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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?

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  • $\begingroup$ Can you show an outline of how you got from $\frac{L(0,0)}{L(x_1,x_2)}$ to $e^{-\frac12(x_1^2+x_2^2)}$? $\endgroup$
    – Glen_b
    Jul 12 '17 at 0:52
  • $\begingroup$ @Glen_b I have added as requested. $\endgroup$
    – user164144
    Jul 12 '17 at 1:03
  • $\begingroup$ I don't think the denominator is correct -- you seem to have treated the denominator as if it were (1,1) $\endgroup$
    – Glen_b
    Jul 12 '17 at 1:05
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    $\begingroup$ That seems okay to me. The issue is substituting $\theta_1=x_1,\theta_2=x_2$ The maximizer on the denominator would be the more likely of $(x_1,0)$ and $(0,x_2)$ wouldn't it? $\endgroup$
    – Glen_b
    Jul 12 '17 at 1:12
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    $\begingroup$ I didn't say the maximizer was two ordered pairs; it's whichever of the two has the higher likelihood. Your maximizer for the denominator is not in the alternative. However I made a mistake in typing my comment .. I should have said "the more likely of (0,1) and (1,0)". $\endgroup$
    – Glen_b
    Jul 12 '17 at 1:47
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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 $\theta_1$

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