 Solution

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