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A Bernstein-von Mises theorem identifies a centered, scaled posterior distribution with the Gaussian law suggested by an efficient likelihood approximation. In an infinite-dimensional model, this conclusion requires control of prior-induced bias; a positive marginal prior density alone does not state enough assumptions.

A concrete calculation shows the issue. In the Gaussian sequence model, let $Y_k={\theta }_{0k}+n^{-1/2}{g}_k$, with ${\theta }_0\in {\ell }^2$, and assign independent priors ${\theta }_k\sim N(0,{\tau }_k^2)$ with every ${\tau }_k^2>0$ and ${\sum }_k{\tau }_k^2<\infty$. Put $a_{nk}=n{\tau }_k^2/(1+n{\tau }_k^2)$. For a nonzero coefficient vector $l\in {\ell }^2$, the linear functional $L(\theta )={\sum }_k{l}_k{\theta }_k$ has the exact posterior lawThe efficient observation statistic $L(Y)$ exists as a Gaussian noise pairing even though $Y$ itself is not in ${\ell }^2$. dominated convergence gives ${\sum }_k{l}_k^2{a}_{nk}\to \Vert l{\Vert }_2^2$. To center at $L(Y)$, however, one also needsThe remaining noise part of the scaled centering error has variance ${\sum }_k{l}_k^2(1-a_{nk}{)}^2\to 0$. Under this displayed bias condition, the conditional law of $\sqrt{n}[L(\theta )-L(Y)]$ converges in total variation distance, in sampling probability, to $N(0,\Vert l{\Vert }_2^2)$. Without bias control, convergence of posterior variances is insufficient. The calculation is specific to conjugate product Gaussian priors; general nonparametric results impose their own prior, truth and functional regularity hypotheses.^[ref]

A Gaussian net test turns a finite covering of an alternative into a test with an explicit covering-number cost. In the Gaussian sequence model $Y=\theta +n^{-1/2}g$, compare a null ${\theta }_0$ with centers ${\theta }_1,\dots ,{\theta }_N$ in L2 space satisfying $\Vert {\theta }_j-{\theta }_0{\Vert }_2\ge r>0$. Put ${\delta }_j={\theta }_j-{\theta }_0$ and reject if anyEach pairing is well-defined as an isonormal Gaussian process pairing even though the noise sequence itself is not in L2 space. Under the null, it is centered normal with variance $\Vert {\delta }_j{\Vert }_2^2/n$ before subtracting the threshold. The Gaussian tail bound and union bound giveIf $\Vert \theta -{\theta }_j{\Vert }_2\le r/4$ for some center, the mean of that center's displayed statistic is at least $\Vert {\delta }_j{\Vert }_2^2/4$, by the Cauchy-Schwarz inequality. ThusA radius-$r/4$ net with centers in the separated alternative supplies such a test for its entire covered set. To make the first error small, the squared separation $n{r}^2$ must dominate $\log N$. This is why metric complexity, rather than just individual Gaussian tails, enters composite testing.

Letso $T(\theta )=a^T({\theta }_1,\dots ,{\theta }_4)$. In the Gaussian sequence model, put $h_k=\sqrt{n}({\theta }_k-Y_k)$. For each relevant coordinate, the posterior density of $h_k$ relative to the standard-normal density $\phi$ isThe log-Lipschitz assumption impliesThese bounds provide Gaussian-integrable domination, while the ratio converges pointwise to one. Dominated convergence, coordinate independence, and the same argument after multiplying by $e^{t{a}^{T}h}$ show that, under the posterior,almost surely. The supplied moment-generating-function criterion therefore gives the finite-functional Bernstein-von Mises theoremwith uniform convergence of distribution functions.

If $z_{1-\alpha }={\Phi }^{-1}(1-\alpha )$, the posterior quantile defining $R_n$ consequently satisfiesin probability. Under $P_{{\theta }_0}^Y$,for every $n$. Quantile convergence and the Slutsky theorem now yield