Let $(X_{1i},X_{2i})$ follow a bivariate normal distribution for $i=1,\dots,n$ with means $(\theta_1,\theta_2)$ and an identity variance matrix. Suppose that the parameter space is restricted to $\theta_1 \geq 0$ and $\theta_2 \geq 0$. Find the log-likelihood ratio statistic $\lambda_n$ for testing $H_0: (\theta_1,\theta_2)=(0,0)$, and show that under $H_0$, $\lambda_n$ will be a mixture distribution.

I have shown that the MLE will be $\hat{\theta_1}=\max(\bar{X_1},0)$ and $\hat{\theta_2}=\max(\bar{X_2},0)$ (due to the restrictions on the parameters). Now, when trying to find the log-likelihood ratio statistic, one has:

$$L((x_i,y_i)|(\theta_1,\theta_2))=(1/2\pi)^n \exp\left(-\frac{1}{2}\left[\sum_{i=1}^{n}(x_{1i}-\theta_1)^2 +\sum_{i=1}^{n}(x_{2i}-\theta_2)^2 \right] \right)$$

Considering only the part that depends on the parameters, the log-likelihood will be:

$$l((\theta_1,\theta_2))=-\frac{1}{2}\left[\sum_{i=1}^{n}(x_{1i}-\theta_1)^2 +\sum_{i=1}^{n}(x_{2i}-\theta_2)^2 \right]$$

When trying to find the log-likelihood statistic, I will obtain that for the parameter space:

$$l((\hat{\theta}_1,\hat{\theta}_2))=-\frac{1}{2}\left[\sum_{i=1}^{n}(x_{1i}-\hat{\theta}_1)^2 +\sum_{i=1}^{n}(x_{2i}-\hat{\theta}_2)^2 \right]$$

And for the null parameter space:

$$l((\theta_{01},\theta_{02}))=-\frac{1}{2}\left[\sum_{i=1}^{n}x_{1i}^2 +\sum_{i=1}^{n}x_{2i}^2 \right]$$

Combining these:

$$\lambda_n=2(l(\hat{\theta}_1,\hat{\theta}_2)-l(\theta_{01},\theta_{02}))=-\sum_{i=1}^{n}(x_{1i}-\hat{\theta}_1)^2 -\sum_{i=1}^{n}(x_{2i}-\hat{\theta}_2)^2+\sum_{i=1}^{n}x_{1i}^2 +\sum_{i=1}^{n}x_{2i}^2$$

$$\lambda_n=2\hat{\theta_1}n\bar{x_1}-n\hat{\theta_1}^2 + 2\hat{\theta_2}n\bar{x_2}-n\hat{\theta_2}^2$$

After having derived this form of $\lambda_n$, under $H_0$, the $x_{1i}$ and $x_{2i}$ would be normal random variables $N(0,1)$. However, I don't know how to transform this into a mixture distribution, as I would not be sure on how to deal with the fact that $\hat{\theta_1}$ and $\hat{\theta_2}$ are maximums (if they were just the sample mean, the answer would be rather simple). Or is my derivation of $\lambda_n$ incorrect? Thanks!


1 Answer 1


Note that because $\hat \theta = \max(\bar x,0) $ you have

$$ \lambda_n^{(1)}= \begin{cases} n\bar x_1^2,& \text{if } \bar x_1\gt 0\\ 0, & \text{otherwise} \end{cases} $$

so $\lambda_n^{(1)}$ has a mixture distribution where one component is a delta function at zero and the other one is the distribution of $n\bar x_1^2$ under $H_0$ (which is a $\chi^2$ distribution with one degree of freedom, since $\bar x_1$ is normal with variance $1/n$).

$$ \lambda_n^{(1)} \sim \frac{1}{2} \delta(0) + \frac{1}{2} \chi_1^2 $$

So $\lambda_n = \lambda_n^{(1)} + \lambda_n^{(2)}$ is a sum of two independent such variables and therefore it has a 4 component mixture distribution

$$ \lambda_n \sim \frac{1}{4} \delta(0) + \frac{1}{2} \chi_1^2 + \frac{1}{4} \chi_2^2$$

  • $\begingroup$ Thank you, that does make sense! I understand the $\chi^2$ component. Where would the delta function term come from? Just from the fact that the probability of $\bar{x_1} \leq 0$ is $1/2$ and that at $\overline{x_1}=0$ we would already have the point mass term considered? I am trying to make sense of it, but I'm having some trouble. $\endgroup$ Feb 11, 2022 at 20:21
  • $\begingroup$ You have a 1/2 probability of getting exactly zero, that is a delta function. Not sure what is causing you trouble $\endgroup$
    – J. Delaney
    Feb 11, 2022 at 20:37
  • $\begingroup$ I think I get it now. Just to see if I'm on the right track, if we were to replace $\hat{\theta}_1$ with $\max(\overline{\x_1},1)$ (rather than $0$) then it would be a $\delta_1$ term for the mixture but with probability $P(\hat{x_1} \leq 1)$? Thank you! $\endgroup$ Feb 11, 2022 at 21:22
  • $\begingroup$ Yes, exactly right. $\endgroup$
    – J. Delaney
    Feb 11, 2022 at 21:39

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