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?
 A: 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.
