I'm trying to follow the princeton review of likelihood theory. They define Fisher’s score function as The first derivative of the log-likelihood function, and they say that the score is a random vector. E.g for the Geometric distribution:

$$ u(\pi) = n(\frac{1}{\pi} - \frac{\bar{y}}{1-\pi} ) $$

And I can see that it is indeed a function (of the parameter $\pi$), and it is random, as it involves $\bar{y}$.

BUT then they say something I don't understand: "the score evaluated at the true parameter value $\pi$ has mean zero" and they formulate it as $E(u(\pi)) = 0$. What does it mean to evaluate it at the "true parameter value" and then find out its mean? And in the Geometric example, if I use the identity $E(y) = E(\bar{y}) = \frac{1-\pi}{\pi}$ won't I immediately get that $E(u(\pi)) = 0$? what does the "true parameter value" has to do with this?

I'm trying to follow the princeton review of likelihood theory. They define Fisher’s score function as The first derivative of the log-likelihood function, and they say that the score is a random vector. E.g for the Geometric distribution:

$$ u(\pi) = n\left(\frac{1}{\pi} - \frac{\bar{y}}{1-\pi} \right) $$

And I can see that it is indeed a function (of the parameter $\pi$), and it is random, as it involves $\bar{y}$.

BUT then they say something I don't understand: "the score evaluated at the true parameter value $\pi$ has mean zero" and they formulate it as $E(u(\pi)) = 0$. What does it mean to evaluate it at the "true parameter value" and then find out its mean? And in the Geometric example, if I use the identity $E(y) = E(\bar{y}) = \frac{1-\pi}{\pi}$ won't I immediately get that $E(u(\pi)) = 0$? what does the "true parameter value" has to do with this?