# Why is this a generalized pivotal quantity?

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?

• $(\bar{x} - \bar{y} - \theta)$ is a constant and not a random variable; ($\bar{X} - \bar{Y} - \theta)$ is the random variable. Therefore, $(\bar{x} - \bar{y} - \theta)$ does not depend on the variance parameters in the sense that it is not a function of the variance parameters. Could that be the meaning?
– Cat
Commented Jul 22, 2021 at 3:35
• @Cat yep that’s it. Distribution is not the representation. I was about to delete the question, but I don’t want to rob you of the points if you wanted to type that up. Commented Jul 22, 2021 at 3:45

$$(\bar{x} - \bar{y} - \theta)$$ 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.