Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 32 / 5

Let a group $G$ act on the sample space. A statistic $T$ is a maximal invariant when it is invariant and separates group orbits:Together these conditions say that $T(x)=T(y)$ exactly when $x$ and $y$ belong to the same orbit.

The basic result is that invariant tests factor through a maximal invariant. Specifically, a test function $\phi (x)\in [0,1]$ is invariant exactly when it can be written $b(T(x))$. To prove the nontrivial direction, define $b(t)=\phi (x)$ for any representative with $T(x)=t$. Two representatives are in the same orbit by maximality, and invariance gives them the same test value, so $b$ is well-defined. Conversely $b(T(gx))=b(T(x))$ proves invariance. Measurability is taken on the quotient statistic space; for the finite rank spaces used below it is automatic.

If a null family is transitive under the group action, the null law of $T$ is parameter-free. Indeed, if $P_{{\theta }^{\prime }}$ is the law of $gX$ when $X\sim P_{\theta }$, then for every measurable set $B$,Thus invariant procedures use exactly the information in $T$, and its common null law calibrates tests without estimating the transformed nuisance distribution. This explains the role of maximal invariants in nonparametric statistics.

For simultaneous strictly increasing bijections of the real line, the labeled ranks of observations are a maximal invariant on samples without ties. Increasing maps preserve every comparison. Conversely, two such samples with the same ranks can be matched by a piecewise linear increasing bijection through their corresponding ordered observations, extended with positive-slope tails. This proves ranks as a maximal invariant under increasing transformations. Under independent sampling from any common continuous distribution, every labeled ordering is equally likely by exchangeability. Hence rank tests have distribution-free null calibration even without a global transitivity assertion for the entire class of continuous distributions.

Use the usual two mutually independent samples, and rank their $N=n+m$ pooled observations increasingly. Continuity gives no ties with probability one. Let $R_i$ be the pooled rank of $X_i$ and define the Wilcoxon rank sum test statisticThe last identity counts the ranks contributed by the other sample after subtracting the within-$X$ ranks.

When $F_X\ge F_Y$, the $X$ observations are stochastically smaller, so reject for small $W_X$, equivalently small $U_X$. For instance,The inequality follows from $F_X\ge F_Y$ and $\int F_{Y}d{F}_Y=1/2$ for continuous $F_Y$. A common-uniform quantile coupling also makes this direction transparent: $F_X^{-1}(u)\le F_Y^{-1}(u)$, and $U_X$ is nondecreasing in each $X_i$.

Under the common-distribution null hypothesis, the pooled observations are exchangeable. Every selection of $n$ ranks for the $X$ labels has probability ${\left(\genfrac{}{}{0ex}{}{N}{n}\right)}^{-1}$. Thus the exact null law isChoose a lower-tail critical value with probability at most the desired size, optionally randomizing at its boundary to attain the size exactly. This law is independent of the unknown common distribution. Common increasing transformations leave the pooled ranks of observations unchanged, so the statistic is a function of the maximal invariant from part (i). Although a rank sum is not itself maximal, it is an invariant test statistic. These facts justify the test within nonparametric statistics.

Sampling ranks without replacement givesso ${\mathbb{E}}_0{U}_X=nm/2$ with the same variance. For $n,m\to \infty$ with $n/N\to \rho \in (0,1)$, the normal approximation for a rank sum statesEquivalently replace $W_X-n(N+1)/2$ by $U_X-nm/2$. The large-sample one-sided test rejects when this standardized statistic is below the lower $\alpha$ normal distribution quantile. This is the two-sample rank-sum test, rather than the paired Wilcoxon signed-rank test.