Folded medians

In the previous post, we discussed the Gastwirth’s location estimator. In this post, we continue playing with different location estimators. To be more specific, we consider an approach called folded medians. Let $x = \{ x_1, x_2, \ldots, x_n \}$ be a random sample with order statistics $\{ x_{(1)}, x_{(2)}, \ldots, x_{(n)} \}$. We build a folded sample using the following form:

If $n$ is odd, the middle sample element is folded with itself. The folding operation could be applied several times. Once folding is conducted, the median of the final folded sample is the folded median. A single folding operation gives us the Bickel-Hodges estimator.

In this post, we briefly check how this metric behaves in the case of the Normal and Cauchy distributions.

Gastwirth's location estimator

Let $x = \{ x_1, x_2, \ldots, x_n \}$ be a random sample. The Gastwirth’s location estimator is defined as follows:

where $Q_p$ is an estimation of the $p^{\textrm{th}}$ quantile (using classic sample quantiles).

This estimator could be quite interesting from a practical point of view. On the one hand, it’s robust (the breakdown point ⅓) and it has better statistical efficiency than the classic sample median. On the other hand, it has better computational efficiency than other robust and statistical efficient measures of location like the Harrell-Davis median estimator or the Hodges-Lehmann median estimator.

In this post, we conduct a short simulation study that shows its behavior for the standard Normal distribution and the Cauchy distribution.

Dynamical System Case Study 1 (symmetric 3d system)

Let’s consider the following dynamical system:

where $f(x) = \alpha / (1+x^m)$ is a Hill function. In this case study, we explore the phase portrait of this system for $\alpha = 18,\; m = 3$.

Beeping Busy Beavers and twin prime conjecture

In this post, I use Beeping Busy Beavers to show that twin prime conjecture could be proven or disproven.

Hodges-Lehmann-Sen shift and shift confidence interval estimators

In the previous two posts (1, 2), I discussed the Hodges-Lehmann median estimator. The suggested idea of getting median estimations based on a cartesian product could be adopted to estimate the shift between two samples. In this post, we discuss how to build Hodges-Lehmann-Sen shift estimator and how to get confidence intervals for the obtained estimations. Also, we perform a simulation study that checks the actual coverage percentage of these intervals.

Statistical efficiency of the Hodges-Lehmann median estimator, Part 2

In the previous post, we evaluated the relative statistical efficiency of the Hodges-Lehmann median estimator against the sample median under the normal distribution. In this post, we extended this experiment to a set of various light-tailed and heavy-tailed distributions.

Statistical efficiency of the Hodges-Lehmann median estimator, Part 1

In this post, we evaluate the relative statistical efficiency of the Hodges-Lehmann median estimator against the sample median under the normal distribution. We also compare it with the efficiency of the Harrell-Davis quantile estimator.

Expected value of the maximum of two standard half-normal distributions

Let $X_1, X_2$ be i.i.d. random variables that follow the standard normal distribution $\mathcal{N}(0,1^2)$. In the previous post, I have found the expected value of $\min(|X_1|, |X_2|)$. Now it’s time to find the value of $Z = \max(|X_1|, |X_2|)$.

Expected value of the minimum of two standard half-normal distributions

Let $X_1, X_2$ be i.i.d. random variables that follow the standard normal distribution $\mathcal{N}(0,1^2)$. One day I wondered, what is the expected value of $Z = \min(|X_1|, |X_2|)$? It turned out to be a fun exercise. Let’s solve it together!

Unbiased median absolute deviation for n=2

I already covered the topic of the unbiased median deviation based on the traditional sample median, the Harrell-Davis quantile estimator, and the trimmed Harrell-Davis quantile estimator. In all the posts, the values of bias-correction factors were evaluated using the Monte-Carlo simulation. In this post, we calculate the exact value of the bias-correction factor for two-element samples.