# Show that power of size $\alpha$-test is larger than power of level $\alpha$-test

I have two power functions of the null hypothesis, $Q(\theta_0)$ and $Q^*(\theta_0)$ where $Q$ represents $\psi(\mathbf{y})$, a size $\alpha$-test, and $Q^*$ represents $\psi^*(\mathbf{y})$, a level $\alpha$-test.

How do I start to show that $Q(\theta_0)\geq Q^*(\theta_0)$?

I previously proved that $0\leq Q(\theta_a)-Q^*(\theta_a)-k(Q(\theta_0)-Q^*(\theta_0))$.

• Is this question from a course or textbook? If so, please add the [self-study] & read its wiki. – gung Apr 14 '15 at 14:36
• It's a small part of a much larger problem that eventually proves the Neyman-Pearson Lemma. I'm working through the proof myself and this is the part that I'm stuck with. So this question represents the previous work that I've accomplished thus far. I'll add the appropriate tag. – user73522 Apr 14 '15 at 14:39

If the test with power function $Q$ has $$\alpha = \sup_{\theta \in \Theta_0} Q(\theta)$$

then it is size $\alpha$; similarly if the test with power function $Q^*$ has

$$\alpha \geq \sup_{\theta \in \Theta_0} Q^*(\theta)$$ then it is level $\alpha$.

In the context of the Neyman-Pearson lemma we have simple hypotheses so $\Theta_0 = \{\theta_0\}$ which means that

$$\alpha = Q(\theta_0)$$ and $$\alpha \geq Q^*(\theta_0).$$

Putting these together we have that

$$Q(\theta_0) = \alpha \geq Q^*(\theta_0).$$