection Time

Four independent inspection times are identically distributed as uniform on the interval from 0 to 10 minutes. Find the probability that the second-smallest inspection time is between 3 and 7 minutes, inclusive.

Method. First calculate the cdf of the second order statistic at each endpoint, then subtract. For a uniform distribution on \([0,10]\), \(F_X(x)=x/10\) on that interval. The observations are continuous, so including the endpoints does not change the probability.

For \(k=2\) and \(n=4\), the second order statistic is at most \(x\) if at least two inspection times are at most \(x\). Therefore,

P(X_{(2)}\le x) = 1-\binom{4}{0}F_X(x)^0[1-F_X(x)]^4 -\binom{4}{1}F_X(x)[1-F_X(x)]^3.

At \(x=7\), \(F_X(7)=0.7\) and \(1-F_X(7)=0.3\). Hence

P(X_{(2)}\le 7) =1-(0.3)^4-4(0.7)(0.3)^3 =1-0.0080-0.0756 =0.9164.

At \(x=3\), \(F_X(3)=0.3\) and \(1-F_X(3)=0.7\). Thus

P(X_{(2)}\le 3) =1-(0.7)^4-4(0.3)(0.7)^3 =1-0.2401-0.4116 =0.3483.

Subtracting gives \(P(3\le X_{(2)}\le 7)=0.9164-0.3483=0.5681\). The probability is 0.5681, or 56.81%. As a check, it lies between zero and one, and the cdf at 7 exceeds the cdf at 3 as required. The common mistake is to use the probability that a single inspection time lies between 3 and 7; the question concerns the second-smallest of all four times.

Worked Example: A System Requiring Two Surviving Components

A hypothetical system has four independent components with identically distributed lifetimes \(X_i\), each exponential with rate \(0.2\) per year. The system operates at time \(t\) if at least two components are still alive. Find the probability that it operates at \(t=\ln(4/3)/0.2\) years, and then find the probability of operation conditional on at least one component being alive.

For an exponential lifetime with rate \(0.2\) per year, \(P(X_i>t)=e^{-0.2t}\). At the stated time, \(e^{-0.2t}=e^{-\ln(4/3)}=3/4\). Thus each component survives with probability \(3/4\), and fails by \(t\) with probability \(1/4\). The time is approximately 1.4384 years, but the exact expression makes the survival probability clear.

The system operates if at least two of the four lifetimes exceed \(t\). Equivalently, \(X_{(3)}>t\): in an ascending list, the third value must exceed \(t\) for at least two values to exceed \(t\). Count the number of survivors:

P(\text{operates}) =\binom{4}{2}\left(\frac34\right)^2\left(\frac14\right)^2 +\binom{4}{3}\left(\frac34\right)^3\left(\frac14\right) +\binom{4}{4}\left(\frac34\right)^4 =\frac{54+108+81}{256} =\frac{243}{256} =0.9492.

At least one component is alive unless all four have failed. Therefore,

P(\text{at least one alive}) =1-\left(\frac14\right)^4 =\frac{255}{256}.

Operation implies that at least one component is alive, so the conditional probability is the joint event probability divided by the probability of the conditioning event:

P(\text{operates}\mid\text{at least one alive}) =\frac{243/256}{255/256} =\frac{243}{255} =0.9529.

The unconditional operating probability is 0.9492; the conditional probability is 0.9529. The latter is larger because conditioning excludes the case in which all components have failed. A common mistake is to confuse “at least two survive” with “the maximum lifetime exceeds \(t\),” which requires only one survivor. The rank \(X_{(3)}\) captures the requirement of two survivors.

Worked Example: A Discrete Threshold With Ties

Six independent sensors each malfunction by a test time with probability \(0.2\). Let \(X_i=1\) if sensor \(i\) malfunctions by that time and \(X_i=0\) otherwise. Find the probability that at least three sensors malfunction.

The variables are iid Bernoulli with \(P(X_i=1)=0.2\) and \(P(X_i=0)=0.8\). In ascending order, \(X_{(4)}>0\) exactly when at least three observations equal 1: with three malfunctions, the ordered sample has three zeros followed by three ones. This is a discrete distribution with atoms at 0 and 1, so the strict inequality in \(X_{(4)}>0\) matters.

Use the complement: fewer than three malfunctions means zero, one, or two malfunctions. The three probabilities are

\begin{aligned} P(\text{0 malfunctions})&=(0.8)^6=0.262144,\\ P(\text{1 malfunction})&=\binom{6}{1}(0.2)(0.8)^5 =6(0.2)(0.32768)=0.393216,\\ P(\text{2 malfunctions})&=\binom{6}{2}(0.2)^2(0.8)^4 =15(0.04)(0.4096)=0.245760. \end{aligned}

Therefore,

P(\text{at least 3 malfunctions}) =1-(0.262144+0.393216+0.245760) =0.098880 =0.0989.

The answer is 0.0989 to four decimal places. As a check, the excluded probabilities sum to \(0.901120\), so the event and its complement total \(1.000000\). Do not replace \(X_{(4)}>0\) with \(X_{(4)}\ge0\): the latter is always true because every \(X_i\) is nonnegative.

Assumptions and Common Traps

The binomial-count argument requires both independence and identical distributions. If the observations are independent but have different cdfs, the count at a threshold is generally not binomial with one common success probability. If the observations are dependent, multiplying their threshold probabilities is not justified without additional information about the dependence. In either situation, the simple powers and binomial sums above may not apply.

Reversing the order. \(X_{(1)}\) is the minimum, not the maximum. An event involving “at least \(r\) observations exceed \(x\)” corresponds to a rank near the upper end of the ascending list.
Using the wrong tail. \(F_X(x)\) is \(P(X_i\le x)\); the probability of a strict exceedance is \(1-F_X(x)\).
Ignoring atoms. For a discrete distribution, \(P(X_i<x)\), \(P(X_i\le x)\), and \(P(X_i>x)\) must be distinguished. Ties at the threshold can have positive probability.
Applying the formula to the wrong model. A shared environmental shock, or components with different lifetime distributions, can invalidate the iid binomial calculation.

For reliability questions, translate the verbal requirement before calculating. “All components survive” is a minimum-lifetime event; “at least one survives” is a maximum-lifetime event; “at least \(r\) survive” is an interior order-statistic event. This translation prevents a rank error before it reaches the arithmetic.

Key takeaway: For iid observations, a threshold event for \(X_{(k)}\) is a statement about a binomial count. In particular, \(X_{(k)}\le x\) means at least \(k\) observations are at most \(x\). Check the order, threshold inequality, and iid assumptions before using the formula.

Check Your Understanding

For each problem, identify the threshold count before calculating.

Three iid observations are uniform on \([0,1]\). What is the probability their maximum is at most \(0.6\)? (Answer: 0.2160)
Five iid component lifetimes have \(F_X(t)=0.2\). What is the probability that the minimum lifetime exceeds \(t\)? (Answer: 0.3277)
For four iid continuous observations, \(F_X(c)=0.4\). Find the probability that the second-smallest observation is at most \(c\). (Answer: 0.5248)
Six independent devices each survive a test threshold with probability \(0.3\). Find the probability that at least four survive. (Answer: 0.0705)
Three iid continuous observations satisfy \(F_X(c)=0.5\). Find the probability that their maximum is at most \(c\), given that at least one observation is at most \(c\). (Answer: 0.1429)

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