A Worked Exact-Probability Workshop for Ranks and Quantiles

From Rank Language to Exact Counts

Exact probabilities for ranks and quantiles often look complicated because the event is described through ordered observations rather than through a direct count. A useful strategy is to identify what must be counted: possible permutations of observation labels, observations on one side of a fixed threshold, or counts in several intervals. In the continuous independent-sampling settings considered here, these descriptions lead to finite calculations rather than approximations.

Throughout, let \(X_1,\ldots,X_n\) be independent and identically distributed from a continuous distribution \(F\). Continuity ensures that ties occur with probability zero, so the order statistics \(X_{(1)}<\cdots<X_{(n)}\) are distinct with probability one. The rank of \(X_i\) is the index \(r\) for which \(X_i=X_{(r)}\). For a fixed threshold \(t\), write \(N(t)=\sum_{i=1}^n \mathbf{1}\{X_i\le t\}\). The empirical distribution function is \(F_n(t)=N(t)/n\).

The condition \(F(q_p)=p\) is important: it makes the probability that one observation is at or below \(q_p\) equal to \(p\). If \(F\) has a jump or the selected quantile does not satisfy this equality, the binomial success probability must instead be \(F(q_p)\); it cannot automatically be replaced by \(p\).

Specified Ranks Are Permutation Events

Because the observations are independent with the same continuous distribution, each of the \(n!\) possible orderings of their labels is equally likely. To see why, permuting the labels leaves the joint distribution unchanged, and the no-ties event divides the sample space into \(n!\) disjoint ordering events. Their probabilities are equal and sum to one. Thus any one ordering has probability \(1/n!\).

For the second statement, fix the ranks of the \(m\) labels. The other \(n-m\) labels can be assigned to the remaining ranks in \((n-m)!\) ways. Each complete ordering has probability \(1/n!\), so adding the probabilities of these disjoint orderings gives \((n-m)!/n!\). The ranks must be distinct; assigning two different labels to the same rank is impossible.

Rank statements involving only relative order can be counted the same way. For example, among three specified labels, each of their \(3!\) relative orders is equally likely, even though the other observations may appear anywhere among them. This distinction between a relative-order event and a specified-ranks event prevents a common counting error.

Quantile Brackets Become Binomial Events

Fix \(q_p\) with \(F(q_p)=p\), and let \(N=N(q_p)\) count observations at or below it. Each indicator \(\mathbf{1}\{X_i\le q_p\}\) equals one with probability \(p\), and the indicators are independent. Consequently, \(N\) has the binomial distribution with parameters \(n\) and \(p\). The position of \(q_p\) among the order statistics is determined exactly by this count.

To prove the event equivalence, \(X_{(k)}\le q_p\) holds precisely when at least \(k\) sample observations are at or below \(q_p\), which is the event \(N\ge k\). Its complement is \(N\le k-1\). The binomial formula follows by choosing which \(r\) of the \(n\) observations are at or below \(q_p\), then multiplying the independent probabilities \(p^r(1-p)^{n-r}\). Continuity gives \(P(X_i=q_p)=0\), so using “below” instead of “at or below” would not change the probability. \(\square\)

A bracket event is handled by imposing both count restrictions at once. For integers \(a<b\), the event \(X_{(a)}\le q_p<X_{(b)}\) is equivalent to \(a\le N\le b-1\). The first inequality requires at least \(a\) observations at or below \(q_p\); the second requires at most \(b-1\). Therefore its probability is the sum of the binomial probabilities over those allowed counts.

Empirical-CDF Thresholds Use the Same Count

At a fixed \(t\), the empirical CDF is not a new source of randomness separate from the observations: it is the fraction \(N(t)/n\). If \(F(t)=p\), then \(N(t)\) is binomial with parameters \(n,p\). In particular, a threshold such as \(F_n(t)\ge c\) is equivalent to \(N(t)\ge\lceil nc\rceil\). The ceiling is needed because \(N(t)\) is integer-valued.

How to Set Up a Finite Exact Calculation

The following workflow applies to rank assignments, quantile brackets, and pointwise empirical-CDF thresholds. It is not a substitute for checking the model assumptions: it organizes the count only after the correct probability law has been identified.

Common Mistakes and Why These Calculations Matter

• Using the wrong integer cutoff. Replacing \(F_n(t)\ge0.7\) by \(N(t)\ge6\) is incorrect when \(n=10\): the equivalent condition is \(N(t)\ge7\). Multiply the threshold by \(n\), then apply the ceiling.
• Reversing an order-statistic condition. \(X_{(k)}\le q_p\) means at least \(k\) observations are at or below \(q_p\), not at most \(k\). Test the statement with a small count before summing.
• Counting labels without counting completions. Fixing \(m\) labels at specified ranks leaves \((n-m)!\) ways to arrange the other labels; the probability is not generally \(1/n^m\).
• Assuming all ranks are equally likely when ties can occur. The permutation argument requires a continuous common distribution, so ties have probability zero. With atoms, tie handling changes the event and the calculation.
• Replacing \(F(q_p)\) by \(p\) without checking. The binomial success probability is \(F(q_p)\). It equals \(p\) under the stated quantile condition, but need not do so for every quantile convention or every distribution.

These calculations matter because exact finite-sample probabilities provide calibration without relying on large-sample approximations. A rank event is controlled by exchangeability, while a threshold event is controlled by a count of independent indicators. Recognizing which structure applies makes the probability both easier to calculate and easier to audit.