# Standard trimmed Harrell-Davis median estimator

Update: this blog post is a part of research that aimed to build a new measure of statistical dispersion called quantile absolute deviation. A preprint with final results is available on arXiv: arXiv:2208.13459 [stat.ME]. Some information in this blog post can be obsolete: please, use the preprint as the primary reference.

In one of the previous posts, I suggested a new measure of dispersion called the standard quantile absolute deviation around the median ($$\operatorname{SQAD}$$) which can be used as an alternative to the median absolute deviation ($$\operatorname{MAD}$$) as a consistent estimator for the standard deviation under normality. The Gaussian efficiency of $$\operatorname{SQAD}$$ is $$54\%$$ (comparing to $$37\%$$ for MAD), and its breakdown point is $$32\%$$ (comparing to $$50\%$$ for MAD). $$\operatorname{SQAD}$$ is a symmetric dispersion measure around the median: the interval $$[\operatorname{Median} - \operatorname{SQAD}; \operatorname{Median} + \operatorname{SQAD}]$$ covers $$68\%$$ of the distribution. In the case of the normal distribution, this corresponds to the interval $$[\mu - \sigma; \mu + \sigma]$$.

If we use $$\operatorname{SQAD}$$, we accept the breakdown point of $$32\%$$. This makes the sample median a non-optimal choice for the median estimator. Indeed, the sample median has high robustness (the breakdown point is $$50\%$$), but relatively poor Gaussian efficiency. If we use $$\operatorname{SQAD}$$, it doesn’t make sense to require a breakdown point of more than $$32\%$$. Therefore, we could trade the median robustness for efficiency and come up with a complementary measure of the median for $$\operatorname{SQAD}$$.

In this post, we introduce the standard trimmed Harrell-Davis median estimator which shares the breakdown point with $$\operatorname{SQAD}$$ and provides better finite-sample efficiency comparing to the sample median.

### Trimmed Harrell-Davis quantile estimator

The concept of this estimator is fully covered in my recent paper [Akinshin2022]. Here I just briefly recall the basic idea.

Let $$x$$ be a sample with $$n$$ elements: $$x = \{ x_1, x_2, \ldots, x_n \}$$. We assume that all sample elements are sorted ($$x_1 \leq x_2 \leq \ldots \leq x_n$$) so that we could treat the $$i^\textrm{th}$$ element $$x_i$$ as the $$i^\textrm{th}$$ order statistic $$x_{(i)}$$. Based on the given sample, we want to build an estimation of the $$p^\textrm{th}$$ quantile $$Q(p)$$.

The classic Harrell-Davis quantile estimator (see [Harrell1982]) suggests the following approach:

$Q_{\operatorname{HD}}(p) = \sum_{i=1}^{n} W_{\operatorname{HD},i} \cdot x_i,\quad W_{\operatorname{HD},i} = I_{i/n}(\alpha, \beta) - I_{(i-1)/n}(\alpha, \beta),$

where $$I_x(\alpha, \beta)$$ is the regularized incomplete beta function, $$\alpha = (n+1)p$$, $$\;\beta = (n+1)(1-p)$$.

When we switch to the trimmed modification of this estimator, we perform summation only within the highest density interval $$[L;R]$$ of $$\operatorname{Beta}(\alpha, \beta)$$ of size $$D$$ (as a rule of thumb, we can use $$D = 1 / \sqrt{n}$$):

$Q_{\operatorname{THD},D}(p) = \sum_{i=1}^{n} W_{\operatorname{THD},D,i} \cdot x_i, \quad W_{\operatorname{THD},D,i} = F_{\operatorname{THD},D}(i / n) - F_{\operatorname{THD},D}((i - 1) / n),$

$F_{\operatorname{THD},D}(x) = \begin{cases} 0 & \textrm{for }\, x < L,\\ \big( I_x(\alpha, \beta) - I_L(\alpha, \beta) \big) / \big( I_R(\alpha, \beta) \big) - I_L(\alpha, \beta) \big) \big) & \textrm{for }\, L \leq x \leq R,\\ 1 & \textrm{for }\, R < x. \end{cases}$

Thus, we use only sample elements with the highest weight coefficients ($$W_{\operatorname{THD},D,i}$$) and ignore sample elements with small weight coefficients. It allows us to get a high statistical efficiency (which is close to the efficiency of the classic Harrell-Davis quantile estimator) and a good robustness level (in most cases, outliers have zero impact on the final result).

