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First fix the photon momentum convention. With $Q^2=-q^2>0$, $\nu =P\cdot q>0$ and $x=Q^2/(2\nu )>0$, an absorbed photon gives final momentum $p_X=P+q$. This is the physical momentum support of a deep-inelastic hadronic tensor. The original PDF instead prints ${\delta }^{(4)}(P-p_X-q)$ while keeping that positive-$x$ convention. Taken literally, it would requirein the Bjorken scaling limit, leaving no physical final states. The sign is therefore inconsistent with the requested nonzero deep-inelastic structure functions. Below, use the physically consistent absorption delta function ${\delta }^{(4)}(P+q-p_X)$. Alternatively all photon-momentum and scaling-variable conventions would have to be reversed together.

Work in the leading parton model: the target is unpolarized; scattering is incoherent from massless collinear quarks and antiquarks; intrinsic transverse momentum and target-mass effects are neglected. The parton distribution functions count constituents and include their colour multiplicity. At this order, neglect strong radiative corrections and the resulting logarithmic scale evolution.

Let a quark carry momentum $p=\xi P$, using a lightlike reference $P$ at leading order, and let $r=p+q$. The spin-averaged electromagnetic trace isIntegrating its one-particle final Lorentz-invariant phase space, with the $1/(4\pi )$ tensor normalization, supplies ${\delta }_{+}(r^2)/2$. Thus the spin-averaged electromagnetic parton tensor isOn physical support, $r^2=0$ gives $\xi =x$ and $p\cdot q=Q^2/2$. Define the transverse tensorsThe bracket in $w_f$ becomesThis follows by writing $p=\xi \tilde{P}-q/2$ on shell and expanding; it also verifies transversality to $q$.

The correct parton convolution normalization for a hadronic tensor isThe $1/\xi$ converts number weighting into covariant target-state normalization: a parton response per constituent is normalized by $2p^0$, whereas the hadron tensor uses $2P^0$, and $P^0/p^0=1/\xi$. Antiquark charge changes sign but its square and symmetric spin trace are the same.

Use the parton on-shell delta identity,Matching the two tensor coefficients now givesIn particular $F_2=2x{F}_1$, the Callan-Gross relation for spin-one-half partons. Scaling here means the naive leading approximation; full Quantum chromodynamics produces scale-dependent distributions and corrections.

Each quark contributes $+1$ and each antiquark contributes $-1$ to its flavour-vector charge. Sea pairs therefore cancel in the quark number sum rule:Here $N_f$ is the net flavour number, not the total number including sea pairs. This condition by itself fixes no antiquark distribution.

Under the separately imposed valence-only approximation, ${\bar{q}}_f=0$ and only up/down distributions remain. The proton has $(N_u,N_d)=(2,1)$ and the neutron has $(1,2)$. With $Q_u=2/3$, $Q_d=-1/3$, the valence-only proton and neutron structure-function moments areNo extra factor of three belongs here: colour is already included in the constituent distributions and the flavour counts. These are conditional valence-model moments, not unrestricted asymptotic statements in a theory whose evolution generates a sea.