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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.

blahblah
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1 Answers1

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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

enter image description here

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

tommik
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