3

it is not clear to me what happens if I want to use the Wald test to test a simple hypothesis vs a one-sided alternative. I know that if I have $H_0: \theta = \theta_0$ vs $H_1: \theta \neq \theta_0$, the statistics $$ W =(\widehat{\theta} - \theta_0)^2I_{x_n}(\widehat{\theta}) \xrightarrow{d} \chi^2_1 $$ where $\widehat{\theta}$ is the MLE and $I_{x_n}(\widehat{\theta})$ is the sample Fisher information.

But what if $H_1:\theta \gt \theta_0$ or $H_1:\theta \lt \theta_0$? Will $W$ still converge to a $\chi^2_1$?

John Giorgio
  • 145
  • 4
  • It depends upon whether you restrict $\hat{\theta}$ to lie in the region implied by $H_0 \cup H_1$ or not. If you don't, then $\hat{\theta}$ is asymptotically Normal and everything carries through regardless of the alternative. If you do, then, as you have suspected, things break down; $\hat{\theta}$ is not Normally distributed, even asymptotically, so $W$ won't converge to a $\chi^2_1$ variate. I'm not going to comment on the appropriateness of the choice mentioned above, which is (at least one reason) why this doesn't qualify as an answer. – jbowman Jan 11 '21 at 01:52

0 Answers0