To construct the UMP test, we have to construct the corresponding MP test. Hence, the LR is given by
\begin{align}
\frac{L_{1}(X_1,..,X_n|\lambda_1)}{L_{1}(X_1,..,X_n|\lambda_0)} = \frac{\prod_{i=1}^n \frac{e^{-\lambda_1} \lambda_1^{x_i}}{x_i!}}{\prod_{i=1}^n \frac{e^{-\lambda_0} \lambda_0^{x_i}}{x_i!}} &= \exp\{n(\lambda_0 - \lambda_1)\}\left(\frac{\lambda_1}{\lambda_0}\right)^{\sum_i^nx_i}\\
&= \exp\{n(\lambda_0 - \lambda_1)\}\left(\frac{\lambda_1}{\lambda_0}\right)^{n\bar{x}_n} > c\,.
\end{align}
This statistic depends on the distribution of $\bar{X}_n$, hence for large enough $n$ we can use the CLT to approximate the rejection region. Note that the MP is $\Psi(\mathrm{X}) = \mathcal{I}\left( \bar{X}_n >c' \right)$.
\begin{align}
\alpha = \mathbb{E}_{\lambda_0}\Psi(\mathrm{X}) &= \mathbb{P}_{\lambda_0}\left( \bar{X}_n >c' \right)\\
&a\approx 1-\phi\left(\frac{c'-\lambda_0}{\sqrt{\lambda_0/n}} \right)\\
&c' = \lambda_0 + Z_{1-\alpha}\sqrt{\lambda_0/n}.
\end{align}
For $\lambda_0 = 1$,
$$
c' = 1 + Z_{1-\alpha}\sqrt{1/n}\,.
$$
- Power function for the parametric space $\Lambda = \mathbb{R}^+$.
\begin{align}
\pi(\Psi(\mathrm{X})|\Lambda) &= \mathbb{E}_{\Lambda}\Psi(\mathrm{X}) = \mathbb{P}_{\Lambda}(\bar{X}_n>c')\\
&=1-\phi\left( \frac{c'-\lambda}{\sqrt{\lambda/n}} \right),& \forall \lambda \in \Lambda.
\end{align}
Before trying to find a UMP test, one needs to first check if there exists one. To do this one needs to find the likelihood ratio function
$$l(x)=f_{\theta_1}(x)/f_{\theta_0}(x)$$
This function must be monotone non-decreasing in $x$ for every $\theta_1\geq \theta_0$. In the given question $\theta_1=2$, and the density function is $$f_{2}(x)=2x.$$ Similarly for $\theta_0\in[1/2,1]$, $$f_{\theta_0}(x)=\theta_0x^{\theta_0-1}$$ Hence, the likelihood ratio function is
$$l_{\theta_0}(x)=\frac{2x}{\theta_0x^{\theta_0-1}}=\frac{2}{\theta_0}x^{2-\theta_0}$$
Since this function is increasing in $x$ for all $\theta_0\in[1/2,1]$, there exists a UMP test of level $\alpha$.
By definition of UMP test, the significance level $\alpha$ is the expected value of the decision rule (which is the likelihood ratio test with a certain threshold $\lambda$), for which the false alarm probability lies below $\alpha$, for every $\theta_0$
$$\alpha=\sup_{\theta_0}\int_{\{x:l_{\theta_0}(x)>\lambda\}}f_{\theta_0}(x)\mathrm{d}x=\sup_{\theta_0}\int_{\{x:l_{\theta_0}(x)>\lambda\}}\theta_0x^{\theta_0-1}\mathrm{d}x$$
Now, we have a nice simplification (Why?) $${\{x:l_{\theta_0}(x)>\lambda\}}\equiv {\{x:x>\lambda^{'}\}}$$
Hence
$$\alpha=\sup_{\theta_0}\int_{\{x:l_{\theta_0}(x)>\lambda\}}\theta_0x^{\theta_0-1}\mathrm{d}x=\sup_{\theta_0}\int_{\lambda^{'}}^1\theta_0x^{\theta_0-1}\mathrm{d}x=\sup_{\theta_0}1-{\lambda^{'}}^{\theta_0}=0.05$$
It is known that $\lambda^{'}\in[0,1]$ and $\theta_0\in[1/2,1]$. Now what value of $\theta_0$ maximizes $1-{\lambda^{'}}^{\theta_0}$ or similarly minimizes ${\lambda^{'}}^{\theta_0}$?
The UMP test is then $$\phi(x)=\begin{cases}1,\quad x>\lambda^{'}\\0,\quad x\leq \lambda^{'}\end{cases}$$
Best Answer
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!