# likelihood ratio test

#### firebio

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

#### gustavodecastro

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...

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firebio

#### firebio

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

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#### gustavodecastro

I've changed my post because I misunderstood your question. Sorry for that.