### Standard trimmed Harrell-Davis median estimator

The highest density interval size $$D$$ defines the portion of the sample that is actually used to estimate quantiles using $$Q_{\operatorname{THD},D}$$. Therefore, the asymptotic breakdown point of $$Q_{\operatorname{THD},D}$$ is $$1-D$$. In order to make $$Q_{\operatorname{THD},D}$$ consistent with $$\operatorname{SQAD}$$ in terms of robustness, we should use $$D=\Phi(1)-\Phi(-1) \approx 0.6827$$. We call such an estimator the standard trimmed Harrell-Davis median estimator and denote by $$Q_{\operatorname{STHD}}$$.

In the scope of this research, we are interested only in the median estimator $$Q_{\operatorname{STHD}}(0.5)$$. For the median estimator, the beta function $$\operatorname{Beta}(\alpha, \beta)$$ becomes symmetric around $$0.5$$ since $$\alpha = \beta = (n + 1) / 2$$; the interval $$[L;R]$$ becomes $$[\Phi(-1); \Phi(1)]$$.

$Q_{\operatorname{STHD}}(0.5) = \sum_{i=1}^{n} W_{\operatorname{STHD},i} \cdot x_i, \quad W_{\operatorname{STHD},i} = F_{\operatorname{STHD}}(i / n) - F_{\operatorname{STHD}}((i - 1) / n),$

$F_{\operatorname{STHD}}(x) = \begin{cases} 0 & \textrm{for }\, x < \Phi(-1),\\ \dfrac{ I_x(\frac{n+1}{2}, \frac{n+1}{2}) - I_L(\frac{n+1}{2}, \frac{n+1}{2}) }{I_R(\frac{n+1}{2}, \frac{n+1}{2}) - I_L(\frac{n+1}{2}, \frac{n+1}{2})} & \textrm{for }\, \Phi(-1) \leq x \leq \Phi(1),\\ 1 & \textrm{for }\, \Phi(1) < x. \end{cases}$

### Finite-sample efficiency of standard trimmed Harrell-Davis median estimator

In order to evaluate the actual finite-sample efficiency of $$Q_{\operatorname{STHD}}(0.5)$$, we perform a simple Monte-Carlo simulation. We enumerate various sample sizes; for each sample size we generate multiple random samples from the standard normal distribution; estimate the mean, the sample median, and $$Q_{\operatorname{STHD}}(0.5)$$; evaluate the relative efficiency of the sample median and $$Q_{\operatorname{STHD}}(0.5)$$ against the mean as the ratio of the corresponding variance values.

Here are the results for $$n \leq 100$$:

As we can see, for small samples, $$Q_{\operatorname{STHD}}(0.5)$$ is noticeably more efficient than the sample median.

And here are the results for $$n \leq 100\,000$$:

As we can see, asymptotically both estimators converge to the same value which is $$2 / \pi \approx 63.66%$$. It makes sense: the Harrell-Davis median estimator is asymptotically consistent with the traditional sample median (see [Yoshizawa1985]). Since $$Q_{\operatorname{STHD}}(0.5)$$ is “between” these two estimators, it is expected that all of them converge to the same value.

### References

• [Akinshin2022]
Andrey Akinshin (2022) Trimmed Harrell-Davis quantile estimator based on the highest density interval of the given width, Communications in Statistics - Simulation and Computation,
DOI: 10.1080/03610918.2022.2050396
• [Yoshizawa1985]
Carl N Yoshizawa, Pranab K Sen, and C Edward Davis. “Asymptotic equivalence of the Harrell- Davis median estimator and the sample median”. In: Communications in Statistics-Theory and Methods 14.9 (1985), pp. 2129–2136.
https://doi.org/10.1080/03610928508829034
• [Harrell1982]
Harrell, F.E. and Davis, C.E., 1982. A new distribution-free quantile estimator. Biometrika, 69(3), pp.635-640.
https://doi.org/10.2307/2335999

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