Let $U_i$ be iid and $U_i \approx U(0,1)$.
$$\begin{align} H_0: X & \approx \min(U_1,U_2,U_3) \\ H_1: X & \approx \min(U_1,U_2) \end{align}$$
Find power of this test if $\alpha = \frac{1}{8}$.
I think that by $H_0$ we have $X \approx F_X(x) = (3x-3x^2+x^3) \cdot 1_{(0,1)}(x) $ and by $H_1$ we have $X \approx F_X(x) = (2x-x^2) \cdot 1_{(0,1)}(x) $. Thus, $\frac{f_1(x)}{f_0(x)}= \frac{-2}{3(x-1)}$. Then $P_0 \left( \frac{-2}{3(X-1)} > k \right) = \cdots = P_0 \left( X < \frac{3k-2}{3k} \right) = \alpha$. Using $F$ from $H_0$ we have $1- \frac{8}{27k^3} = \alpha$. So now consider power of statistical test we have: $P_1(X<\frac{3k-2}{3k})$ and using $F$ from $H_1$ we have $P_1(X<\frac{3k-2}{3k}) = 1-(\frac{2}{3k})^2 = 1-(1-\alpha)^{\frac{2}{3}}$
But in answer I have that power is equal to $\frac{1}{4}$.
