[proofplan] The proof has two parts. First, we expand the two deviances and observe that the saturated log-likelihood is the same for both models, so it cancels in the difference. The remaining statistic is exactly the likelihood-ratio statistic comparing the constrained model $M_0$ with the larger model $M_1$. Since the true parameter lies in the constrained model and the constraint has rank $q$, [Wilks' theorem](/theorems/1864) gives the limiting $\chi^2_q$ distribution. [/proofplan]

[step:Cancel the saturated log-likelihood in the deviance difference] For $i \in \{0,1\}$, the deviance of $M_i$ is \begin{align*} D_{i,n} = 2\{\ell_{\mathrm{sat},n}-\ell_{i,n}(\hat{\beta}_{i,n})\}. \end{align*} Therefore \begin{align*} D_{0,n}-D_{1,n} &= 2\{\ell_{\mathrm{sat},n}-\ell_{0,n}(\hat{\beta}_{0,n})\} - 2\{\ell_{\mathrm{sat},n}-\ell_{1,n}(\hat{\beta}_{1,n})\} \\ &= 2\{\ell_{1,n}(\hat{\beta}_{1,n})-\ell_{0,n}(\hat{\beta}_{0,n})\}. \end{align*} This proves the asserted identity between the deviance difference and twice the maximized log-likelihood difference. [/step]

[step:Identify the deviance difference as the likelihood-ratio statistic] Define the likelihood-ratio statistic \begin{align*} \Lambda_n := 2\left\{ \sup_{\beta \in \Theta_1}\ell_{1,n}(\beta) - \sup_{\beta \in \Theta_0}\ell_{0,n}(\beta) \right\}. \end{align*} Since $\hat{\beta}_{1,n}$ is a maximum likelihood estimator over $\Theta_1$ and $\hat{\beta}_{0,n}$ is a maximum likelihood estimator over $\Theta_0$, we have \begin{align*} \sup_{\beta \in \Theta_1}\ell_{1,n}(\beta) &= \ell_{1,n}(\hat{\beta}_{1,n}), \\ \sup_{\beta \in \Theta_0}\ell_{0,n}(\beta) &= \ell_{0,n}(\hat{\beta}_{0,n}). \end{align*} Hence \begin{align*} \Lambda_n = 2\{\ell_{1,n}(\hat{\beta}_{1,n})-\ell_{0,n}(\hat{\beta}_{0,n})\} = D_{0,n}-D_{1,n}. \end{align*} Thus the deviance difference is exactly the likelihood-ratio statistic for testing the constrained model $M_0$ against the larger model $M_1$. [/step]

[step:Apply Wilks theorem with $q$ independent restrictions]The hypotheses place the true parameter $\beta_*$ in $\Theta_0$, and the restriction map \begin{align*} r: \Theta_1 \to \mathbb{R}^q \end{align*} has Jacobian matrix $Jr_\beta$ of rank $q$ along $\Theta_0$ near $\beta_*$. Thus the null model is a regular codimension-$q$ submodel of the larger regular model. The stated regularity assumptions are exactly the regularity assumptions required by [Wilks' theorem](/theorems/1431). Therefore, by Wilks' theorem (citing a result not yet in the wiki: Wilks theorem), \begin{align*} \Lambda_n \xrightarrow{d} \chi^2_q. \end{align*}[/step]

[guided]We now use the asymptotic distribution theorem for likelihood-ratio statistics. The statistic already identified is \begin{align*} \Lambda_n = 2\left\{ \sup_{\beta \in \Theta_1}\ell_{1,n}(\beta) - \sup_{\beta \in \Theta_0}\ell_{0,n}(\beta) \right\}. \end{align*} Wilks' theorem applies to nested regular parametric models when the smaller model is the true model and is obtained from the larger model by imposing a fixed number of independent smooth restrictions. We verify these hypotheses. First, the models are nested because $\Theta_0 \subset \Theta_1$. Second, the theorem assumes that the true parameter satisfies $\beta_* \in \Theta_0$, so the constrained model $M_0$ is correctly specified. Third, the restriction map \begin{align*} r: \Theta_1 \to \mathbb{R}^q \end{align*} defines the constrained space by \begin{align*} \Theta_0 = \{\beta \in \Theta_1 : r(\beta)=0\}, \end{align*} and the rank condition $\operatorname{rank} Jr_\beta=q$ means that these are $q$ independent restrictions rather than redundant equations. Finally, the statement assumes the standard maximum-likelihood regularity hypotheses required for Wilks' theorem. Therefore Wilks' theorem (citing a result not yet in the wiki: Wilks theorem) gives \begin{align*} \Lambda_n \xrightarrow{d} \chi^2_q. \end{align*} This is the only asymptotic input in the proof; everything before this point was an algebraic identification of the deviance difference with the likelihood-ratio statistic.[/guided]

[step:Transfer the limiting distribution back to the deviance difference] From the identity proved above, \begin{align*} D_{0,n}-D_{1,n}=\Lambda_n. \end{align*} Combining this equality with \begin{align*} \Lambda_n \xrightarrow{d} \chi^2_q \end{align*} gives \begin{align*} D_{0,n}-D_{1,n} = 2\{\ell_{1,n}(\hat{\beta}_{1,n})-\ell_{0,n}(\hat{\beta}_{0,n})\} \xrightarrow{d} \chi^2_q. \end{align*} This is precisely the claimed asymptotic deviance comparison for the nested regular GLMs. [/step]