Testing $\beta_2=\beta_3=0$ in an asymptotic normal distribution

Question

Let $\beta= (\beta_1, \beta_2, \cdots, \beta_n)$ be a vector of random variables, asymptotically distributed as the normal $N(b, \Sigma)$ where $b=(b_1, \cdots, b_n)$ and $\Sigma$ is a symmetric positively defined matrix.

We want to test $\beta_2=\beta_3=0$.

How can we do this?

My guess is that $(\beta_2,\beta_3)$ is also distributed as a normal $N((b_2,b_3), \Sigma_x)$ where $\Sigma_x$ is the $2\times 2$ matrix with entries $\Sigma(2,2)$, $\Sigma(2,3)$,$\Sigma(3,2)$ and $\Sigma(3,3)$. Now we can use $((\beta_2,\beta_3) - (0,0))^T\Sigma_x ((\beta_2,\beta_3) - (0,0)) \simeq \chi^2_2$, where $\chi^2_2$ is the chi squared with $2$ degrees of freedom.

Is this correct?

Thanks!

Answer 1

Following your comments - the usual approach in this case (logistic regression, or any other GLM) is using minus twice the difference of the log-likelihood (Deviance difference), i.e., $$ -2(\log L(RM) - \log L(FM)) \sim \chi^2_q, $$ where $\log L(RM)$ and $\log L(FM)$ are the log-likelihood of the reduced\restricted model (RM) and full model (FM), respectively. $q = 2$ in your case.