Let $X$ be a random variable with pdf $f(x;\theta)=\theta x^{\theta-1}$, $0<x<1$ and $\theta>0$. Consider $H_0: \theta=1/2$ versus $H_A: \theta=1/4$
a) Derive the most powerful test using $\alpha=0.05$
b) Compute the power of this test.
I have shown the first part of the exercise, but I don't know how to show the second part.
Answer a): $$T = \frac{L\left(\theta = \frac{1}{2}\right)}{L\left(\theta = \frac{1}{4}\right)} = 2x^{1/4}$$ $$2x^{1/4}<c$$ $$x < \frac{c^4}{16}$$ $$c_2 = \frac{c^4}{16}$$ $$\alpha = P\left( \text{Type I error} \right) = P\left( \text{reject} \space \space H_0\mid H_0 \space \space \text{true} \right) = 0.05$$ $$\alpha = P\left(x < c \mid \theta = 1/2\right)$$ $$=\int_0^{c_2}{f\left(x;1/2\right)dx}$$ $$\sqrt{c_2} = 0.05$$ $$c_2 = 0.0025$$ $$x < 0.0025$$
I don't know how to compute the power of this test or how to start. The only thing I know, is that I have to find the following:
$$1-\beta = P(\text{Reject} \space H_0 \mid H_A \space \text{true})$$ where $\beta$ is the probability of a Type II error.
Thanks in advance
simply using the definition that is
$$\gamma=\mathbb{P}[X \in C|H_1]$$
Where $C$ is the rejection region. Thus
$$\gamma=\int_0^{0.0025} 0.25 x^{0.25-1}dx=\frac{1}{2\sqrt{5}}\approx22.36\%$$