Likelihood Ratio Test for Bivariate Normal in a Restricted Parameter Space 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!

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