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Suppose we have a random sample $X_1$, $X_2$ from the Beta($\theta$, $1$) distribution and we want to test $H_{\theta} :\theta \leq 1$ against $H_1:\theta > 1$. The following test issued: “Reject $H_0$ if and only if $3X_1 \leq 4X_2$.” How to show that the power function of the test is given by $$\beta(\theta) = 1-\frac12\left(\frac34\right)^\theta$$

My try :

This is not related to question, but I know how to solve if it is only one observation: let's say $X_1$ ~ Beta($\theta$, $1$) and the condition is $X_1 > \frac{1}{2}$ for same $H_0$ and $H_1$. To get the power function we have to solve for :

$$\beta(\theta) = P_\theta(X > 1/2) = \int_{1/2}^1 \frac{\Gamma(\theta + 1)}{\Gamma(\theta)\Gamma(1)}x^{\theta-1}(1-x)^{1-1}\mathrm dx$$

I do not know how to go about the original problem I mentioned in the question.

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The test will reject if $X_{1} < 4X_{2}/3$. Therefore

$$\beta(\theta) = P(X_{1} < 4X_{2}/3 \ | \ \theta) $$

I'm assuming that $X_{1}, X_{2}$ are independent. By smoothing,

$$\beta(\theta) = E_{X_{2}} \left( P(X_{1} < 4X_{2}/3 \ | \ \theta, X_{2} = x) \right ) = E_{X_{2}} ( F_{\theta}(4x/3) ) = \int_{0}^{1} F_{\theta}(4x/3) p_{\theta}(x) dx $$

where $F_{\theta}, p_{\theta}$ are the ${\rm Beta}(\theta,1)$ CDF and PDF, respectively. This is not a very fun integral to compute but when you do, you will find it equals $1 - (1/2)(3/4)^{\theta}$.

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