How can I find a level $\alpha$ most powerful test? Let $X_1,X_2,\dots ,X_n$ be a random sample from a distribution with pdf $f(x,\theta)$. Find a level $\alpha$ most powerful test of $H:\theta=\theta_0$ against $K:\theta = \theta_1$ when 
$$f(x,\theta)=\theta x^2I_{(\theta, \infty)}(x),\theta_0 \neq \theta_1$$
I consider $$\frac{p_1(x)}{p_0(x)}=\dfrac{\theta_1^n(\Pi x_i)^2 \Pi(I_{(\theta_1, \infty)}(x_i))}{\theta_0^n(\Pi x_i)^2 \Pi(I_{(\theta_0, \infty)}(x_i))}\\=(\theta_1/\theta_0)^n\dfrac{\Pi(I_{(\theta_1, \infty)}(x_i))}{ \Pi(I_{(\theta_0, \infty)}(x_i))}$$ Then I need to consider $\frac{p_1(x)}{p_0(x)}>k$ for a Neyman-Pearson test but I am stuck. Please help.
 A: First of all, $f(x;\theta) = \theta x^2 I_{(\theta, \infty)}(x)$ is not a valid probability density function (it doesn't integrate to unity). I suspect that you actually meant $f(x;\theta) = \theta x^{-2} I_{(\theta, \infty)}(x)$ which does integrate to unity for any $\theta \in \mathbb{R}$. So I will assume this from now on (and assuming $\theta_0 < \theta_1$).
To perform Neymann-Pearson, as you stated, you need to consider the how the ratio $\frac{p_1(X)}{p_0(X)}$ varies as the random variable $X$ varies, where $X$ follows the null hypothesis, in this case, we have that the $X_i \sim f(x, \theta_0)$. The intuition is that we reject $H_0$ if a realisation of the random ratio is large (it's kind of like a probabilistic 'proof' by contradiction - we know $H_0$ to be true, but find $H_1$ appears more likely than even $H_0$, which can't possibly have occurred because $H_0$ is true - this is a contradiction, therefore, $H_1$ must be true...).
Note that we only care how the ratio behaves as a function of the $x$'s, and note that $\theta_0$ and $\theta_1$ are fixed constants. Let me take off from where you left off:
$\begin{align}
\frac{p_1(x)}{p_0(x)} &= \left(\frac{\theta_1}{\theta_0}\right)^n \frac{\prod_i I_{(\theta_1, \infty)}(x)}{\prod_i I_{(\theta_0, \infty)}(x)}\\
&= \left(\frac{\theta_1}{\theta_0}\right)^n \frac{ \mathbb{1}_{x_i \geq \theta_1, \forall i}}{\mathbb{1}_{x_i \geq \theta_0, \forall i}}\\
&= \left(\frac{\theta_1}{\theta_0}\right)^n \frac{\mathbb{1}_{\text{min}(x_i) \geq \theta_1}}{\mathbb{1}_{\text{min}(x_i) \geq \theta_0}}\\
\end{align}$
Note that the indicator function in the denominator is always unity since the $x_i$'s obey $H_0$. Therefore, looking at the ratio $\frac{p_1(x)}{p_0(x)}$, we see that it is an increasing function of $\text{min}(x_i)$. Thus, our rejection criterion is equivalent to rejecting $H_0$ if $\text{min}(X_i)$ is large. I will leave it at that for you to complete...
