For a median from n = 20, how does an order-statistic interval get its coverage?

## The trick in one sentence Turn a question about a value into a question about a count, because the count has a distribution you know exactly. ## Setting it up Take 20 independent observations from a continuous distribution and sort them: `X_(1) <= X_(2) <= ... <= X_(20)`. These sorted values are the order statistics. Let `m` be the unknown population median, defined by `P(X < m) = 0.5`. Now define `B` as the number of observations that land below `m`. Each observation is below `m` with probability 0.5 by the definition of the median, and the observations are independent, so `B ~ Binomial(20, 0.5)` Notice what is *not* in that statement: any mention of the shape, spread, skewness or density of the underlying distribution. That is the whole point. Continuity (so ties have probability zero) and independence are the only assumptions. ## From a count to an interval Consider the candidate interval `[X_(r), X_(s)]` with `r < s`. When does it fail to contain `m`? - It sits entirely above `m` when fewer than `r` observations are below `m`, that is `B <= r - 1`. - It sits entirely below `m` when at least `s` observations are below `m`, that is `B >= s`. So the coverage probability is `P(r <= B <= s - 1)`. Choose the endpoints symmetrically, `s = n + 1 - r`, and this becomes `P(r <= B <= n - r)`. ## Doing the arithmetic for n = 20 Take `r = 6`, so `s = 15` and the interval is `[X_(6), X_(15)]`. Coverage is `P(6 <= B <= 14) = 1 - 2 * P(B <= 5)` by the symmetry of the Binomial(20, 0.5) distribution. `P(B <= 5)` sums the binomial coefficients 1, 20, 190, 1140, 4845 and 15504, giving 21700 out of `2^20 = 1048576`, or about 0.0207. Coverage is therefore about `1 - 0.0414 = 0.959`. Step one rank inward, `r = 7`, and the interval `[X_(7), X_(14)]` covers with probability about 0.885. There is nothing between 0.885 and 0.959, so at n = 20 there is no interval of this family with exactly 95% coverage. The convention is to take the conservative one: 95.9% coverage from the 6th and 15th sorted values. ## Why the coverage is exact, not asymptotic Every step above used a finite binomial probability. No central limit theorem was invoked, no density was estimated, no normal approximation was made. The stated coverage holds at n = 20 exactly, which is unusual and valuable - most small-sample intervals only promise approximate coverage. This is the clean counterpoint to the fact that a sample quantile's standard error depends on an unknown density: here you never need that density, because you traded the value question for a count question. ## The costs **Discreteness.** Achievable levels jump. You cannot dial in 95% precisely, and at very small n the gaps are large - the family may offer only a few levels at all. **Endpoints are data points.** The interval can only start and end at observed values, so it inherits their granularity and cannot extend beyond the sample range. **Ties break it.** The Binomial(20, 0.5) argument assumes `P(X = m) = 0` . With heavily rounded or discrete data, observations land exactly on the median and the probability of being strictly below is no longer 0.5, so the stated coverage is wrong - typically conservative, but no longer exact. **Efficiency.** Because it makes no assumptions, it cannot exploit any. When you genuinely know the distributional family, a parametric interval will be narrower at the same level. ## Generalising past the median The same construction works for any quantile `q_p`: the count below `q_p` is Binomial(n, p), and you pick ranks whose binomial tail probabilities sum to your target error. For extreme `p` and small `n` the achievable ranks run off the end of the sample, which is another way of seeing that a p99 from a few hundred points is barely supported by data at all. ## The one-line version Rank order carries distribution-free information: how many points beat the true median is binomial no matter what the data look like, and that count is enough to build an exact interval from two sorted values.