= Solution
Choose real <eigenfunctions> with $\int_0^L W_nW_m\,dx=\delta_{nm}$. For positive modes, <self-adjointness> and <integration by parts> diagonalize the energy:
$$
\mathcal E=\frac12\sum_n\mu_na_n^2,\qquad \mu_n=A k_n^4.
$$
At <temperature> $T$, the <canonical ensemble> is a product of centered <Gaussian distributions>. The <equipartition theorem>, with <Boltzmann constant> $k_B$, gives
$$
\langle a_n\rangle=0,\qquad\langle a_na_m\rangle=\frac{k_BT}{\mu_n}\delta_{nm}.
$$
The resulting <thermal covariance of an elastic filament> is
$$
\boxed{\operatorname{Cov}(h(x),h(y))=\frac{k_BT}{A}\sum_{n:\,k_n>0}\frac{W_n(x)W_n(y)}{k_n^4},\qquad
\operatorname{Var}(h(x))=\frac{k_BT}{A}\sum_{n:\,k_n>0}\frac{W_n(x)^2}{k_n^4}.}
$$
For an unnormalized <eigenfunction> of squared <L2 norm> $N_n$, divide its summand by $N_n$. This normalization factor cannot be absorbed silently into the modal <variance>.
For <clamped boundary conditions> at both ends, the inverse of $d^4/dx^4$ has <Green function>, for $x\leq y$,
$$
G(x,y)=\frac{x^2(L-y)^2}{6L^3}[3Ly-(L+2y)x],\qquad G(x,y)=G(y,x)\ \text{for }x\geq y.
$$
It is cubic on each side of $y$, satisfies the four clamped conditions, has continuous first two derivatives, and has unit jump in its third derivative. Thus $\partial_x^4G=\delta(x-y)$, and its <eigenfunction expansion> is the sum above. In particular,
$$
\boxed{\operatorname{Var}(h(x))=\frac{k_BT}{3A L^3}x^3(L-x)^3,\qquad
\operatorname{Var}(h(L/2))=\frac{k_BT L^3}{192A}.}
$$
These finite <variances> require removal of every <zero-energy filament mode>. An unconstrained free-free <elastic filament> can translate and tilt at no energy cost; a torqued-torqued <elastic filament> can translate. Their unrestricted <Boltzmann distributions> are not normalizable, so the full displacement <variance> is undefined. Fix those rigid degrees of freedom before applying the positive-mode formula.
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