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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?

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    $\begingroup$ 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. $\endgroup$ Commented Nov 28, 2016 at 14:59

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The implied result requires assuming the null hypothesis is true. We need $\theta^*$ 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 $n$. $\endgroup$ Commented Nov 28, 2016 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. $\endgroup$ Commented Nov 28, 2016 at 23:56

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