I'm trying to obtain $\alpha$-level LR Test where $(X_1, … X_n)$ are from Beta($\theta$, 1) with $H_0 = {\theta_0}$ and $H_0 \neq \theta_0$.
I'm looking for
$$
\lambda(X) = \frac{\sup_{\theta \in \Theta_0}l(\theta)}{\sup_{\theta \in \Theta}l(\theta)}
$$
Suppose $T := \sum^n_{i=1}\ln{X_i}$ and we know that MLE of Beta$(\theta, 1)$ equals
$$
\hat{\theta}=\frac{-n}{T}
$$
Our $\lambda(X)$ is then:
$$
\lambda(X) = \frac{\theta_{0}^{n} (X_1 \cdot … \cdot X_n)^{\theta_0 – 1}}{(\frac{-n}{T})^n(X_1 \cdot … \cdot X_n)^{\frac{-n}{T} – 1}} = \left(\frac{- \theta_0 T}{n}\right)^n(X_1 \cdot … \cdot X_n)^{\theta_0 + \frac{n}{T}}
$$
We want to find $\lambda(X) < c$ but we may as well look for $\ln\lambda(X) < \ln c = \hat c $. Taking the logarithm:
$$
\ln\lambda(X) = n \ln{\frac{- \theta_0 T}{n}}+\left({\theta_0 + \frac{n}{T}}\right)T =n \ln{\frac{- \theta_0 T}{n}} + \theta_0T + n
$$
This has to be lesser than some $\hat c$
$$
n \ln{\frac{- \theta_0 T}{n}} + \theta_0T + n < \hat c $$
We define $f(x) = n \ln{\frac{- \theta_0 x}{n}} + \theta_0x + n$ and see how it behaves:
$$
f'(x) = \frac{n}{\frac{\theta_0x}{n}} + \theta_0 \\
f'(x) = \frac{n^2}{\theta_0x} + \theta_0
$$
$f'(x) > 0$ iff:
$$
\frac{n^2}{\theta_0x} + \theta_0 > 0 \\
\frac{n^2}{\theta_0x} > – \theta_0 \\
\theta_0x < -\frac{n^2}{\theta_0} \\
x < -\left(\frac{n}{\theta_0}\right)^2
$$
This makes x negative, but it does not raise my concern because $x=\sum \ln X_i$ where $X_i \in (0,1)$ so $\ln X_i < 0$ it can be indeed negative.
I'd then get the LR test looking like:
$$
\varphi(X) = \begin{cases}
1 & T < d_1 \text{ or } T > d_2 \\
\gamma_1 & T = d_1 \\
\gamma_2 & T = d_2 \\
0 & T \in (d_1, d_2)
\end{cases}
$$
where $d_1 < -\frac{n^2}{\theta_0^2} < d_2$ for $d_1, d_2$ calculated to meet $\alpha$.
Is my solution correct so far? I have noticed the question has been answered here, but I could not fully get the grasp of it and wanted to go step by step.
Best Answer
I didn't check your counts but they look good.
This is the graph of your $ln\lambda(\mathbf{x})$ where I just set $T=-\sum_{i}ln X_i$ and fixed some parameters
As you can see, say
$ln\lambda(\mathbf{x})<c$ is equivalent to
$T<c_1$ and $T>c_2$
As suggested, I think it is better to set $T=-\sum_{i}ln X_i$
Now it is easy to check that $Y=-lnX\sim Exp(\theta)$ so $T\sim Gamma(n;\theta)$.
Thus under null hypothesis $2\theta_0 T\sim \chi_{(2n)}^2$
and you can now solve the problem with the tables