Why is the pseudomedian better than the median in a Wilcoxon test?

Question

The wilcox.test function in R calculates the pseudomedian and a confidence interval when conf.int=TRUE. In this question, Wilcoxon signed rank test - help on interpretation of pseudo median for example, there is a description about the calculation of the pseudomedian, but I don't understand why the confidence interval cannot be done only with the median, but with the median of the means.

Answer 1

I assume you are asking about the Wilcoxon signed-rank test.

So consider a single sample location problem where $X_1,X_2,\ldots,X_n$ are i.i.d with distribution $F(x-\theta)$, under the assumption that $F$ is continuous, $\theta$ is the unique population median and $F(\cdot)$ is symmetric about $0$.

In this setup, the sample pseudomedian is the Hodges-Lehmann estimator of $\theta$. It is a consistent and median-unbiased estimator. Note that the population pseudomedian coincides with the population median because $F$ is symmetric.

The Wilcoxon signed-rank statistic for testing $H_0:\theta=0$ is

$$T_n=\sum_{i=1}^n I(X_i>0)R_i^+\,,$$

where $R_i^+$ is the rank of $|X_i|$ among $\{|X_1|,|X_2|,\ldots,|X_n|\}$.

This can be rewritten in the form

$$T_n=\sum_{1\le i\le j\le n}I\left(\frac{X_i+X_j}{2}>0\right)$$

In other words, $$T_n=\sum_{k=1}^{m}I(Z_k>0)\,,$$

where $Z_1=X_1,\,Z_2=\frac{X_1+X_2}{2},\,Z_3=\frac{X_1+X_3}{2},\,\ldots,Z_m=X_n$ with $m=n+\binom{n}{2}=\frac{n(n+1)}{2}$.

Suppose the alternative hypothesis is $H_1:\theta>0$, so that a right-tailed test based on $T_n$ is appropriate.

Now, under $H_0$, distribution of $T_n=T_n(X_1,\ldots,X_n)$ is symmetric about $E_{H_0}(T_n)=\frac{n+\binom{n}{2}}{2}=\frac{n(n+1)}{4}$.

So, under $\theta$, distribution of $T_n(X_1-\theta,\ldots,X_n-\theta)$ is also symmetric about $\frac{n(n+1)}{4}$.

While constructing the Hodges-Lehmann estimator of location, we estimate $\theta$ by $\hat\theta$ such that

$$T_n(X_1-\hat\theta,\ldots,X_n-\hat\theta)\approx \frac{n(n+1)}{4}$$

It can be shown that this leads to the sample pseudomedian

$$\hat\theta=\operatorname*{med}_{1\le i\le j\le n}\left\{\frac{X_i+X_j}{2}\right\}$$