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!