Proof: Let $\LR = 2\log\lbrace L(\bth)/L(\bth_0)\rbrace = 2\lbrace\ell(\bth) - \ell(\bth_0)\rbrace$ and let $\fI = \fI(\bts)$ throughout. Under $H_0$, both $\bth$ and $\bth_0$ are $\sqrt{n}$-consistent for $\bts$ by the asymptotic normality of the MLE and of the restricted MLE, respectively (the first block of $\bth_0$ is exactly $\bt_0 = \bts_1$). Thus, both lie in a neighborhood of $\bts$ contained in $\bTs$ with probability tending to 1, which is all we need to take Taylor expansions there; furthermore, as interior maximizers, $\u(\bth) = \zero$ and $\u_2(\bth_0) = \zero$.

First, we expand the log-likelihood about $\bth$:

\[\begin{alignat*}{2} \ell(\bth_0) &= \ell(\bth) + \u(\bth) \tp (\bth_0-\bth) - \tfrac{1}{2}(\bth_0-\bth) \tp \oI_n(\bar{\bt}) (\bth_0-\bth) &\hspace{4em}& \href{taylor-expansion.html#lagrange-form-1}{\text{Taylor}}; \bar{\bt} \in LS(\bth_0, \bth) \\ \LR &= \sqrt{n}(\bth_0-\bth) \tp \{\tfrac{1}{n}\oI_n(\bar{\bt})\} \sqrt{n}(\bth_0-\bth) && \u(\bth) = \zero \\ \tag*{$\tcirc{1}$} &= \sqrt{n}(\bth_0-\bth) \tp \fI \sqrt{n}(\bth_0-\bth) + o_p(1) && \href{information-continuity.html}{\text{Info continuity}}; \href{o-notation.html}{O \text{ notation}} \end{alignat*}\]

In the last step, $\bar{\bt}$ is consistent because it lies between two consistent estimators, and $\sqrt{n}(\bth_0-\bth) = O_p(1)$ because both estimators are $\sqrt{n}$-consistent. Next, we express this in terms of the score at $\bth_0$:

\[\begin{alignat*}{2} \tfrac{1}{\sqrt{n}}\u(\bth_0) &= \tfrac{1}{\sqrt{n}}\u(\bth) - \fI \sqrt{n}(\bth_0-\bth) + o_p(1) &\hspace{4em}& \href{score-linearization.html#corollary}{\text{Score linearization}} \\ \tag*{$\tcirc{2}$} &= -\fI \sqrt{n}(\bth_0-\bth) + o_p(1) && \u(\bth) = \zero \\ \tag*{$\tcirc{3}$} \LR &= \tfrac{1}{\sqrt{n}}\u(\bth_0) \tp \fI^{-1} \tfrac{1}{\sqrt{n}}\u(\bth_0) + o_p(1) && \tcirc{1}, \tcirc{2}; \href{o-notation.html}{O \text{ notation}} \end{alignat*}\]

Finally, we relate the score at $\bth_0$ to the score at $\bts$. Letting

\[\H = \begin{bmatrix} \zero & \zero \\ \zero & \fI_{22}^{-1} \end{bmatrix},\]

we have

\[\begin{alignat*}{2} \tag*{$\tcirc{4}$} \tfrac{1}{\sqrt{n}}\u(\bth_0) &= \tfrac{1}{\sqrt{n}}\u(\bts) - \fI \sqrt{n}(\bth_0-\bts) + o_p(1) &\hspace{4em}& \href{score-linearization.html}{\text{Score linearization}} \\ \zero &= \tfrac{1}{\sqrt{n}}\u_2(\bts) - \fI_{22} \sqrt{n}(\bth_2(\bt_0)-\bts_2) + o_p(1) && \text{Block 2 of } \tcirc{4}; \u_2(\bth_0) = \zero, \bt_0 = \bts_1 \\ \sqrt{n}(\bth_2(\bt_0)-\bts_2) &= \fI_{22}^{-1} \tfrac{1}{\sqrt{n}}\u_2(\bts) + o_p(1) && \href{regularity-conditions.html}{\fI \text{ positive definite}} \\ \sqrt{n}(\bth_0-\bts) &= \begin{bmatrix} \sqrt{n}(\bt_0-\bts_1) \\ \sqrt{n}(\bth_2(\bt_0)-\bts_2) \end{bmatrix} && \href{restricted-mle.html}{\text{Def } \bth_0} \\ &= \begin{bmatrix} \zero \\ \fI_{22}^{-1} \tfrac{1}{\sqrt{n}}\u_2(\bts) \end{bmatrix} + o_p(1) && \bt_0 = \bts_1; \text{ line above} \\ \tag*{$\tcirc{5}$} &= \H \tfrac{1}{\sqrt{n}}\u(\bts) + o_p(1) && \text{Def } \H \\ \tag*{$\tcirc{6}$} \tfrac{1}{\sqrt{n}}\u(\bth_0) &= (\I - \fI\H) \tfrac{1}{\sqrt{n}}\u(\bts) + o_p(1) && \tcirc{4}, \tcirc{5} \end{alignat*}\]

Together, $\tcirc{3}$, $\tcirc{6}$, and the distribution of the score, $\tfrac{1}{\sqrt{n}}\u(\bts) \inD \Normal(\zero, \fI)$, show that $\LR$ is asymptotically a quadratic form in a multivariate normal random vector. Showing that this quadratic form follows a $\chi_r^2$ distribution is a homework problem.