Size Control for Conditional Stochastic Dominance
Why should a critical value computed from uniforms control a test of conditional dominance? The nearest-observation argument of (Bugni, Canay, and Kim 2025) first gives a finite limit experiment. Within that experiment, a quantile coupling bounds every null statistic by the uniform statistic. The remaining task is to transfer that bound back to the observed sample when outcomes may tie.
1. Main result and its scope
Use the independent samples and fixed retained sample sizes from the nearest-observation construction. At the target \(z_0\), write \(F_Y(t)\) and \(F_X(t)\) for the conditional cdfs. The statistic and critical value are
where \(U\) contains \(q_y+q_x\) independent uniforms, split into groups of sizes \(q_y,q_x\).
Theorem 1. Suppose each neighborhood of the target has positive covariate probability, both conditional cdfs vary continuously at the target in the uniform norm, and each target conditional cdf has only finitely many discontinuities. For every fixed distribution satisfying \(F_Y(t)\leq F_X(t)\) for all \(t\),
In the limit experiment, the largest null rejection probability is attained when both groups have the same continuous cdf. Its value is
The asymptotic assertion is pointwise in the data-generating distribution. It does not, as stated, put a supremum over distributions inside the limit. Also, \(\bar\alpha\) need not equal \(\alpha\): fixed sample sizes give a discrete rank statistic, even with continuous outcomes.
2. The useful calculation: couple the two samples
Let \(Q_Y,Q_X\) be generalized quantile functions and construct the limit observations as \(Y_j=Q_Y(U_j)\) and \(X_j=Q_X(U_{q_y+j})\). Define
The quantile identity gives, simultaneously in \(t\) up to null sets,
The inequality uses only the null and the monotonicity of \(A\). Taking suprema gives the pathwise bound
This proves size control in the limit experiment. If \(F_Y=F_X=F\) and \(F\) is continuous, its range contains \((0,1)\), so both inequalities become equalities. A continuous common distribution is therefore least favorable. With atoms, the cdf can skip intervals of \(u\), which can make the test conservative.
3. Why ties need a separate argument
At tied vectors, a rank statistic can jump under an arbitrarily small perturbation. Thus \(S_n\Rightarrow S\) alone does not justify convergence of its rejection probability.
An almost-sure coupling deals with strict inequalities: if two limit coordinates are separated, their order eventually agrees. At an atom \(d\), one additionally needs the selected outcome to hit \(d\), rather than merely approach it. Uniform convergence of the marginal cdfs supplies this stronger conclusion.
To see the distinction, if \(V_n\to V\) in probability and \(\|F_{V_n}-F_V\|_\infty\to0\), then
For fixed \(\varepsilon\), the second probability is bounded by the corresponding punctured-neighborhood probability for \(V\), plus a term tending to zero. Send \(n\) to infinity and then \(\varepsilon\) to zero. Summing over the finite set of atoms preserves all relevant ties with probability tending to one. Together with order preservation away from ties, this transfers the limit-experiment bound back to the original test.
4. The permutation connection
Without ties, the rank patterns from label permutations have the same law as those from independent uniforms. This explains the shared critical value. Under strict dominance, the groups themselves need not be exchangeable; validity comes from the coupling inequality.
With ties, the equivalence can fail. If \(q_y=q_x=1\) and both observed outcomes are equal, every permutation has statistic zero. For two independent continuous uniforms, the statistic is zero or one, each with probability one half. At level \(0.1\), the uniform critical value is one, while the tied-sample permutation critical value is zero. The dominance proof above remains valid; exchangeability is not a substitute for it.
5. Filling in the atom calculation
The distinction between approaching an atom and hitting it is easy to miss. Take \(V=d\) almost surely and \(V_n=d+1/n\). Then \(V_n\to V\), but \(V_n\) never equals \(d\). Moreover,
This example converges in value but not uniformly in cdf. The uniform-cdf assumption rules out moving an entire atom a vanishing distance away.
Let \(\eta_n=\|F_{V_n}-F_V\|_\infty\). The same bound holds for the left limits of the cdfs. For a fixed \(\varepsilon>0\),
The punctured intervals shrink to the empty set as \(\varepsilon\downarrow0\), so their probability tends to zero even when \(d\) has positive mass. Combining this display with the coupling inequality above proves that the probability of losing a specified atomic value tends to zero.
