Pareto distribution: is there a uniformly most powerful test (UMP) at some level $\beta$? I have that $X_1, X_2, \dots, X_n$ is a random sample from a distribution with density
$$f_{\varphi}(x) = \dfrac{3\varphi^3}{x^4} \ \ \ x \ge \varphi > 0$$
Is there a uniformly most powerful test (UMP) at some level $\beta$ for testing $H_0 : \varphi \le \varphi_0 \ \ \ \text{vs} \ \ \ H_1 : \varphi > \varphi_0$?
This exact same question has been asked in previous years here and here, but no one has provided an answer, so I think it would be useful to finally have this answered.
 A: Solution. The sample likelihood is $$f_n(\textbf x)=\frac{3^n \varphi^{3n}}{\prod_{i=1}^n x_i^4}\mathbb 1_{\min \textbf x >\varphi}$$
and hence the likelihood ratio is
$$\frac{f_n(\textbf x|\varphi_1)}{f_n(\textbf x|\varphi_0)}=\left(\frac{\varphi_1}{\varphi_0}\right)^{3n}\frac{\mathbb 1_{\min \textbf x > \varphi_1}}{\mathbb 1_{\min \textbf x > \varphi_0}}$$
This is monotonically non-decreasing in the test statistic $T(\textbf x)=\min \textbf x$ (e.g. it is $0$ for $\varphi_0<\min \textbf x<\varphi_1$ and then a positive constant for $\min \textbf x>\varphi_1$). This allows us to use the Karlin-Rubin theorem from the UMP Wikipedia page, which states that the test
$$\delta(T(\textbf x))=\begin{cases}1&\text{if } T(\textbf x)>c\\0&\text{if }T(\textbf x)<c\end{cases}$$
where $c$ is chosen such that $E_{\varphi_0}\delta (T(\textbf x))=\alpha$ is the UMP test of size $\alpha$ for testing
$$H_0:\varphi\le \varphi_0\\
H_1:\varphi>\varphi_0$$
It remains to compute $c$. Before doing this we need the cdf of $\text{Pareto}(x_0, \alpha)$ which has density $\frac{\alpha x_0^\alpha}{x^{\alpha+1}}, x>x_0$. The cdf is $\int_{x_0}^x \frac{\alpha x_0^\alpha}{t^{\alpha+1}}dt=\alpha x_0^\alpha \frac{t^{-\alpha}}{-\alpha}\bigg |_{t=x_0}^x=-x_0^\alpha \left(x^{-\alpha}-x_0^{-\alpha}\right)=1-\left(\frac{x_0}{x}\right)^\alpha$. We also need the distribution of the minimum of $n$ iid Pareto random variables. $\Pr(\min \textbf X>x)=\Pr(X_1>x)\dots \Pr(X_n>x)=\left(\frac{x_0}{x}\right)^\alpha\dots \left(\frac{x_0}{x}\right)^\alpha=\left(\frac{x_0}{x}\right)^{n\alpha}\sim\text{Pareto}(x_0, n\alpha)$. So the minimum of $n$ iid $\text{Pareto}(\varphi_0, 3)$ random variables is $\text{Pareto}(\varphi_0, 3n)$. We have $$\begin{split}\alpha&=E_{\varphi_0}\delta (T(\textbf x))\\
&=P(T(\textbf x)>c|\varphi_0)\\
&=\left(\frac{\varphi_0}{c}\right)^{3n}\end{split}$$
Solving for $c$ finds that $c=\frac{\varphi_0}{\alpha^{\frac 1 {3n}}}$. Thus the UMP test is
$$\delta(\min \textbf x)=\begin{cases}1&\text{ if } \min \textbf x > \frac{\varphi_0}{\alpha^{\frac 1 {3n}}}\\
0&\text{ if } \min \textbf x < \frac{\varphi_0}{\alpha^{\frac 1 {3n}}}\end{cases}$$
That is we reject the null for sufficiently large minimum statistic. Hope this helps!
