Is the Hodges-Lehmann estimator $\hat\theta_{HL}=\operatorname{median}\limits_{1\le i\le j\le n}\left\{\frac{X_i+X_j}{2}\right\}$ in some sense 'optimal' for estimating the location parameter $\theta$ in a $\text{Logistic}(\theta,1)$ distribution?

The asymptotic relative efficiency (ARE) of $\hat\theta_{HL}$ with respect to the sample mean $\overline X_n$ based on a sample of size $n$ is known to be $\operatorname{ARE}(\hat\theta_{HL},\overline X_n)=\frac{\pi^2}9(>1)$. On the other hand, ARE of the sample median $\widetilde{X_n}$ with respect to $\overline X_n$ is $\operatorname{ARE}(\widetilde{X_n},\overline X)=\frac{\pi^2}{12}(<1)$. This perhaps shows that $\hat\theta_{HL}$ is 'better' than sample median when compared to $\overline X_n$; it is certainly more efficient than $\overline X_n$ unlike the sample median.

If we consider the $\text{Laplace}(\theta,1)$ distribution, then $\operatorname{ARE}(\hat\theta_{HL},\overline X_n)=1.5$ whereas $\operatorname{ARE}(\widetilde{X_n},\overline X_n)=2$. Here sample median performs better than $\hat\theta_{HL}$ (both are more efficient than $\overline X_n$) and is perhaps optimal. I can somewhat justify this based on the fact that sample median is MLE of $\theta$.

The full set of order statistics is minimal sufficient for both the Logistic and Laplace location families, but I am not sure what a 'good' estimator for location looks like in the Logistic family.


1 Answer 1


The Hodges-Lehmann estimator is asymptotically efficient, and so is asymptotically optimal. The same is true for the MLE, which doesn't have any particularly helpful formula. The MLE is not the Hodges-Lehman estimator, but they are asymptotically equivalent when the distribution truly is logistic.

This paper talks about the relationship between efficiency of M-estimators (like the MLE) and R-estimators (like the Hodges-Lehmann estimator)


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