Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having trouble with this exercise from Bain and Engelhardt's textbook:

Consider independent random samples of size $n_1$ and $n_2$ from respective exponential distributions $X_i \sim EXP(\theta_1)$ and $Y_i \sim EXP(\theta_2)$. Derive the Generalize Likelihood Ratio test of $H_0:\theta_1=\theta_2$ versus $H_1:\theta_1\neq\theta_2$.

The Generalized Likelihood Ratio is defined by


where $\hat{\vec{\theta}}$ denotes the usual Maximum Likelihood Estimator of $\vec{\theta}$ and $\hat{\vec{\theta_0}}$ denotes the MLE under the restriction that $H_0$ is true.

One is then supposed to apply the Neyman-Pearson lemma.

I've thought about this exercise for some time now, unsuccesfully.

Thank you for any help given.

share|cite|improve this question
up vote 2 down vote accepted

When $\theta_1 \neq \theta_2$, the likelihood is (given the independence assumption) $\mathcal{L}(\theta_1,\theta_2)=\theta_1 ^ {n_1} \exp [-\theta_1 (x_1 + \dots + x_{n_1})] \theta_2 ^ {n_2} \exp [-\theta_2 (y_1 + \dots + y_{n_2})]$.

When $\theta = \theta_1 = \theta_2$, the likelihood is \begin{align} \mathcal{L}(\theta)&=\theta ^ {n_1} \exp [-\theta (x_1 + \dots + x_{n_1})] \theta ^ {n_2} \exp [-\theta (y_1 + \dots + y_{n_2})] \\ &= \theta^{n_1+n_2} \exp [ -\theta ( x_1 + \dots + x_{n_1} + y_1 + \dots + y_{n_2} )] \end{align}

I leave the rest to you.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.