= Solution
Put
$$
\overline X=\frac1n\sum_{i=1}^nX_i,
\qquad
S=\sum_{i=1}^n(X_i-\overline X)^2,
\qquad
\widehat\sigma^2=\frac Sn.
$$
The <log-likelihood>, apart from an additive constant, is
$$
\ell_n(\mu,\sigma^2)
=-\frac n2\log\sigma^2
-\frac1{2\sigma^2}\sum_{i=1}^n(X_i-\mu)^2.
$$
Over the full parameter space its <maximum likelihood estimation>[maximum-likelihood estimates] are $(\widehat\mu,\widehat\sigma^2)=(\overline X,S/n)$. Under $H_0$, maximizing over the nuisance parameter $\mu$ gives $(\widehat\mu_0,\sigma_0^2)=(\overline X,1)$. Thus the generalized <likelihood-ratio test> statistic, in the convention $2(\sup_\Theta\ell_n-\sup_{\Theta_0}\ell_n)$, is
$$
\begin{aligned}
\Lambda_n(\Theta,\Theta_0)
&=2\{\ell_n(\overline X,S/n)-\ell_n(\overline X,1)\}\\
&=S-n-n\log(S/n)\\
&=n\{\widehat\sigma^2-1-\log\widehat\sigma^2\}.
\end{aligned}
$$
Under $H_0$, the normal-sample decomposition gives
$$
S\sim\chi^2_{n-1}.
$$
Writing $Z_n=(S-(n-1))/\sqrt{2(n-1)}$, the <central limit theorem> for a sum of squared independent standard normal variables yields $Z_n\xrightarrow dN(0,1)$. Hence <Slutsky theorem> gives
$$
\sqrt n(\widehat\sigma^2-1)
=\frac{S-n}{\sqrt n}
=\sqrt{\frac{2(n-1)}n}\,Z_n-\frac1{\sqrt n}
\xrightarrow dN(0,2),
$$
and in particular $\widehat\sigma^2\xrightarrow P1$.
For $h(x)=x-1-\log x$, <Taylor theorem> at $x=1$ gives
$$
h(x)=\frac12(x-1)^2+o((x-1)^2).
$$
Consequently
$$
\Lambda_n
=\frac12\{\sqrt n(\widehat\sigma^2-1)\}^2+o_P(1)
\xrightarrow d Z^2,
\qquad Z\sim N(0,1),
$$
by the <continuous mapping theorem> and Slutsky's theorem. Since the square of a standard normal variable has the <chi-squared distribution> with one degree of freedom,
$$
\boxed{\Lambda_n(\Theta,\Theta_0)\xrightarrow d\chi_1^2.}
$$
Back to article page
