# Deriving confidence interval by inverting LRT statistic

Consider a random sample of size $n$ from the distribution with pdf $$f(y;\theta)=\theta y^{\theta-1}, 0<y<1, \theta > 0.$$ I want to find a $1-\alpha$ confidence interval for $\theta$ by using the likelihood ratio test. I have already derived the MLE of $\theta$: With the likelihood function equal to $L(\theta)=\theta^n\prod_{i=1}^ny_i^{\theta-1}$ and log-likelihood equal to $\log L(\theta)=n\log(\theta)+(\theta-1)\sum_{i=1}^n\log{y}_i$, the MLE is given by $\hat\theta=-n/\sum_{i=1}^n\log{y}_i$.

I have derived an approximate $1-\alpha$ confidence interval by using the MLE and the asymptotical properties of it, but now I want to create a confidence interval by using the LRT. As I have understood it, it is possible to obtain a confidence interval by inverting the LRT. Looking at the LRT statistic,

$$\lambda=\frac{\sup_{\Theta_0}L(\theta|\mathbf y)}{\sup_ {\Theta}L(\theta|\mathbf y)},$$ where $\Theta$ denotes the entire parameter space, and $\Theta_0$ denotes the parameter space under $H_0: \theta\in\Theta_0$. So I have tried to plug in the parameter value under $H_0$ in the likelihood function in the numerator (just some arbitrary $\theta_0)$ and the MLE of $\theta$ in the likelihood function in the denominator, but I don't manage to go any further and set up a confidence interval. I guess I don't know what to do next, plus that the expression gets large and seems messy to work with. I was hoping that someone here would have some idea of how to construct the interval by using the LRT, with my efforts so far as help.

• You should probably mark this with the self-study tag. – kjetil b halvorsen Jan 29 '17 at 17:07

1. You can write the loglikelihood function as $$\ell(\theta)=n\log\theta -(\theta-1)T$$ where $$T=-\sum \log y_i$$ is sufficient.
2. Show that $$-\log y_i$$ has an exponential distribution.
3. Hence show that $$T$$ has a gamma distribution.
4. Then you can write the likelihood ratio $$\lambda$$ as function of the sufficient statistic $$T$$, so a test could equally be based directly on $$T$$.
5. Now you can show that $$\frac{\theta T}{n}$$ is a pivot, find its distribution and construct the CI.