Show that distribution of function of LRT statistic for normal mean hypothesis testing is normally distributed Suppose $X_1 ... X_n$ ~$^{iid}$ N($\mu, \sigma$), with $\sigma$ known. What is the distribution of $-2ln(\lambda)$ where $\lambda$ is the LRT statistic for testing $H_0:\mu = \mu_0, H_1:\mu \neq \mu_0$?
So we know $\lambda = \left(\frac{\hat{\sigma^2}}{\hat{\sigma^2_0}}\right)^{\frac{n}{2}},$ with $\hat{\sigma^2} = \frac{1}{n}\sum(X_i-\bar{X})^2, \hat{\sigma^2_0} = \frac{1}{n}\sum(X_i-\mu_0)^2$.
Correct answer: $N(\mu=0, \sigma=\sigma)$
My work: 
$-2ln(\lambda) = -nln\left(\frac{\sum(X_i-\bar{X})^2}{\sum(X_i-\mu_0)^2}\right) = -nln\left(\frac{\sum(X_i-\bar{X})^2}{\sigma^2}\frac{\sigma^2}{\sum(X_i-\mu_0)^2}\right) = -nln\left(\chi_{n-1}^2 \cdot \frac{1}{\sum\left(\frac{X_i-\mu_0}{\sigma}\right)^2}\right)$, 
since the square of a standard normal variable $Z$ is $\chi_1^2$ distributed, and we have each $X_i$ iid,
$ = -nln\left(\chi_{n-1}^2 \cdot \frac{1}{\sum{Z_i^2}}\right) = -nln\left(\chi_{n-1}^2 \cdot \frac{1}{\sum{\chi_1^2}}\right) = -nln\left(\frac{\chi_{n-1}^2}{\chi_n^2} \right)$. 
From here I fail to see the relation to the normal distribution?
 A: For unknown $\sigma^2$
$$\lambda=\left(\frac{\sum(X_i-\bar{X})^2}{\sum(X_i-\mu_0)^2}\right) =\left(\frac{\sum(X_i-\bar{X})^2}{\sum(X_i-\bar{X})^2+n(\bar{X}-\mu_0)^2}\right)$$
$$=\left(\frac{1}{1+\frac{n(\bar{X}-\mu_0)^2}{\sum(X_i-\bar{X})^2}}\right)$$
$$=\left(\frac{1}{1+\frac{n(\bar{X}-\mu_0)^2}{(n-1)\frac{1}{n-1}\sum(X_i-\bar{X})^2}}\right)$$
$$=\left(\frac{1}{1+\frac{T^2}{n-1}}\right)$$
where 
 $T^2=\frac{n(\bar{X}-\mu_0)^2}{\frac{1}{n-1}\sum(X_i-\bar{X})^2}$
Now reject $H_0$ if $\lambda \leq \lambda_0 $ $\Leftrightarrow$ $T^2>c$ $\Leftrightarrow$ $|T|>k$, $T\sim t(n-1)$
For known $\sigma^2$
$$\lambda=\left(\frac{(2\pi \sigma^2)^{-n/2} e^{-\frac{1}{2\sigma^2}\sum (X_i -\mu_0)^2}}{((2\pi \sigma^2)^{-n/2} e^{-\frac{1}{2\sigma^2}\sum (X_i -\bar{X})^2}}\right) $$
$$=e^{-\frac{1}{2\sigma^2} \left( \sum (X_i -\mu_0)^2  -\sum (X_i -\bar{X})^2     \right)}$$
$$=e^{-\frac{1}{2\sigma^2} \left( n(\bar{X}-\mu)^2)     \right)}$$
$$=e^{-\frac{1}{2} \left( (\frac{\bar{X}-\mu}{\sigma / \sqrt{n}})^2)     \right)}$$
now $\lambda  \leq \lambda_0$ $\Leftrightarrow$ 
$$(\frac{\bar{X}-\mu}{\sigma/ \sqrt{n}})^2 >c$$
$\Leftrightarrow$ 
$$|Z|=|\frac{\bar{X}-\mu}{\sigma/ \sqrt{n}}| >c$$
$Z=\frac{\bar{X}-\mu}{\sigma/ \sqrt{n}}$
