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.

  • 1
    $\begingroup$ The expression $(\hat\theta-\theta_0)^2\sum x_i^2$ is a function of only $Y$'s and constants. [Further, keep in mind that you're dealing with the case where $H_0$ is true here, so you know the distribution of the $Y$'s] $\endgroup$
    – Glen_b
    Oct 16, 2018 at 4:01
  • $\begingroup$ Yes, but since this leads to $Pr(\sum Y_i x_i <(\sqrt{(c/\sum x_i^2}) + \theta_0)\sum x_i^2$, do I know the distribution of that? I cannot seem to find it. Whatever I try, I can't find a way to express it as $Pr(\sum Y_i < k^*)$. Either way, thanks for your help, I made progress at least. $\endgroup$
    – user569579
    Oct 16, 2018 at 6:03
  • $\begingroup$ You may want to see en.wikipedia.org/wiki/Wilks%27_theorem $\endgroup$ Oct 16, 2018 at 8:00
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    $\begingroup$ @user569579 You have a linear combination of (presumably independent) Gaussians on the left. You can compute the distribution of $\sum Y_ix_i$ immediately. The stuff on the right is just some other constant, $c^\prime$. $\endgroup$
    – Glen_b
    Oct 16, 2018 at 23:02

1 Answer 1


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}


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