Normal location fitted under the wrong variance

Let $Z\sim N(0,1)$, so . Direct integration gives $E\mathrm{\mid }Z\mathrm{\mid }=2(2\pi {)}^{-1\mathrm{/}2}{\int }_0^{\mathrm{\infty }}z{e}^{-z^2\mathrm{/}2}dz=\sqrt{2\mathrm{/}\pi }<\mathrm{\infty }$, so symmetry gives $EZ=0$. Since ${\varphi }^{\mathrm{\prime }}(z)=-z\varphi (z)$ and $z^k\varphi (z)\to 0$ at both tails, integration by parts gives $E{Z}^2=\int z^2\varphi (z)dz=1$ and $E{Z}^4=\int z^4\varphi (z)dz=3\int z^2\varphi (z)dz=3$. Hence $E_{P}X=0$, $E_P{X}^2={\tau }^2$, and $E_P{X}^4=3{\tau }^4$. The fitted log-density is ${\mathrm{\ell }}_{\theta }(x)=-\frac{1}{2}\log (2\pi )-\frac{1}{2}(x-\theta {)}^2$. The functions $1$, $X$, and $X^2$ are integrable by the finite absolute first and second moments just established. Thus the expanded log-density is a finite linear combination of integrable functions, and [F9] gives $E_P(X-\theta {)}^2=E_P{X}^2-2\theta E_{P}X+{\theta }^2$. Therefore $M(\theta )=E_P{\mathrm{\ell }}_{\theta }(X)=-\frac{1}{2}\log (2\pi )-\frac{1}{2}({\tau }^2+{\theta }^2)$, so ${\theta }^{\ast }=0$ is the unique separated maximizer, and $H=E_P{\mathrm{\partial }}_{\theta }{S}_{{\theta }^{\ast }}(X)=-1$.

The uniform criterion error is bounded by ${\sup }_{\theta \in [-1,1]}\mathrm{\mid }{P}_n{\mathrm{\ell }}_{\theta }-M(\theta )\mathrm{\mid }\le \frac{1}{2}\mathrm{\mid }{P}_n{X}^2-{\tau }^2\mathrm{\mid }+\mathrm{\mid }{P}_{n}X\mathrm{\mid }$. Put $Y_i=X_i$ and $W_i=X_i^2-{\tau }^2$. All terms of the following single-variable expansions are integrable by the finite first, second, and fourth moments in step 1.1. Linearity [F9] therefore gives ${\mathbb{E}}_P{Y}_i={\mathbb{E}}_P{W}_i=0$, ${\mathbb{E}}_P{Y}_i^2={\tau }^2$, and ${\mathbb{E}}_P{W}_i^2=3{\tau }^4-2{\tau }^2{\mathbb{E}}_P{X}_i^2+{\tau }^4=2{\tau }^4$. Moreover ${\mathbb{E}}_P\mathrm{\mid }{Y}_i\mathrm{\mid }=\tau \mathbb{E}\mathrm{\mid }Z\mathrm{\mid }<\mathrm{\infty }$ and ${\mathbb{E}}_P\mathrm{\mid }{W}_i\mathrm{\mid }\le {\mathbb{E}}_P{X}_i^2+{\tau }^2=2{\tau }^2$, so [F8] applies to these variables. By independence and [F8], all cross expectations ${\mathbb{E}}_P[Y_i{Y}_j]$ and ${\mathbb{E}}_P[W_i{W}_j]$ vanish for $i\mathrm{\ne }j$. For either centered family $V_i=Y_i$ or $V_i=W_i$, $\mathrm{\mid }{V}_i{V}_j\mathrm{\mid }\le (V_i^2+V_j^2)\mathrm{/}2$ makes every mixed product integrable. Thus [F9] permits termwise expectation of each finite square expansion and gives ${\mathbb{E}}_P[(P_{n}X{)}^2]={\tau }^2\mathrm{/}n$ and ${\mathbb{E}}_P[(P_n{X}^2-{\tau }^2{)}^2]=2{\tau }^4\mathrm{/}n$. Applying [F5] to these squared deviations gives $P_{n}X{\to }_{P}0$ and $P_n{X}^2{\to }_P{\tau }^2$ by [F7], proving the required uniform law.

The exact constrained MLE is ${\widehat{\theta }}_n={\mathrm{\Pi }}_{[-1,1]}({\bar{X}}_n)$, the projection of the sample mean onto $[-1,1]$. step 2.1 and [F7] give ${\bar{X}}_n{\to }_{P}0$, hence ${\widehat{\theta }}_n{\to }_{P}0$ and the separated-criterion consistency condition holds. On $\{\mathrm{\mid }{\bar{X}}_n\mathrm{\mid }\le 1\}$ the score equation is exact: $P_n{S}_{{\widehat{\theta }}_n}={\bar{X}}_n-{\widehat{\theta }}_n=0$. The complementary event has probability tending to zero, so $\sqrt{n}{P}_n{S}_{{\widehat{\theta }}_n}{\to }_{P}0$. The score is $S_{\theta }(x)=x-\theta$, its second moment at ${\theta }^{\ast }=0$ under $P$ is $J={\tau }^2$, and the sample Hessian is identically $-1$, so the Hessian law holds with $H=-1$.

All hypotheses of [F4] now hold. It gives $\sqrt{n}{\widehat{\theta }}_n\Rightarrow N(0,H^{-1}J{H}^{-T})=N(0,{\tau }^2)$. The fitted model's score second moment at $\theta =0$ is $E_{N(0,1)}[(X-0{)}^2]=1$, so the sandwich variance ${\tau }^2$ differs from the inverse fitted-model information $1$ because ${\tau }^2\mathrm{\ne }1$.