Question

If $X_1,X_2,\dots ,X_n$ is a random sample from a distribution having pdf of the form \(f(x ; \theta)=\theta x^{\theta-1}, 0

Short answer

Following the steps derived from likelihood ratio test, and by properly comparing the critical region criterion, the given set $C=\{(x_1,x_2,\dots ,x_n):c\le \prod _{i=1}^n{x}_i\}$ shown on the problem indeed is a best critical region when testing the hypothesis $H_0:\theta =1$ against $H_1:\theta =2$.

Step-by-step solution

Defining Things

First, let's define a few key components. Let's say our null hypothesis $H_0:\theta =1$ and alternative hypothesis $H_1:\theta =2$. The likelihood function for a chosen hypothesis, $H_i$, can be expressed as $L(\theta ;x)=\prod _{i=1}^{n}f(x_i;\theta )$ which is simply the product of the pdf, $f(x;\theta )$, for all observed samples.

Compute likelihood ratio test

Computing the likelihood ratio entails that we divide the likelihood under the null hypothesis by that of the alternative hypothesis as: $LR(x)=\frac{L(\theta =1;x)}{L(\theta =2;x)}$

Apply pdf into likelihood ratio test

We substitute the given pdf into the likelihood ratio to get: $LR(x)=\frac{1\cdot \prod _{i=1}^n{x}_i^{1-1}}{2\cdot \prod _{i=1}^n{x}_i^{2-1}}=\frac{1}{2^n\cdot \prod _{i=1}^n{x}_i}$

Applying critical region criterion

A best critical region for a likelihood ratio test is a criterion that accepts $H_0$ if and only if $LR(x)\ge k$ for some constant $k$. Thus, by substituting the expression we got above, $H_0$ will be accepted if and only if $\frac{1}{2^n\cdot \prod _{i=1}^n{x}_i}\ge k$. This simplifies to $\prod _{i=1}^n{x}_i\le \frac{1}{2^n\cdot k}$.

Given critical region comparison

The given critical region is of the shape $C=\{(x_1,x_2,\dots ,x_n):c\le \prod _{i=1}^n{x}_i\}$. We can see from the inequality above that such a shape can indeed exist, and thus the given set is a best critical region.