likelihood ratio test

Jul 2009
69
6
Let x have the distribution with pdf \(\displaystyle f = \frac {1}{2} \exp {-|x-\theta|} \)
Let \(\displaystyle X_1...X_5 \) be ordered statistic, x between -inf and inf

Find the likelihood ratio for testing the hypothesis \(\displaystyle H_0: \theta=\theta_0 \) against\(\displaystyle H_a : \theta not = \theta_0 \)

likelihood function= \(\displaystyle \Pi \frac {1}{2} \exp {-|x-\theta|} \)

Not really sure how to continue?

Any help will be appreciated

Thanks in advance
 
Aug 2009
27
7
Not sure here.

The Variable

\(\displaystyle D = -2\left( ln L(\theta_0) - ln L(\hat{\theta} \right) \)

follows a chi-square distribution with \(\displaystyle df_1 - df_2 \) degrees of freedown of each model. The likelihoods are maximized under the null hypothesis (\(\displaystyle \theta_0 \)) and under the alternative (MLE).

Since you already the likelihood, you know \(\displaystyle L(\theta_0) \), as \(\displaystyle \theta_0 \) is given. Finally, \(\displaystyle \hat{\theta} \) is just the MLE (the point where the likelihood is a maximum) and then the variable \(\displaystyle D \) can be obtained.

Again, not 100% sure...
 
Last edited:
  • Like
Reactions: firebio
Jul 2009
69
6
Should i split the likelihood ratio into 2 regions, 1 is \(\displaystyle x > \theta \) and other one is \(\displaystyle x < \theta \) ?

for \(\displaystyle x > \theta \), MLE is \(\displaystyle \theta = X_{(1)} \)
and
for \(\displaystyle x < \theta \), MLE is \(\displaystyle \theta = X_{(5)} \)?


Thanks
 
Last edited:
Aug 2009
27
7
I've changed my post because I misunderstood your question. Sorry for that.
About your post, I think your are right, but perhaps someone can be certain! :)
 
Last edited: