A correctly specified regular normal mean model

The prior density is continuous and strictly positive at ${\theta }^{\ast }$, and ${\theta }^{\ast }$ is interior to $\mathrm{\Theta }=\mathbb{R}$. The log-likelihood ratio at ${\theta }^{\ast }+h\mathrm{/}\sqrt{n}$ is exactly $h{\mathrm{\Delta }}_n-h^2\mathrm{/}2$, where ${\mathrm{\Delta }}_n=\sqrt{n}(\bar{X}-{\theta }^{\ast })$. The standard-normal density is even, so its mean is zero; integration by parts using ${\varphi }^{\mathrm{\prime }}(x)=-x\varphi (x)$ gives $E{Z}^2=\int x^2\varphi (x)dx=\int \varphi (x)dx=1$, with boundary term $x\varphi (x)\to 0$. By [F2], the sample mean has variance $1\mathrm{/}n$, hence $E{\mathrm{\Delta }}_n^2=1$. Since $R^2{\mathbf{1}}_{\{\mathrm{\mid }{\mathrm{\Delta }}_n\mathrm{\mid }>R\}}\le {\mathrm{\Delta }}_n^2$, $P(\mathrm{\mid }{\mathrm{\Delta }}_n\mathrm{\mid }>R)\le R^{-2}$, so ${\mathrm{\Delta }}_n=O_P(1)$. The compact-uniform LAN condition holds with $V=1$.

Multiplying the likelihood by the prior gives a kernel proportional to $\exp \{-\frac{1}{2}[n(\theta -\bar{X}{)}^2+{\tau }^{-2}{\theta }^2]\}$. Completing the square yields posterior variance $v_n=(n+{\tau }^{-2}{)}^{-1}$ and posterior mean $m_n=n\bar{X}\mathrm{/}(n+{\tau }^{-2})$. Thus the scaled posterior of $H_n=\sqrt{n}(\theta -{\theta }^{\ast })$ is $N(a_n,b_n)$, where $a_n=\sqrt{n}(m_n-{\theta }^{\ast })$ and $b_n=n\mathrm{/}(n+{\tau }^{-2})$.

We have $b_n\to 1$ and $a_n-{\mathrm{\Delta }}_n=-{\tau }^{-2}\sqrt{n}\bar{X}\mathrm{/}(n+{\tau }^{-2})=o_P(1)$ because $\sqrt{n}\bar{X}=\sqrt{n}{\theta }^{\ast }+O_P(1)$. Translation invariance and the triangle inequality reduce the total-variation distance to a centered scale change and a location shift. For $b\in [1\mathrm{/}2,2]$, the centered densities ${\varphi }_b(x)=(2\pi b{)}^{-1\mathrm{/}2}{e}^{-x^2\mathrm{/}(2b)}$ are dominated by $C{e}^{-x^2\mathrm{/}4}$ and converge pointwise to ${\varphi }_1$ as $b\to 1$, so dominated convergence gives $\mathrm{\parallel }{\varphi }_b-{\varphi }_1{\mathrm{\parallel }}_1\to 0$. For a shift $u$, the fundamental theorem of calculus gives $\mathrm{\parallel }{\varphi }_b(\cdot -u)-{\varphi }_b{\mathrm{\parallel }}_1\le \mathrm{\mid }u\mathrm{\mid }\mathrm{\parallel }{\varphi }_b^{\mathrm{\prime }}{\mathrm{\parallel }}_1=C\mathrm{\mid }u\mathrm{\mid }\mathrm{/}\sqrt{b}$. These two bounds prove $\mathrm{\parallel }N(a_n,b_n)-N({\mathrm{\Delta }}_n,1){\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}=o_P(1)$ under the convention [F3].

The scaled posterior $N(a_n,b_n)$ is tight in probability because $a_n=O_P(1)$ and $b_n\le 1$. More explicitly, on $\mathrm{\mid }{a}_n\mathrm{\mid }\le R$, its mass outside $[-M_n,M_n]$ is at most $P(\mathrm{\mid }Z\mathrm{\mid }>M_n-R)$ for $Z\sim N(0,1)$; first choose $R$ by boundedness of $a_n$, then let $M_n\to \mathrm{\infty }$. Thus every-diverging-radius contraction holds, and step 2.1 directly proves the asserted posterior normal approximation.