# Wald test: Why do we assume normality of estimate

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?

• 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. Commented Nov 28, 2016 at 14:59

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 $$N(m,1)$$.
• Both $\theta^*$ and $\theta_0$ are fixed values, which have nothing to do with sample size $n$. Commented Nov 28, 2016 at 15:57