Intrinsic parity of fermions and anti-fermions: why are they opposite? Is there any reason as to why we assign opposite parities to fermions and anti-fermions?
As far as I know, the only theoretical reason is that this is what Dirac's equation predicts, but this seems very limited in my opinion. I was wondering if, on a more fundamental level, this could be deduced from the properties of the Lorentz Group or the spin-statistic theorem$^{[1]}$.
In the same spirit, is there any reason to assign the same parity to bosons and anti-bosons? Again, is this just a convention, or does it follow from some fundamental requirement, like Lorentz invariance?

$^{[1]}$ or even the CPT theorem. This would be slightly contrived and roundabout in my opinion, but it is certainly more fundamental than Dirac's equation.
 A: If you consider the Dirac field $\psi$ as a linear combination of the particle annihilation part $\psi^+$ and the antiparticle creation part $\psi^{-c}$, then we have the requirement that $P\psi(x)P^{-1}$ should be proportional to $\psi(Px)$, where $P$ is the parity/space inversion operator. (I commit a slight abuse of notation here - the $P$ is both the unitary operator representing parity acting on the space of states and the space inversion operator on Minkowski space sending $(t,\vec x)$ to $(t,-\vec x)$ in the argument.
In general, we have also that this should hold for $\psi^+$ and $\psi^{-c}$, i.e.
\begin{align}
P \psi^+(x)P^{-1} & = \eta^\ast b_u \psi^+(Px) \\
P \psi^{-c}(x)P^{-1} & = \eta^c b_v \psi^{-c}(Px)
\end{align}
where the $\eta^\ast$ and $\eta^c$ are the phases by which the anti-creation /annihilation operators transform and the $b_{u/v}$ are the phases by which the fundamental spinors traditionally denoted as $u_s,v_s$ transform. These formulae come about because e.g. $\psi^+(x) = \sum_s \int u_s(\vec p) \exp(-\mathrm{i}px)a_s(\vec p)\mathrm{d}^3p$ schematically, and likewise for $\psi^{-c}$. One can then show that $b_u = -b_v$, and for $P\psi(x)P^{-1}$ to be propertional to $\psi(Px)$ we must then have that $\eta^\ast = -\eta^c$ because otherwise $\psi(x) = c_1 \psi^+(x) + c_2 \psi^{-c}$ cannot be proportional because the relative sign would change under parity.
For a more detailed derivation, see "The quantum theory of fields" by Weinberg, volume 1, section 5.5 as suggested in a comment by TwoBs.
