Rejection Region for Likelihood Ratio Test I have $((Y_1,x_1),(Y_2,x_2),\ldots,(Y_n,x_n))$ where $Y_i$ is distributed as $N(\theta x_i,1)$. I want to find the rejection region $[0, c]$ associated with $\lambda$ for a test with significance level $\alpha$. $\lambda$ is the log likelihood ratio for the test: $H_0: \theta=\theta_0$ or $H_1: \theta\neq\theta_0$. I have determined this to be $$-2\log \left(\frac{\exp (-\frac{1}{2}(\hat\theta-\theta_0)^2\sum x_i^2)}{\exp (-\frac{1}{2}(\hat\theta-\hat\theta)^2\sum x_i^2)}\right) = -2\log \left(\exp \left(-\frac{1}{2}(\hat\theta-\theta_0)^2\sum x_i^2\right)\right),$$ which simplifies to $(\hat\theta-\theta_0)^2\sum x_i^2$. In this equation, $\hat\theta$ denotes the maximum likelihood estimator, which I have determined to be $\sum Y_ix_i/\sum x_i^2$. My question is: How do I find the rejection region for this test? I know that I need to find $\Pr((\hat\theta-\theta_0)^2\sum x_i^2<c)=\alpha$ and solve this for $c$. The problem is that the expression does not follow a distribution, as far as I'm aware. How do I then determine the rejection region? Thanks in advance.
 A: If I simply take the likelihood ratio $\Lambda$ for testing $H_0$ versus $H_1$, I get
\begin{align}
\Lambda(y_1,\ldots,y_n)&=\frac{\exp\left[-\frac{1}{2}\sum (y_i-\theta_0 x_i)^2\right]}{\exp\left[-\frac{1}{2}\sum (y_i-\hat\theta x_i)^2\right]}
\\&=\exp\left[-\frac{1}{2}\left\{\theta_0\sum x_i^2-2\theta_0\sum x_iy_i-\hat\theta^2\sum x_i^2+2\hat\theta \sum x_iy_i\right\}\right]
\\&=\exp\left[-\frac{1}{2}(\theta_0-\hat\theta)\left\{(\theta_0+\hat\theta)\sum x_i^2-2\sum x_iy_i\right\}\right]
\end{align}
Putting the value of $\hat\theta$, I finally get
$$\Lambda(y_1,\ldots,y_n)=\exp\left[-\frac{1}{2\sum x_i^2}\left(\theta_0\sum x_i^2-\sum x_i y_i\right)^2\right]$$
So, for some positive $k$,
$$\Lambda(y_1,\ldots,y_n)<c\implies \left|\theta_0\sum x_i^2-\sum x_i y_i\right|>k$$
Keeping in mind the level restriction, we need to find $k$ so that 
$$P_{H_0}\left[\left|\theta_0\sum x_i^2-\sum x_i Y_i\right|>k\right]=\alpha$$
Or, $$P_{H_0}\left[\theta_0\sum x_i^2-\sum x_i Y_i>k\right]+P_{H_0}\left[\theta_0\sum x_i^2-\sum x_i Y_i<-k\right]=\alpha$$
I think all you need to note now is that for $i=1,2,\ldots,n$,
\begin{align}
Y_i\stackrel{\text{ind}}{\sim}\mathcal N(\theta x_i,1)&\implies x_iY_i\stackrel{\text{ind}}{\sim}\mathcal N(\theta x_i^2,x_i^2)
\\&\implies \sum_{i=1}^n x_i Y_i\stackrel{H_0}{\sim}\mathcal N\left(\theta_0\sum_{i=1}^n x_i^2,\sum_{i=1}^n x_i^2\right)
\end{align}
