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I was viewing the solution provided by StubbornAtom in the following question: https://stats.stackexchange.com/questions/81151/likelihood-ratio-for-two-sample-exponential-distribution

Specifically, some of the steps aren't clear to me:

  • When computing $\lambda(\bf{x}) = \frac{sup_{\theta_1 = \theta_2} L(\theta_1, \theta_2)}{sup_{\theta_1,\theta_2} L(\theta_1, \theta_2)}$, as outlined by the user, one can show that the MLE under $H_0$ is given by $\hat{\theta} = \frac{\sum_{i = 1}^{n_1}x_i + \sum_{i = 1}^{n_2} y_i}{n_1 + n_2} = \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2}$, and the MLEs of $\theta_1$ and $\theta_2$ are given by $\hat{\theta_1} = \bar{x}$, $\hat{\theta_2} = \bar{y}$, so that one gets likelihood ratio given by $\lambda(\bf{x}) = \frac{L(\hat{\theta}, \hat{\theta})}{L(\hat{\theta_1}, \hat{\theta_2})}$. However, the user then claims that, after simplification, this likelihood ratio becomes $\lambda(\bf{x}) = \text{constant} \cdot(\frac{n_1\bar{x}}{n_1\bar{x} + n_2\bar{y}})^{n_1}(\frac{n_2\bar{y}}{n_1\bar{x} + n_2\bar{y}})^{n_2}$. How does the user get this simplified version of the likelihood ratio ?
  • How does the user see that $2n_1\bar{X}/\theta_1 \sim \chi_{2n_1}^2$ and $2n_2\bar{Y}/\theta_2 \sim \chi_{2n_2}^2$ ?
  • Where does the conclusion that $\lambda(\bf{x}) < c \iff$ $v < c_1$ or $v > c_2$ come from ? Also, in a formal hypothesis test, does this give us our rejection region ?

Edit: To expand upon my first question, here is what I get when computing the likelihood ratio :

$$\lambda(\bf{x}) = \frac{\bar{x}^{n_1}\bar{y}^{n_2}}{\left( \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2}\right)^{n_1} \left( \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2}\right)^{n_2}}$$

Thank you for your time!

Michael Hardy
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michiganbiker898
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    Have you actually tried to write down the likelihood ratio? What do you get? Forget the answer for a moment and try to derive the test your way. – StubbornAtom Feb 03 '20 at 08:23
  • Hello sir! Thanks for your comment. I edited my question to include what I get when I compute the likelihood ratio. It looks similar to what you've gotten, but not quite there. I would include all my steps, but with all of the notation, it gets too cluttered and is hard to pick apart. (= – michiganbiker898 Feb 03 '20 at 18:24
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    Please don't say 'sir'. There is also a similar answer here. Since your question is about the specific answer, you could have just left a comment below the answer. – StubbornAtom Feb 03 '20 at 18:41
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    Does this answer your question? GLR Test for 2 samples from exponential distributions For your last question, try to find the region of $\mathbf x$ for which $\lambda(\bf{x}) < c$. That would give the rejection region. – StubbornAtom Feb 03 '20 at 18:50
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    Part of the question seems to amount to this: How does one show that the exponential distribution with expected value $2$ is a chi-square distribution with $2$ degrees of freedom? Thus if $Z_1,Z_2\sim\text{i.i.d.}\operatorname N(0,1),$ then for $x\ge0$ one has $\Pr(Z_1^2 + Z_2^2>x) = e^{-x/2}.$ One way to show that involved polar coordinates. $\qquad$ – Michael Hardy Feb 03 '20 at 19:30

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