Recall that to say that
\delta(X) = \sum_{p\le X}g(X)
possesses a limiting distribution $\mu_\delta$ with respect to the multiplicative measure $dX/X$
means that for continuous bounded functions $f$ on $\mathbf{R}$ we have:
\lim_{X \to {\infty}}\ {\frac{1}{\log(X)}}\int_0^Xf(\delta(x))\frac{dx}{x} = \int_{\mathbf{R}}f(x)d\mu_\delta(x).
The
mean of $\delta(X)$ is by definition:
{\mathcal E} : = \lim_{X \to {\infty}}\ {\frac{1}{\log(X)}}\int_0^X\delta(x)\frac{dx}{x} = \int_{\mathbf{R}}d\mu_\delta(x).
In the work of Sarnak and Fiorilli, another measure for understanding "bias behavior" is given by what one might call
the percentage of positive support (relative to the multiplicative measure $dX/X$). Namely:
\begin{align*}
{\mathcal P} & := \lim {\rm inf}_{X\to \infty}{\frac{1}{\log(X)}}\int_{2\le x \le X; \delta(x)\le 0}dx/x\\
\quad &= \lim {\rm sup}_{X\to \infty}{\frac{1}{\log(X)}}\int_{2\le x \le X; \delta(x)\le 0}dx/x
\end{align*}
It is indeed a conjecture, in specific instances interesting to us, that these limits ${\mathcal E} $ and ${\mathcal P}$ exist.
(Discuss a beautiful result of Fiorilli about ${\mathcal P}$)