= Solution
Here a smooth family is understood as a <smooth proper family of compact complex manifolds>: $f:\mathcal X\to S$ is a proper holomorphic submersion, with compact Kähler fibres. Properness and compactness are essential for the usual statement. Indeed, the projection $(\mathbb C_t\times\mathbb C_z)\setminus\{(0,0)\}\to\mathbb C_t$ is a holomorphic submersion with Kähler fibres, but its fibres are $\mathbb C$ for $t\ne0$ and $\mathbb C^*$ for $t=0$. Their first <Betti numbers> are zero and one, so their <cohomology> cannot give a fixed-rank variation. This supplies a counterexample if “smooth family” is read without the usual proper compact-fibre hypothesis. The <Ehresmann fibration theorem> gives local differentiable trivializations, so
$$
\mathbb H_{\mathbb Z}=R^kf_*\mathbb Z/\text{torsion}
$$
is a <local system> with fibre $H^k(X_s,\mathbb Z)/\text{torsion}$. Its associated flat bundle has the <Gauss–Manin connection>, given by differentiating <cohomology> classes in topologically constant frames.
The <Kähler Hodge decomposition> gives
$$
H^k(X_s,\mathbb C)=\bigoplus_{p+q=k}H^{p,q}(X_s).
$$
The <Betti numbers> are locally constant by differentiable triviality. The dimensions of $H^q(X_s,\Omega^p_{X_s})$ are upper semicontinuous, and their sums for fixed $p+q=k$ equal the fixed <Betti number>. Near any fixed fibre, every summand can only decrease, while their sum cannot change; hence none decreases. Thus the <Hodge numbers> are locally constant. The <relative de Rham complex> and the <Grauert base change theorem> then make
$$
F^p_s=\bigoplus_{r\ge p}H^{r,k-r}(X_s)
$$
the fibres of holomorphic subbundles $F^p\mathcal H$. More intrinsically these are the images of the <hypercohomology> of the truncated <relative de Rham complex> $\Omega_{\mathcal X/S}^{\ge p}$ in the full relative de Rham <cohomology>. Kähler degeneration of the spectral sequence ensures that these images have the required ranks.
For the transversality proof, let $A^2$ denote the ideal in the absolute holomorphic de Rham complex generated by two pulled-back base one-forms. Keeping zero or one base differential gives a <short exact sequence of cochain complexes>, with the usual shifted differential in the left-hand complex
$$
0\longrightarrow f^*\Omega_S^1\otimes
\Omega_{\mathcal X/S}^{\bullet-1}
\longrightarrow
\Omega_{\mathcal X}^{\bullet}/A^2
\longrightarrow
\Omega_{\mathcal X/S}^{\bullet}
\longrightarrow0.
$$
The connecting morphism in degree $k$ is the <Gauss–Manin connection>
$$
\nabla:\mathcal H\longrightarrow\mathcal H\otimes\Omega_S^1.
$$
To compute it, lift a relative de Rham cocycle to the middle complex, apply the absolute differential, and retain the terms with one base differential. A cocycle in filtration degree $p$ can be represented using absolute forms of degree at least $p$. In a term containing one base differential, removing that factor leaves a relative form of degree at least $p-1$. The connecting construction therefore lands in $F^{p-1}$:
$$
\boxed{\nabla F^p\subseteq F^{p-1}\otimes\Omega_S^1.}
$$
This is the filtration shift behind <Griffiths transversality>. It explains the inequality rather than merely naming a theorem. The unfiltered construction agrees with topological parallel transport by the de Rham comparison, and its square is zero, as required for the flat connection.
Consequently the fibre <cohomology> gives a pure <variation of Hodge structure>. To claim a polarized integral variation one needs the additional compatible flat polarization. In a smooth projective family, a relative ample <line bundle> supplies a flat integral <Kähler class>; its primitive pieces are polarized by the <Hodge–Riemann bilinear relations>, and the full <cohomology> is polarized using the appropriate primitive-piece signs. For a general compact Kähler family, the unpolarized variation above is automatic, but a flat integral polarization is not supplied merely by knowing that each fibre is Kähler. This distinction is particularly relevant to unrestricted K3 families.
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