4.7 Practice Problems
Problem 4.7.1. Define an estimable parameter of degree \(k\), a kernel, and the \(U\)-statistic generated by a kernel. Show that every kernel may be taken symmetric without loss of generality, and give the symmetrising formula. Where to start: the definitions open Section 3.1; the symmetrising formula is the average of the kernel over all \(k!\) permutations of its arguments.
Problem 4.7.2. Show that \(\gamma = \sigma ^{2}\) is estimable of degree \(2\), that \(h(x_1,x_2)=\tfrac 12(x_1-x_2)^{2}\) is a symmetric kernel for it, and that the resulting \(U\)-statistic is the usual unbiased sample variance \(S^{2}\).
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Solution. Take \(h(x_1,x_2)=x_1^{2}-x_1x_2\). Then \(E(h)=E(X_1^{2})-E(X_1)E(X_2)=\sigma ^{2}\), so \(\gamma \) is estimable of degree at most \(2\); it is not of degree \(1\) because no function of a single observation has expectation \(\sigma ^{2}\) for all \(F\). Symmetrising, \[h^{*}(x_1,x_2)=\tfrac 12\left [(x_1^{2}-x_1x_2)+(x_2^{2}-x_1x_2)\right ] =\tfrac 12(x_1-x_2)^{2}.\] Then \[U=\frac {1}{\binom {n}{2}}\sum _{i<j}\tfrac 12(X_i-X_j)^{2} =\frac {1}{n-1}\sum ^{n}_{i=1}\left (X_i-\overline {X}\right )^{2}=S^{2},\] using the identity \(\sum _{i<j}(X_i-X_j)^{2} = n\sum _i\left (X_i-\overline {X}\right )^{2}\).
Problem 4.7.3. Derive the general expression for the variance of a one-sample \(U\)-statistic in terms of the quantities \(\zeta _c\), and deduce that \[\lim _{n\to \infty } n\operatorname {var}\left (U\right ) = k^{2}\zeta _1 .\] Where to start: expand the square of the double sum, classify pairs of terms by the number \(c\) of shared observations, and count how many pairs fall in each class.
Problem 4.7.4. Prove that a \(U\)-statistic converges in quadratic mean to \(\gamma \), and hence in probability. State the moment condition required. Where to start: unbiasedness makes the mean squared error equal the variance, which the previous problem shows is \(O(1/n)\). The condition is a finite second moment of the kernel.
Problem 4.7.5. State Hoeffding’s one-sample \(U\)-statistic theorem, and prove it using the projection principle. Identify precisely the step at which the projection is shown to be asymptotically equivalent to the \(U\)-statistic itself. Where to start: Section 3.3. The equivalence step is the demonstration that \(n\,E\{(U_n-\gamma -V^*_n)^2\}\to 0\).
Problem 4.7.6. Let \(X_1,\dots ,X_m\) and \(Y_1,\dots ,Y_n\) be independent samples from continuous distributions \(F\) and \(G\). Show that \(\gamma = P(X<Y)\) is estimable of degree \((1,1)\), identify the kernel, and show that the resulting \(U\)-statistic is the Mann–Whitney statistic divided by \(mn\). Where to start: Example 3.4.3.
Problem 4.7.7. Using the two-sample variance formula, verify that under \(H_0:F=G\) \[E\left (U_{XY}\right )=\frac {mn}{2},\qquad \operatorname {var}\left (U_{XY}\right )=\frac {mn(m+n+1)}{12},\] and explain where the factor \(\tfrac {1}{12}\) originates. Where to start: Theorem 3.4.6 with \(r=s=1\); the \(\tfrac {1}{12}\) is the variance of \(F(X)\), uniform by Theorem 2.2.1.
Problem 4.7.8. Explain, with reference to the parameter actually estimated, why a significant Mann–Whitney test does not by itself establish a difference in medians. State the additional assumption under which it does. Where to start: the note following Example 3.4.3. The parameter estimated is \(P(X<Y)\), not a difference of centres.
Problem 4.7.9. Define the Pitman asymptotic relative efficiency and explain why the alternatives must be allowed to approach the null at rate \(n^{-1/2}\). What goes wrong if the alternative is held fixed?
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Solution. Under a fixed alternative the power of any consistent test tends to \(1\) as \(n\to \infty \). Two consistent tests therefore both have limiting power \(1\), and the ratio of sample sizes required for a common power ceases to distinguish them: the comparison degenerates. If instead the alternative recedes faster than \(n^{-1/2}\), the power of both tests tends to \(\alpha \) and the comparison degenerates in the other direction. At the rate \(n^{-1/2}\) the limiting power is a constant strictly between \(\alpha \) and \(1\), which is the only regime in which a non-trivial limit of \(n_2/n_1\) exists.
Problem 4.7.10. Show that the efficacy of the sign test under location shift is \(4f^{2}(0)\), and hence that \(\mathrm {ARE}(S,t)=2/\pi \) for normal data and \(2\) for the double exponential.
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Solution. Under the shift model the sign statistic \(S\) counts positive observations, so \(E_\theta (S)=nP(X>-\theta )=n\left [1-F(-\theta )\right ]\) and \(\mu '(0)=nf(0)\). The null variance is \(n/4\). Hence \[\mathrm {eff}(S)=\lim _n\frac {\left [nf(0)\right ]^{2}}{n\cdot n/4}=4f^{2}(0).\] For \(N(0,\sigma ^{2})\), \(f(0)=1/(\sigma \sqrt {2\pi })\), so \(\mathrm {eff}(S)=2/(\pi \sigma ^{2})\) and, dividing by \(\mathrm {eff}(t)=1/\sigma ^{2}\), \(\mathrm {ARE}(S,t)=2/\pi \approx 0.637\). For the double exponential with scale \(b\), \(f(0)=1/(2b)\) and \(\sigma ^{2}=2b^{2}\), giving \(\mathrm {eff}(S)=1/b^{2}\) and \(\mathrm {ARE}(S,t)=2\).
Problem 4.7.11. Verify the entry \(\mathrm {ARE}(W,t)=1\) for the uniform distribution, and \(\pi ^{2}/9\) for the logistic. Where to start: Corollary 4.4.1; for the uniform on \((0,a)\), \(\int f^2 = 1/a\) and \(\sigma ^2 = a^2/12\).
Problem 4.7.12. The Hodges–Lehmann theorem states that \(\mathrm {ARE}(W,t)\geq 108/125\) for every continuous \(F\) with finite variance. Explain in practical terms what this bound guarantees, and why the absence of any corresponding upper bound is the more important half of the result. Where to start: Theorem 4.5.1 and the note following it; express the bound as the extra observations required per hundred.
Problem 4.7.13. A colleague proposes to test the data for normality first and then choose between the \(t\)-test and the Wilcoxon test according to the outcome. Using the results of this chapter, give two distinct objections to this procedure.
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Solution. First, the two-stage procedure has no valid level. The choice of test is made using the same data as the test itself, so the nominal significance level of whichever test is finally applied is not its actual level; the preliminary test also has its own error rates, and a failure to reject normality is not evidence of normality, particularly at the small sample sizes where the choice matters most.
Second, the gains do not justify it. By Theorem 4.5.1 the most that can be lost by always using the Wilcoxon test is about \(13.6\%\) in efficiency, and under normality only \(4.5\%\); the most that can be lost by wrongly using the \(t\)-test is unbounded. A procedure that risks an unbounded loss to chase a gain of at most a few percent is not a good trade, quite apart from its invalidity.
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