Solved – Deriving confidence interval by inverting LRT statistic

confidence intervalhypothesis testinglikelihoodlikelihood-ratiopivot

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.

Best Answer

Since this looks like a self-study question some hints:

  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.

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