Why is this a generalized pivotal quantity?

Question

I am reading "Generalized Confidence Intervals" by Weerahandi, and I'm trying to get my head around the definition of a generalized pivotal quantity. I understand what a (regular) pivotal quantity is.

This is definition 2.1 in the paper:

---

Let $ = r(\mathbf{X}; \mathbf{x}, \nu)$ be a function of $\mathbf{X}, \mathbf{x}, \nu$ (but not necessarily a function of all), where $\nu = (\theta, \delta)$. If $R$ has the following two properties, then it is said to be a generalized pivotal quantity:

1. $R$ has a probability distribution free of unknown parameters.
2. The observed pivotal, defined as $r_{\text{obs}} = r(\mathbf{x}; \mathbf{x}, \nu)$, does not depend on the nuisance parameter $\delta$.

---

I don't really have an intuition for the purpose of two independent copies of the data, yet. So, I started looking at examples, hoping that would help me. They give the following example of a pivotal quantity for the Behrens-Fisher problem:

$$ R = (\bar{X} - \bar{Y} - \theta)\left(\frac{\sigma_x^2 s^2_x / (mS_X^2) + \sigma^2_ys_y^2/(nS_Y^2)}{\sigma^2_x/m + \sigma^2_Y/n} \right)^{1/2}. $$

Why is this a generalized pivotal quantity? When you set $\mathbf{x} = \mathbf{X}$ and $\mathbf{y} = \mathbf{Y}$, then you get $(\bar{x} - \bar{y} - \theta)$, and that has a distribution that is dependent on the unknown nuisance variance parameters. Is it because the observed values are nonrandom now?

Answer 1

$(\bar{x} - \bar{y} - \theta)$ is a constant and not a random variable. $(\bar{X} - \bar{Y} - \theta)$ is the random variable. Therefore, $r_\text{obs}$ in this case does not "depend on" the variance parameters in the sense that it is not a function of the variance parameters. Hence, condition 2 --- despite being poorly worded --- is satisfied.