Suppose one sample has distribution $F(x)$ and a second sample has distribution $F(y-\Delta)$ and assume that $F(x)$ has a density $f(x)$. Furthermore, suppose the sample sizes are $m$ and $n$ with $m+n=N$ where $\frac{m}N$ converges to $\rho$ and $\frac{n}N$ converges to $1-\rho$. Let $\hat{\Delta}$ denote the Hodges-Lehmann estimate associated with the Wilcoxon-Mann-Whitney test; that is, the median of the pairwise differences between the elements in the two samples. Equation (4.3.36) of Lehmann's Elements of Large-Sample Theory states that $$ \sqrt{N} (\hat{\Delta}-\Delta) \xrightarrow{L} N(0,\tau^2) $$ where $$\tau^2 = \frac{1}{\rho (1-\rho)} \left(\int f^2(x) dx \right)^{-2}$$ It is left as an exercise (Problem 3.8(ii)) to prove without any hints about how to prove it. Can someone provide a proof or a hint?

  • $\begingroup$ Since you are looking at pairwise differences, may we assume you have $m=n$ (i.e., both samples are of the same size)? If not, can you explain how you are "pairing" the sample values? $\endgroup$ – Ben Dec 30 '20 at 9:57
  • $\begingroup$ @Ben $m$ does not have to be the same as $n$. Here is an example: one sample is 10, 11, 12 and the other is 3, 5. All of the ways to make paired differences between the samples are : 10-3, 11-3, 12-3, 10-5, 11-5, 12-5. $\endgroup$ – John L Dec 30 '20 at 15:03

Consider $P_\Delta (\sqrt{N}(\hat{\Delta}-\Delta)\leq a)$ for some fixed $a$. We want to show that this converges to $\Phi(a/\tau)$.

Now $\hat{\Delta}-\Delta$ is the median of a set of elements of the form $Y_j-X_i-\Delta$. However, the distribution of $Y-X-\Delta$ does not depend on $\Delta$ ($Y$ is assumed to be a 'shift' of $X$ by $\Delta$) so we can reduce to the case $\Delta=0$. Therefore $P_\Delta (\sqrt{N}(\hat{\Delta}-\Delta)\leq a)=P_0(\hat{\Delta}\leq a/\sqrt{N})$ and from now on we assume that $\Delta=0$.

Note that $\hat{\Delta}\leq a/\sqrt{N}$ iff the Mann-Whitney statistic (defined at (3.2.7) in Lehmann's book) $W_{X,Y+a/\sqrt{N}}$ exceeds $nm/2$. Since $\Delta=0$, $X$ and $Y$ are identically distributed, so the distribution of $Y+a/\sqrt{N}$ is just that of $X$ shifted by $a/\sqrt{N}$. This observation allows us to make use of the results in Example 3.3.7 on the Wilcoxon two sample test. The distribution of the second sample $Y+a/\sqrt{N}$ changes as $N\to\infty$ so we can't use results on the asymptotic distribution of $W$ such as (3.2.9). Instead, we aim to use the asymptotic power result (3.3.51). In fact, $P(\hat{\Delta}\leq a/\sqrt{N})=P(W_{X,Y+a/\sqrt{N}}\geq nm/2)$ is the probability of rejection for the Wilcoxon two-sample test when $\alpha=0.5$ and the true parameter value is $a/\sqrt{N}$. So, this probability converges to $\Phi(a/\tau)$ by (3.3.51), and this is what we wanted.

The comments at the end of Example 4.3.7 in the book suggested this approach to me.

  • $\begingroup$ asymptotic results about $W_{X,Y}$ apply only when distributions of two samples are fixed and $N->\infty$. In this case, the distribution of $Y+a/\sqrt{N}$ is changing with $N$. $\endgroup$ – John L Jan 4 at 16:18
  • $\begingroup$ In (3.3.51) the distribution of one of the samples is changing as $k\to\infty$ with $\theta_k=\Delta/\sqrt{N_k}$. I'm just replacing (3.3.47) with $\theta_N=a/\sqrt{N}$. $\endgroup$ – S. Catterall Jan 4 at 16:35

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