Suppose, I am using Wald test to test following hypothesis. $$H0:\theta = \theta_0$$ $$H1:\theta \neq \theta_0$$

Given the MLE estimate $\hat{\theta}$ , Wald test makes the following normality assumption: $$\frac{\hat{\theta}-\theta_0}{\hat{se}}\rightsquigarrow \mathcal{N}(0,1)$$

Suppose $\theta = \theta^*$ is the true value of parameter, then for MLE estimate $\hat{\theta}$ we know: $$\frac{\hat{\theta}-\theta^*}{\hat{se}}\rightsquigarrow \mathcal{N}(0,1)$$

Why can we make the assumption of normality around $\theta_0$, when we know $\hat{\theta}$ is normal around $\theta^*$ and still trust the results?

  • $\begingroup$ You are misinterpreting the arrow symbol. It means that the quantity approaches the N(0,1) distribution as n tends to infinity. It is a central limit result. Now I will answer the question based on the proper interpretation. $\endgroup$ – Michael R. Chernick Nov 28 '16 at 14:59

The implied result requires assuming the null hypothesis is true. We need theta star to equal theta 0 and this will happen if the null hypothesis is true. Otherwise it will be a non-zero constant say m. In that case the asymptotic distribution will be N(m,1).

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    $\begingroup$ Both $\theta^*$ and $\theta_0$ are fixed values, which have nothing to do with sample size $n$. $\endgroup$ – user2329744 Nov 28 '16 at 15:57
  • $\begingroup$ Sorry I meant that theta star would have to equal theta 0 i.e. the true theta is the null value of theta. I will edit my post. $\endgroup$ – Michael R. Chernick Nov 28 '16 at 23:56
  • $\begingroup$ If you down voted my answer please note my correction. $\endgroup$ – Michael R. Chernick Nov 28 '16 at 23:58
  • $\begingroup$ Thanks for your answer. I didn't down vote it. $\endgroup$ – user2329744 Nov 29 '16 at 6:45

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