# Find supremum of Type II error in Neyman-Pearson framework

Let $$X_1,\dots,X_n$$ be an iid sample from an $$N(\theta,1)$$ distribution. We want to test $$H_0:\:\theta=0$$ against the alternative $$H_1\:\theta \neq 0$$ using the test statistic $$T_n(X_1,\dots,X_n) = \bar{X}_n$$ and the corresponding test function $$\delta(X_1,\dots,X_n) = \begin{cases} 1, & \text{if } |T_n(X_1,\dots,X_n| \geq Q, \\ 0, & \mbox{otherwise}\end{cases}$$ where $$Q\geq0$$.

The goal is now to find the smallest value of $$Q$$ for which the significance level is $$\alpha = 0.05$$ and then to find the supremum over the paramater space of the probability of type II error for that value of $$Q$$.

For the Type I error, I got $$\mathbb{P}_\theta(\delta = 1) = 2\cdot\Phi(-\sqrt{n}\cdot Q), \quad \theta \in \Theta_0$$ In order to fix this error at level $$\alpha = 0.05$$, we need $$2\cdot\Phi(-\sqrt{n}\cdot Q) = 0.05$$, i.e. $$Q=1.96/\sqrt{n}$$, since $$2\cdot\Phi(1.96)=2\cdot 0.025 = 0.05$$

For the Type II error, I got $$\mathbb{P}_\theta(\delta = 0) = \Phi(\sqrt{n}\cdot (Q-\theta))-\Phi(-\sqrt{n} \cdot (Q+\theta)), \quad \theta \in \Theta_1$$

I now need to find the supremum of this error over the parameter space. Intuitively, the Type II error should increase the closer $$\theta$$ gets to 0, since that is the $$\theta$$-value of $$H_0$$. This would mean that $$\text{sup}_\theta \mathbb{P}_\theta(\delta = 0) = \Phi(\sqrt{n}\cdot Q)-\Phi(-\sqrt{n} \cdot Q), \quad \theta \in \Theta_1$$ Is this correct?