# Untied quantile absolute deviation

The below text contains an intermediate snapshot of the research and is preserved for historical purposes.

In the previous posts, I tried to adapt the concept of the quantile absolute deviation
to samples with tied values so that this measure of dispersion never becomes zero for nondegenerate ranges.
My previous attempt was the *middle non-zero quantile absolute deviation*
(modification 1, modification 2).
However, I’m not completely satisfied with the behavior of this metric.
In this post, I want to consider another way to work around the problem with tied values.

## Untied quantile absolute deviation

First of all, let’s recall the definition of the simple quantile absolute deviation $\operatorname{QAD}$:

$$ \operatorname{QAD}(x, p, q) = \operatorname{Q}(|x - \operatorname{Q}(x, p)|, q), $$where $\operatorname{Q}(x, p)$ is the estimation of the $p^\textrm{th}$ quantile for the given sample $x$.

The most popular example of $\operatorname{QAD}$ is the median absolute deviation $\operatorname{MAD}$:

$$ \operatorname{MAD}(x) = \operatorname{QAD}(x, 0.5, 0.5) = \operatorname{median}(|x - \operatorname{median}(x)|). $$We may have a problem when more than half of the sample elements are the same. In this case, $\operatorname{MAD}$ becomes zero regardless of the other values. For example,

$$ \operatorname{MAD}(\{ 1, 1, 1, 10, 20 \}) = 0. $$To work around this problem, let’s introduce an operator $\operatorname{U}$ that removes all duplicated values from the given sample. For example,

$$ \operatorname{U}(\{ 1, 1, 1, 10, 20 \}) = \{ 1, 10, 20 \}, $$$$ \operatorname{U}(\{ 1, 1, 1, 1, 1, 2, 2, 4, 7, 7 \}) = \{ 1, 2, 4, 7 \}. $$Now, let’s define the *untied quantile absolute deviation* $\operatorname{UQAD}$ as follows:

Similarly, we can define the *untied median absolute deviation* $\operatorname{UMAD}$:

It’s easy to see that this approach resolves the problem:

$$ \operatorname{UMAD}(\{ 1, 1, 1, 10, 20 \}) = \operatorname{median}(\operatorname{U}(|\{ 1, 1, 1, 10, 20 \} - 1|)) = \operatorname{median}(\operatorname{U}(|\{ 0, 0, 0, 9, 19 \}|)) = \operatorname{median}(\{ 0, 9, 19 \}) = 9. $$## UQAD problems

While $\operatorname{UQAD}$ solves the problem with zero dispersion on some samples, it brings other problems. Let’s consider the two following samples:

$$ x_1 = \{ 1, 1, 1, 2, 2, 3, 4, 4, 5, 5, 5 \}, $$$$ x_2 = \{ 1, 2, 2, 3, 3, 3, 4, 4, 5 \}. $$While we may expect that the dispersion of $x_1$ should be larger than the dispersion of $x_2$, we have $\operatorname{UMAD}(x_1) = \operatorname{UMAD}(x_2) = 1$. In this particular case, my second modification of $\operatorname{MNZQAD}$ has the same problem: $\operatorname{MNZQAD}(x_1) = \operatorname{MNZQAD}(x_2) = 2$.

While the $\operatorname{UQAD}$ approach could be useful in some cases, I’m still not satisfied with its behavior in the general case. An additional investigation of possible options is required.