Finding the Most Powerful Test for bivariate data Let $X_1\sim N(\theta_1,1)$ and $X_2\sim N(\theta_2,1)$ be independent RVs. It is of interest to test the following:
$H_o: (\theta_1,\theta_2)=(0,0)$
$H_a: (\theta_1,\theta_2)=(1,0) \text{ or } (0,1)$
The first objective is to find the generalized likelihood ratio test (GLRT) while the second is to find the MP test. My first question is if I was able to find the GLRT correctly with the following procedure:
First, I defined the joint distribution of $(X_1,X_2)$ as having the following pdf:
$f(x_1,x_2|\theta_1,\theta_2)=\frac1{2\pi}e^{-\frac12[(x_1-\theta_1)^2+(x_2-\theta_2)^2]}$
Then I applied the LRT, $\lambda(x_1,x_2)=\frac{L_{Ho}(\theta_1,\theta_2)}{L_{Ho \cup Ha}(\theta_1,\theta_2)}=\frac{L(0,0)}{L(x_1,x_2)}=e^{-\frac12(x_1^2+x_2^2)}<c$
$\frac{L(0,0)}{L(x_1,x_2)}=\frac{\frac1{2\pi}e^{-\frac12[(x_1-0)^2+(x_2-0)^2]}}{\frac1{2\pi}e^{-\frac12[(x_1-x_1)^2+(x_2-x_2)^2]}}=e^{-\frac12(x_1^2+x_2^2)}$
I reasoned that in the denominator, the values of $\theta_1$ and $\theta_2$ that will maximize the likelihood are $x_1$ and $x_2$ respectively. This was consistent with results from taking derivatives.
$L_{Ho \cup Ha}(\theta_1,\theta_2)=\frac1{2\pi}e^{-\frac12[(x_1-\theta_1)^2+(x_2-\theta_2)^2]}$
$ln(L_{Ho \cup Ha}(\theta_1,\theta_2))=ln(\frac1{2\pi})-{\frac12[(x_1-\theta_1)^2+(x_2-\theta_2)^2]}$
$\frac{d}{d\theta_1}ln(L_{Ho \cup Ha}(\theta_1,\theta_2))=x_1-\theta_1$
$\frac{d}{d\theta_2}ln(L_{Ho \cup Ha}(\theta_1,\theta_2))=x_2-\theta_2$
Equating both to 0 and solving for the $\theta's$, I got $\theta_1=x_1$ $\theta_2=x_2$
Is this the correct GLRT?
If it is, how do I proceed to finding the most powerful test?
 A: Ok. So after the helpful comments on my question, I improved my answer to the GLRT as follows:
$\lambda(x_1,x_2)=\frac{L_{Ho}(\theta_1,\theta_2)}{L_{Ho \cup Ha}(\theta_1,\theta_2)}=\frac{L(0,0)}{L(x_1,x_2)}=\frac{e^{-\frac12(x_1^2+x_2^2)}}{L(x_1,x_2)}$
For the denominator, considering that the only possibilities in the alternative are (1,0) or (0,1), then it is either $e^{-\frac12((x_1-1)^2+x_2^2)}$ or $e^{-\frac12(x_1^2+(x_2-1)^2)}$
The first expression is larger than the second when $x_1>x_2$ So, my GLRT is
$\frac{e^{-\frac12(x_1^2+x_2^2)}}{e^{-\frac12((x_1-1)^2+x_2^2)}}=e^{\frac12-x_1}<c$ when $x_1>=x_2$ for some $0<=c<=1$
and
$\frac{e^{-\frac12(x_1^2+x_2^2)}}{e^{-\frac12((x_2-1)^2+x_1^2)}}=e^{\frac12-x_2}<c$ when $x_1<x_2$ for some $0<=c<=1$
Is this ok? The problem then asks to find the power of the most powerful test. Is this just the power when c=0?
Here is my attempt at finding the power of the most powerful test.
When $x_1>x_2$
$e^{\frac12-x_1}<c$ which implies that
$x_1>1/2-ln(c)$
The power function is $\beta(\theta)=Pr_{\theta}(x \epsilon Rejection)=Pr(x_1>1/2-ln(c))$
Since $0<=c<=1$, $-ln(c)$ is always positive except when it is undefined (c=0) or 0 (c=1). So the most powerful test is when c=1. The power of this test is $P(x_1>1/2)$. Here is where I get stuck because I know that $x_1$ is normal but I do not know what value I should use for the parameter $\theta_1$
