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I am having problems trying to solve the following problem:

Let $X = (X_{1}, X_{2}, ..., X_{n})$ be a random sample, where $X_{1}$ has pdf given by

$f(X_{1};\theta) = \theta x_{1}^{\theta -1} \mathbb{1}[ x_{1} \in (0,1 )]$, where $\theta \in \left\{ \theta_{0}=0 ,\theta_{1}=2 \right\}$ is unknown.

Show that the best test of $H_{0}:\theta = 1$ is given by a critical function of the form:

$\phi(x)= \mathbb{1} \bigg[ \prod_{i=1}^{n} x_{i} \geq c \bigg] $

What I took the ratio of the likelihood functions replacing $\theta$ with the values $1$ and $2$ and then set that ratio to be less or equal to $k$ and then got $\prod_{i=1}^{n} x_{i} \geq 1 - \frac{k}{2^{n}}$.

I am not sure if this is correct, can somebody please tell me if I'm doing something wrong?

thanks!

juajua
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1 Answers1

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I assume you are starting with this:

$$ \frac {\displaystyle \prod_{i=1}^n \theta_0 x_i^{\theta_0-1}} {\displaystyle \prod_{i=1}^n \theta_1 x_i^{\theta_1-1} } \leq k \iff \left(\frac {\theta_0} {\theta_1}\right)^n\left(\prod_{i=1}^nx_i\right)^{\theta_0-\theta_1} \leq k $$

Of course when the question put $\theta_0 = 1, \theta_1 = 2$ into it, it will greatly simplify the expression. Your result seems not quite matching with this, but the direction of the inequality is right (which is the most important part).

In general, when $\theta_0 < \theta_1$, the function $f(x) = x^{\theta_0 - \theta_1}$ is a decreasing function in $x$, therefore the above inequality becomes

$$ \prod_{i=1}^nx_i \geq \left[\left(\frac {\theta_1} {\theta_0}\right)^nk\right]^{\frac {1} {\theta_0-\theta_1}} $$

So as long as you arrive at this step, you can claim that you reject $H_0: \theta = \theta_0$ when $$ \prod_{i=1}^nx_i \geq c$$

where $c$ is some constant depends on the significance level of the test. So this is exactly what the question want you to prove.

BGM
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