For this question, both parameters are unknown.
My idea of doing the test is to find the MLE of $\sigma^2$ in the null parameter space where $\mu$ = $\sigma^2$ and substitute it into $\mathit L$($\hat{\hat{\theta}}$). Then, I would find the LRT-statistic $\lambda$ = $\cfrac{\mathit L(\hat{\hat{\theta}};\tilde x_s)}{\mathit L(\hat\theta;\tilde x_s)}$.
The following is what I have come up with:
I approached the log-likelihood as per normal and eventually substituted $\mu$ = $\sigma^2$.
log$\mathit L$($\hat\theta$) = -$\frac{n}{2}$log2$\pi$ - $\frac{n}{2}$log$\sigma^2$ - $\frac{1}{2\sigma^2}$$\sum^n_{j=1}(x_j-\sigma^2)$
= -$\frac{n}{2}$log2$\pi$ - $\frac{n}{2}$log$\sigma^2$ - $\frac{1}{2\sigma^2}\sum^n_{j=1}x_j^2 + n\sum^n_{j=1}x_j - \frac{n\sigma^2}{2}$
I then substituted $\sigma^2$ = S and $\sum^n_{j=1}x_j^2$ = k, where k is a constant (I did this to make the equation look simpler). To get the MLE of $\sigma^2$, i differentiated log$\mathit L$($\hat\theta$) and equated the results to zero. This gives me the following equation:
$-\frac{n}{2S}-\frac{k}{2S^2}-\frac{n}{2} = 0$, simplifying this gives me
$S^2 + S + \frac{k}{n} = 0$
I solved the quadratic equation and chose the positive answer since $\sigma^2$ > 0.
Thus, $\hat\sigma^2$ = $\frac{1+\sqrt {1-\frac{4k}{n}}}{2}$
At this point, I obtained ${\mathit L({\hat{\theta}};\tilde x_s)}$ as per normal, which gives me
$(2\pi)^{-\frac{n}{2}}\cdot({{\hat\sigma}^2})^{-\frac{n}{2}}\cdot e^{-\frac{n}{2}}$
However, the confusing part is where I try to get $\mathit L(\hat{\hat{\theta}};\tilde x_s)$ as it reduces down to
$(2\pi)^{-\frac{n}{2}}\cdot({{\hat\sigma}^2})^{-\frac{n}{2}}\cdot e^-(\frac{1}{2\hat{\hat{\sigma^2}}}) \cdot exp(\sum^n_{j=1}x_j - \frac{1+\sqrt {1-\frac{4k}{n}}}{2})^2$.
The equation seemed too complicated and I had no idea how I should continue to approach this as trying to achieve a solution with $\lambda$ doesn't seem to work with what I have. However, I also have a contradicting thought that I am on the right track, and that I am able to somehow relate it to -2log$\lambda > \chi_d^2$, were d=1 in this question.
Can anyone help me out and point me in the right direction maybe?
Thank you.