Weighted Mann-Whitney U test, Part 1

Previously, I have discussed how to build weighted versions of various statistical methods. I have already covered weighted versions of various quantile estimators and the Hodges-Lehmann location estimator. Such methods can be useful in various tasks like the support of weighted mixture distributions or exponential smoothing. In this post, I suggest a way to build a weighted version of the Mann-Whitney U test.

We consider the one-sided Mann-Whitney U test that compares two samples $\mathbf{x}$ and $\mathbf{y}$:

$$ \mathbf{x} = ( x_1, x_2, \ldots, x_n ), \quad \mathbf{y} = ( y_1, y_2, \ldots, y_m ). $$

The U statistic for this test is defined as follows:

$$ U(x, y) = \sum_{i=1}^n \sum_{j=1}^m S(x_i, y_j),\quad

S(a,b) = \begin{cases} 1, & \text{if } a > b, \ 0.5, & \text{if } a = b, \ 0, & \text{if } a < b. \end{cases} $$

It is easy to see that $U(x, y) \in [0; n \cdot m]$. For further discussion, it is convenient to also consider a “normalized” version of $U$ that we denote by $U_\circ$:

$$ U_\circ(x, y) = \frac{U(x, y)}{n \cdot m} \in [0; 1]. $$

For the weighted version of the test, we assign vectors of weights $\mathbf{w}$ and $\mathbf{v}$ for $\mathbf{x}$ and $\mathbf{y}$ respectively.

$$ \mathbf{w} = ( w_1, w_2, \ldots, w_n ), \quad \mathbf{v} = ( v_1, v_2, \ldots, v_m ), $$

where $w_i \geq 0$, $v_j \geq 0$, $\sum w_i > 0$, $\sum v_j > 0$.

Let us consider the normalized versions of the weight vectors:

$$ \overline{\mathbf{w}} = ( \overline{w}_1, \overline{w}_2, \ldots, \overline{w}_n ), \quad \overline{\mathbf{v}} = ( \overline{v}_1, \overline{v}_2, \ldots, \overline{v}_m ), $$$$ \overline{w}_i = \frac{w_i}{\sum_{k=1}^n w_k},\quad \overline{v}_j = \frac{v_j}{\sum_{k=1}^m v_k}. $$

Let us introduce the weighted version of the $U$ statistic. Similarly to the weighted Hodges-Lehmann location estimator, I suggest using the following approach for the normalized weighted version:

$$ U_\circ^\star(\mathbf{x}, \mathbf{y}, \mathbf{w}, \mathbf{v}) = \sum_{i=1}^n \sum_{j=1}^m S(x_i, y_j) \cdot \overline{w}_i \cdot \overline{v}_j. $$

It seems that denormalizing $U_\circ^\star$ does not make a lot of sense. Firstly, the sample size in the weighted case is not unambiguously defined.¹ Secondly, the denormalized version is not much more useful than the normalized one since we cannot reuse the distribution of the classic non-weighted $U$ statistic for the weighted case. Therefore, we can continue with the normalized statistic $U_\circ^\star$.

To convert the statistic value to the p-value, we can approximate the distribution of $U_\circ^\star$ via bootstrap.