There are only finitely many coordinates in the retained vector. For any pair whose limit values differ, convergence under the coupling eventually preserves its order. For a pair whose limit values agree, independence of the limit coordinates implies that equality can have positive probability only at a common atom. Indeed, for independent \(V,W\),
A finite union bound over common atoms and coordinate pairs now preserves all ties with probability tending to one. Strict orders are preserved by the coupling as well. On that event, the grouped ranks and hence the statistic agree exactly. This proves convergence of rejection probabilities even when the critical value is an atom of the statistic’s law.
6. A complete critical-value calculation with four observations
Suppose \(q_y=q_x=2\). Under independent continuous uniforms, the six possible label sequences in increasing outcome order are equally likely. Reading a \(Y\) raises the empirical-cdf difference by \(1/2\), and reading an \(X\) lowers it by \(1/2\). The statistic is the largest partial sum, including zero.
| Increasing label order | Partial sums | \(T\) |
|---|---|---|
| \(YYXX\) | \(1/2,1,1/2,0\) | \(1\) |
| \(YXYX\) | \(1/2,0,1/2,0\) | \(1/2\) |
| \(YXXY\) | \(1/2,0,-1/2,0\) | \(1/2\) |
| \(XYYX\) | \(-1/2,0,1/2,0\) | \(1/2\) |
| \(XYXY\) | \(-1/2,0,-1/2,0\) | \(0\) |
| \(XXYY\) | \(-1/2,-1,-1/2,0\) | \(0\) |
Counting the six paths gives
For \(\alpha=0.2\), strict rejection above \(1/2\) has probability \(1/6\). For \(0.1\), the critical value is \(1\) and strict rejection never occurs. Increasing the source sample cannot change this discreteness in the fixed retained sample.
One can obtain exact least-favorable size by randomizing at the boundary. If \(c\) is a critical value, reject above \(c\) and, using an independent uniform random variable, reject at \(c\) with probability
For the \(0.2\) example this is \((1/5-1/6)/(1/2)=1/15\). The rejection probability as a function of \(T\) is nondecreasing. Applying the same extra uniform to the coupled statistics therefore preserves the dominance bound. This is an optional randomized version; the theorem above uses strict, nonrandomized rejection.
6.1. Count the reference paths for arbitrary retained sizes
The four-observation calculation generalizes without simulating any outcome values. A continuous pooled sample has \(q_y\) labels \(Y\) and \(q_x\) labels \(X\), with all \(\binom{q_y+q_x}{q_y}\) label orders equally likely. After reading \(i\) labels \(Y\) and \(j\) labels \(X\), the cdf difference is \(i/q_y-j/q_x\).
For \(c\geq0\), let \(N_c(i,j)\) count label paths reaching \((i,j)\) while keeping every partial difference at most \(c\). Initialize \(N_c(0,0)=1\), and assign zero to counts with negative indices. At every other grid point,
A path reaches the grid point by appending either \(Y\) or \(X\). The indicator keeps it only if the new partial difference stays below the boundary. Dividing the count by the number of label orders gives
For \(q_y=q_x=2\) and \(c=1/2\), this count is five, recovering the cdf value \(5/6\) from the table. At \(c=0\), only two paths survive. Integer counting avoids Monte Carlo error and keeps the test’s strict inequality visible when choosing a critical value.
The event \(T=1\) requires every \(Y\) to appear before every \(X\), so its reference probability is exactly \(\binom{q_y+q_x}{q_y}^{-1}\). If the nominal level is smaller than this probability, the critical value is one and strict rejection is impossible. This supplies a direct check when selecting fixed retained sample sizes. The source sample can grow indefinitely without changing this rank resolution.
The construction combines two arguments. Localization approximates conditional distributions at the specified covariate values; the coupling bounds the composite-null statistic by a finite rank experiment. Exchangeability of the two conditional outcome groups is unnecessary.
7. Why size and power must be discussed separately
Imagine an alternative where every \(Y\) outcome is smaller than every \(X\) outcome. Then \(T=1\) with probability one: the ordering is always \(YYXX\). At level \(0.2\) the test rejects with probability one. At level \(0.1\) the same nonrandomized test has power zero because its critical value is also one. Even complete separation cannot overcome a rank resolution that makes rejection impossible.
For less extreme alternatives, limiting power may lie strictly between zero and one. The fixed-sample experiment continues to govern the limit as the source sample grows. Retained sample sizes therefore determine both the resolution of the test and its limiting power.