There are $X_1, X_2$ where $X_i \sim N(\mu_i,1), i=1,2$. They are independent. The question is

Find the likelihood ratio test with $H_0:(\mu_1,\mu_2)=(0,0), H_1:(\mu_1,\mu_2) \neq (0,0)$. The significance level is $\alpha (0< \alpha <1)$ and parameter space $\Omega$ is $$\Omega = \left\{ (\mu_1,\mu_2) : \mu_1 \geq 0, \mu_2 \geq 0\right\}$$

My solution is $X_1^2 \geq \chi_p^2(1)$ or $X_2^2 \geq \chi_q^2(1)$ or $X_1^2+X_2^2 \geq \chi_r^2(2)$ where $p+q+r=\alpha$. Is it right?

Detail of my solution :

Let $\mu = (\mu_1, \mu_2)^T$. Then $\hat{\mu}^{\Omega_0}=(0,0)$ and $\hat{\mu}^{\Omega} = \left(\max\{x_1, 0\}, \max\{x_2,0\}\right)$ because parameter space is not $\mathbb{R}^2$.

Then I calculated $\Lambda = 2[l(\hat{\mu}^{\Omega}) - l(\hat{\mu}^{\Omega_0})]$ to find the rejection region from $\Lambda \geq \lambda (\lambda > 0)$.

After some algebra, I got $\Lambda = x_1^2I_{(x_1>0, x_2<0)} + x_2^2I_{(x_1<0, x_2>0)} + (x_1^2+x_2^2)I_{(x_1>0, x_2>0)}$.

Under the null hypothesis, $X_i^2 \sim \chi^2(1)(i=1,2)$ so $X_1^2+X_2^2 \sim \chi^2(2)$.

Finally I got the above rejection region.

  • $\begingroup$ Thanks. I added the tag and detail. $\endgroup$
    – flossy
    Commented Oct 25, 2020 at 4:29
  • $\begingroup$ Can you say something about how do you get the MLE of $(\mu_1, \mu_2)$? $\endgroup$
    – Hermi
    Commented Jan 15, 2022 at 21:23

1 Answer 1


Your solution seems to be correct. The strange shape of your parameter space (it's not a open subset of $\mathbb R^2$) creates this ambiguity in the final result: each combination of $(p,q,r)$ gives a different LRT. Some are more powered for $\mu_1=0$, some are more powered for $\mu_2=0$ and some are more powered for $\mu_1\neq0,\mu_2\neq0$, but all of them are valid LRTs with significance level $\alpha$.

  • $\begingroup$ Thank you for your comment! I understand as follows: My rejection region is valid LRT but not MP(most powerful) for all hypotheses $\endgroup$
    – flossy
    Commented Oct 26, 2020 at 16:07
  • $\begingroup$ @flossy that's correct. also, there is no uniformly most powerful test in this setting, since the LRT with $p=1$ is MP if $\mu_2=0$ and the LRT with $q=1$ is MP if $\mu_1=0$. $\endgroup$
    – PedroSebe
    Commented Oct 27, 2020 at 4:00

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