Question

Let \(X\) have a pdf of the form \(f(x ; \theta)=\theta x^{\theta-1}, 0

Short answer

The power function of a test represents the probability of correctly rejecting the null hypothesis for each possible true value of the parameter. To compute it, integrate over the critical region, switching out the distribution according to which hypothesis you're considering. In this case, under \(H_0\), the power is computed using the distribution \(f(x;1)\), and under \(H_1\), using \(f(x;2)\).

Step-by-step solution

Identify the Distributions

As given, under \(H_0: \theta=1\), the probability density function (pdf) is \(f(x; 1)=x^{1-1}=1\), for \(0<x<1\). Under \(H_1: \theta=2\), the pdf is \(f(x; 2)=2x^{2-1}=2x\), for \(0<x<1\). We'll use these in the next steps.

Compute the Power Under \(H_0\)

Using the definition of the critical region, we find the area under the \(H_0\) distribution that corresponds to the rejection region. Here, \(C=\{(x_{1}, x_{2}): \frac{3}{4} \leq x_{1}x_{2} \}\). To solve for the power, double integrate the density under \(H_0\) over the values of \((x_{1},x_{2})\) that fall within \(C\), using the limits \(x_1^2 \leq 4/3\) and \(0 < x_{2} < 1\). The integral represents the probability of a type I error, or rejecting \(H_0\) when it's true.

Compute the Power Under \(H_1\)

Now we find the power under \(H_1\). We do the same as in Step 2, but with the density under \(H_1\). The integral represents the probability of correctly rejecting \(H_0\), or the power of the test against the specific alternative \(H_1\).