Why Quantile Brackets Reduce to Counts
A pair of sample order statistics can bracket a population quantile with a probability that is calculated exactly, without estimating a standard error or using a large-sample approximation. The key is to count how many observations fall on one side of the quantile. For an independent sample, that count has a binomial distribution.
This connection is useful when choosing order-statistic endpoints for a finite-sample interval. It also makes an important distinction clear: the probability that a particular sample bracket contains a fixed population quantile is a coverage probability, not a probability distribution for the quantile itself.
Definition and Exact Bracketing Result
Let \(X_1,\ldots,X_n\) be independent and identically distributed from a continuous distribution function \(F\). Write their order statistics as \(X_{(1)}\le\cdots\le X_{(n)}\). For \(0<p<1\), define the population \(p\)-quantile by \(q_p=\inf\{x:F(x)\ge p\}\). Continuity gives \(F(q_p)=p\).
The result depends on the sample being independent and identically distributed and on the population having no atom at \(q_p\). A continuous distribution function ensures the latter. No particular parametric form for \(F\) is needed.
Proof. Each indicator \(\mathbf{1}\{X_i\le q_p\}\) is a Bernoulli random variable with success probability \(P(X_i\le q_p)=F(q_p)=p\). The indicators are independent because the observations are independent. Their sum therefore has the stated binomial distribution.
If \(X_{(r)}\le q_p\), at least \(r\) observations are at most \(q_p\), so \(N_p\ge r\). If \(q_p\le X_{(s)}\), at most \(s-1\) observations are strictly below \(q_p\). Since the distribution has no atom at \(q_p\), the probability that any sampled observation equals \(q_p\) is zero; consequently, the number strictly below \(q_p\) equals \(N_p\) almost surely. Thus the second inequality is equivalent, almost surely, to \(N_p\le s-1\). Combining both conditions gives \(r\le N_p\le s-1\), and summing the binomial probabilities over those values proves the formula. \(\square\)
Calculating Coverage and Its Tail Errors
The theorem shows that the bracket covers precisely when the count is neither too small nor too large. The two ways coverage can fail are disjoint: fewer than \(r\) observations fall at or below the quantile, or at least \(s\) do.
This form is convenient when binomial tail probabilities are available from a table or software. It also helps when selecting endpoints: the lower-tail noncoverage is \(P(N_p\le r-1)\), and the upper-tail noncoverage is \(P(N_p\ge s)\). Their sum is the total noncoverage. Because the binomial distribution is discrete, a desired coverage level may not be attainable exactly with any pair of integer indices.
Specify \(p\), so the count below or at \(q_p\) has success probability \(p\).
For \([X_{(r)},X_{(s)}]\), coverage requires a count from \(r\) through \(s-1\).
Calculate \(P(r\le N_p\le s-1)\), or subtract the two excluded tails from one.
Worked Example: A Bracket for the Median
For an independent sample of size \(n=10\) from a continuous population, find the probability that \([X_{(3)},X_{(8)}]\) contains the population median.
The median corresponds to \(p=1/2\), so \(N_{1/2}\sim\operatorname{Bin}(10,1/2)\). The bracket covers when \(3\le N_{1/2}\le7\). By symmetry, the excluded counts \(0,1,2\) and \(8,9,10\) have equal total probability.
For one excluded tail, the binomial probabilities have common denominator \(2^{10}=1024\), and the numerator is \(\binom{10}{0}+\binom{10}{1}+\binom{10}{2}=1+10+45=56\). Hence
Thus this bracket has exact coverage \(89.0625\%\). It does not have \(95\%\) coverage merely because it is centered around the sample median; the order-statistic indices determine the exact probability.
Worked Example: A Lower-Quantile Bracket
For \(n=8\), calculate the coverage of \([X_{(1)},X_{(4)}]\) for the population first quartile, \(q_{1/4}\).
Here \(N_{1/4}\sim\operatorname{Bin}(8,1/4)\). The bracket covers when \(1\le N_{1/4}\le3\). Using denominator \(4^8=65{,}536\), the three required binomial probabilities have numerators
Therefore,
The coverage is about \(78.61\%\). The relatively low probability reflects the particular sample size and endpoint choices; it is not changed by the shape of the continuous population distribution.
Worked Example: Unequal Binomial Tails
Suppose \(n=6\), \(p=0.8\), and the proposed bracket is \([X_{(4)},X_{(6)}]\). Find its coverage of \(q_{0.8}\).
The coverage event is \(4\le N_{0.8}\le5\), where \(N_{0.8}\sim\operatorname{Bin}(6,0.8)\). The probability at \(k=4\) is
The probability at \(k=5\) is
Adding the two possible covering counts gives \(0.24576+0.393216=0.638976\). The lower excluded tail is \(P(N_{0.8}\le3)\), while the upper excluded tail is \(P(N_{0.8}=6)\); these are not generally equal. This example shows why symmetry-based shortcuts are valid only in suitable cases, such as a median count with a symmetric binomial law.
Why the Probability Is Distribution-Free
The calculation uses the population distribution only through \(F(q_p)=p\). At the fixed quantile, every observation has probability \(p\) of falling at or below it, regardless of the distribution's scale or shape. Independence then makes the count binomial. This is why the same bracket indices have the same coverage for all continuous distributions.
This property can be used to choose an interval for a desired confidence level. For a proposed \(r,s\), calculate the two excluded binomial tails and check whether their sum is at most the permitted noncoverage. If not, change the indices and recalculate. When a nominal level cannot be achieved exactly, a conservative choice has coverage at least the target; reporting the exact value avoids implying that the coverage equals the target when it does not.
Common Pitfall
A tempting but incorrect argument is: “The two endpoints lie on opposite sides of the quantile in expectation, so the bracket must have the desired coverage.” Expected positions do not determine a probability of simultaneous bracketing. Nor does taking endpoints that appear balanced guarantee a particular confidence level.
The correction is to translate the event into a count before making any coverage claim. The lower endpoint requires at least \(r\) observations at or below \(q_p\), while the upper endpoint requires at most \(s-1\). The exact coverage is the binomial probability between those bounds. It may be above or below a proposed nominal level.
Check Your Understanding
Use the binomial-count translation to answer these questions.
- For \(n=12\), what range of values of \(N_p\) corresponds to coverage by \([X_{(2)},X_{(9)}]\)?
- Write the coverage probability for \([X_{(3)},X_{(7)}]\) as a binomial sum for a general \(p\).
- For a median bracket with \(n=8\), explain why the probabilities of the two excluded tails are equal when the indices are symmetric around the center.
- Why does continuity of the population distribution matter when translating the event \(q_p\le X_{(s)}\) into a bound on \(N_p\)?
- Does a bracket with the same order-statistic indices have the same coverage for every continuous population distribution? Explain using the binomial count.