This question is inspired from Quantum Optics. There one has a nonnegative normalized function, "Husimi function" such that
$$ \int\int dx dp Q(x,p) = 1, $$
where the integrals are done over the whole complex plane. However, in the literature it is insisted time and again that this is only a quasi-distribution, in the sense that the marginals are not obtained from it, e.g.
$$ \int dp Q(x,p) \neq P(x), $$
where $P(x)$ is the probability function over the $x$-axis. However, it might be that $\int dp Q(x,p)= R(x) $ such that $\int R(x)=1$, positive definite and at the end $R(x)$ "resembles" a probability density function in its own right.
With this $R(x)$, is it still possible to say that $Q(x,p)$ is a probability function instead of a "quasi-probability"? Does it exist a probability function $Q'(x,p)$ nonnegative and normalized to one such that $\int dp Q'(x,p) = R'(x)$ but $\int R(x)\neq 1$?
It might help you to refer to this PSE question and its answer. Your texts should have emphasized why all these phase-space distributions, semi-definite-positive or not, cannot be probability distributions.
The reason is they all fail Kolmogorov's third axiom that different points of the plane represent mutually exclusive logical alternatives, which, you might know from physics, contiguous points within an $\hbar$ distance do not, by dint of the uncertainty principle.
The positive semi-definiteness itself of Q is a red herring, ritually confusing the novice. Suppose one has a function f(x), whose expectation value one wishes to find. Your texts should have described why $$ \langle f(x)\rangle \neq \int\int\!\!dpdx ~ Q(x,p) f(x) = \int\!\! dx ~ R(x) f(x). $$
The easiest way to see this is to refer Q to the Wigner distribution W, (the Cartesian coordinate system of these distributions, so to speak!), whose marginal is a bona-fide probability distribution P(x), $$ \langle f(x)\rangle = \int\int\!\!dpdx ~ W(x,p) f(x) = \int\!\! dx ~ P(x) f(x). $$
You thus utilize the fact that the Hussimi is the Weierstrass transform of the Wigner distribution, $$ Q(x,p) = \exp \left({ 1\over 4}(\partial_x^2 +\partial_p^2 ) \right) W(x,p) \\ \propto \int \int dx' dp' \exp \left( -{(x'-x)^2-(p'-p)^2 } \right) ~ W(x',p') , $$ to see that its marginal $$ \int \int \int dp dx' dp' \exp \left( -{(x'-x)^2-(p'-p)^2 } \right) ~ W(x',p') , $$ is not the probability distribution P(x) we already saw.
Instead, you'd need something like
$$ \langle f(x)\rangle = \int\!\!dx ~ \left (\int dp \exp \left(-{ 1\over 4}(\partial_x^2 + \partial_p^2) \right) Q(x,p)\right )~~ f(x) = \int\!\! dx ~ P(x) f(x). $$
Damning the torpedoes to go full speed ahead, and using plain Q s without Weierstrass kernels produces Wehrl quasi-entropies , uncontrolled approximants of the von Neumann entropy.