How to prove the two sample Hodges-Lehmann estimate of difference in location is asymptotically normal and find its variance

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?

• 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? – Ben Dec 30 '20 at 9:57
• @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. – 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.
• 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 $N$. – John L Jan 4 at 16:18
• 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 $\theta_N=a/\sqrt{N}$. – S. Catterall Jan 4 at 16:35