# Median of the shifts vs. shift of the medians, Part 2: Gaussian efficiency

In the previous post, we discussed the difference between shifts of the medians and the Hodges-Lehmann location shift estimator. In this post, we conduct a simple numerical simulation to evaluate the Gaussian efficiency of these two estimators.

### Estimators

We consider two samples of equal size $$n$$: $$x = \{ x_1, x_2, \ldots, x_n \}$$, $$y = \{ y_1, y_2, \ldots, y_n \}$$. We define the shifts of the medians as

$\newcommand{\DSM}{\Delta_{\operatorname{SM}}} \DSM = \operatorname{median}(y) - \operatorname{median}(x).$

and the Hodges-Lehmann location shift estimator as

$\newcommand{\DHL}{\Delta_{\operatorname{HL}}} \DHL = \operatorname{median}(y_j - x_i).$

We also consider the classic estimator that estimates the difference of means:

$\newcommand{\Dbase}{\Delta_{\operatorname{0}}} \Dbase = \operatorname{mean}(y) - \operatorname{mean}(x).$

The Gaussian efficiency of $$\DSM$$ and $$DHL$$ can be defined as follows:

$e(\DSM) = \frac{\mathbb{V}[\Dbase]}{\mathbb{V}[\DSM]},\quad e(\DHL) = \frac{\mathbb{V}[\Dbase]}{\mathbb{V}[\DHL]}.$

### Numerical simulations

We conduct the following simulation:

• Enumerate the sample size $$n$$ from $$3$$ to $$100$$.
• For each $$n$$, generate $$100\,000$$ pairs of random samples from $$\mathcal{N}(0, 1)$$.
• For each pair of samples, estimate the shift between them using $$\Dbase$$, $$\DSM$$, and $$\DHL$$.
• Calculate the Gaussian efficiency of $$\DSM$$ and $$DHL$$ using the above equations.

Here are the results:

As we can see, the Hodges-Lehmann location shift estimator is much more efficient than the shift of the medians.

### References

• [Hodges1963]
Hodges, J. L., and E. L. Lehmann. 1963. Estimates of location based on rank tests. The Annals of Mathematical Statistics 34 (2):598–611.
DOI:10.1214/aoms/1177704172

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