# Q: Determine the Rejection Region of LRT for two Uniform Distributed Samples

Assume we have two independent random samples $(X_i)_{i=1}^{n}$ and $(Y_i)_{i=1}^{m}$ where $X_i\sim\mathrm{Uniform}(0,\theta_1)$ and $Y_j\sim\mathrm{Uniform}(0,\theta_2)$. To test, at the significant level $\alpha$, $H_0: \theta_1 = \theta_2$ against $H_1:\theta_1\ne\theta_2$ using Likelihood Ratio Test, denoting by $Z$ the ratio of the $n$-th ($m$-th resp.) ordered statistics of two samples, i.e., $Z = \frac{X_{(n)}}{Y_{(m)}}$, we may find:

$$\Lambda = \frac{ \left( \max{X_i} \right) ^ {n} \left( \max{Y_j} \right) ^ m }{ \left( \max \left(X_i, Y_j \right) \right) ^ {n + m} } = \begin{cases} Z ^ {-m} \quad X_{(n)} \ge Y_{(m)} \\ Z^{n} \quad X_{(n)} < Y_{(m)} \end{cases} \mathrm{.}$$

The form of the rejection region $C$ is $C = \{Z\le c_1 <1\}\cup\{Z\ge c_2 > 1\}$ by considering the cases where $X_{(n)} < Y_{(m)}$ or otherwise respectively. To further determine the constants $0 < c_1 \le 1$ and $0 < c_2 \le 1$, I am not sure whether to set:

$$\mathrm{Pr} \left( Z \ge c_2 \vert Z \ge 1\right) = \mathrm{Pr} \left (Z \le c_1 \vert Z < 1 \right) \overset{\mathrm{set}}{=} \alpha\mathrm{,}$$

or: $$\mathrm{Pr}\left( Z \ge c_2 \right) = \mathrm{Pr}\left( Z \le c_1 \right) \overset{\mathrm{set}}{=} \frac{\alpha}{2} \mathrm{.}$$ Since for two independent samples with size $n$ and $m$ following the same uniform disribution $\mathrm{Uniform}(0,\beta)$, with the maxima of the observations being $V$ and $W$ resepectively, we have $\mathrm{Pr}(V\le kW) = \frac{mk^n}{n+m}$ for some constant $0<k<1$. If I follow the first idea, I can get:

$$C = \left\{ Z \le \alpha ^ {\frac{1}{n}} \right\} \cup \left\{ Z \ge \alpha ^ {-\frac{1}{m}} \right\} \mathrm{.}$$

Since \begin{align} \mathrm{Pr} \left (Z \ge c_2 \vert Z \ge 1 \right) & = \frac{ \mathrm{Pr} \left( Z \ge c_2, Z \ge 1 \right) }{ \mathrm{Pr} \left ( Z \ge 1 \right)} \\\\ & = \frac{ \mathrm{Pr} \left( Y_{(m)} \le \frac{1}{c_2} X_{(n)} \right) } { \mathrm{Pr} \left( Y_{(m)} \le X_{(n)} \right) } \\\\ & = \frac{ \frac{n}{n + m} \left( \frac{1}{c_2} \right) ^ {m} } { \frac{n}{n + m} } \\\\ & = c_{2}^{-m} \\\\ & \overset{ \mathrm{set} }{=} \alpha \end{align} Hence $c_2 = \alpha ^ {-\frac{1}{m}}$, and similarly $c_1 = \alpha ^ {\frac{1}{n}}$.

But if I follow the second idea, since $\frac{m}{m+n}$ and $\frac{n}{m+n}$ are both less than $1$, I get:

$$C = \left\{ Z \le \left( \frac{n+m}{m} \frac{\alpha}{2} \right) ^ {\frac{1}{n}} \right\} \cup \left\{ Z \ge \left( \frac{n+m}{n} \frac{\alpha}{2} \right)^{-\frac{1}{m}} \right\} \mathrm{.}$$

Apparently, I have made a mistake since those two do not agree if two samples are not of the equal size. Nevertheless, I have trouble in locating where did my argument go wrong.

Thank you very much in advance.

EDIT:

1. I found a good discussion in this thread on finding the joint p.d.f. of $X_{(n)}$ and $Y_{(n)}$ (and more) for this test. Nevertheless, it remains a question for me to determine the rejection region.
2. I think this is similar to the the cases where we perform a two-sided LRT where the likelihood ratio is a step function. For example, in a two-sided $F$ test for the variances of two normal samples ($H_0: \sigma_{0}^2=\sigma_{1}^2$ against $H_1: \sigma_{0}^2\ne\sigma_{1}^2$), the standard procedure follows the second idea above. Also of interests, in DeGroot (p. 603, 4th Ed.), it is noted while deriving the two sided $F$ test from LRT that:

Unfortunately, it is usually tedious to compute the necessary values $c_1$ and $c_2$. For this reason, people often abandon the strict likelihood ratio criterion in this case and simply let $c_1$ and $c_2$ be the $\frac{\alpha_0}{2}$ and $1 - \frac{\alpha_0}{2}$ quantiles of the appropriate $F$ distribution.

Might it also be theoretically justifiable to derive the rejection region using the first idea?

3. This question is also posted on Quora.