Likelihood Ratio Test statistic for the exponential distribution I need to test null hypothesis $\lambda = \frac12$ against the alternative hypothesis $\lambda \neq \frac12$ based on data $x_1, x_2, ..., x_n$ that follow the exponential distribution with parameter $\lambda > 0$.  The sample mean is $\bar{x}$.  
The log likelihood is $\ell(\lambda) = n(\log \lambda - \lambda \bar{x})$
The MLE of $\lambda$ is $\hat{\lambda} = 1/\bar{x}$.
What is $2\log(\text{LR})$?
This is a past exam paper question from an undergraduate course I'm hoping to take. 
My first attempt is:
The likelihood ratio is the test of the null hypothesis against the alternative hypothesis with test statistic 
$L(\theta_1)/L(\theta_0)$
I get as far as 
$2\log(\text{LR}) = 2\{\ell(\hat{\lambda})-{\ell(\lambda})\}$
but get stuck on which values to substitute and getting the arithmetic right.  
 A: This is one of the cases that an exact test may be obtained and hence there is no reason to appeal to the asymptotic distribution of the LRT. To see this, begin by writing down the definition of an LRT,
$$L = \frac{ \sup_{\lambda \in \omega} f \left( \mathbf{x}, \lambda \right) }{\sup_{\lambda \in \Omega} f \left( \mathbf{x}, \lambda \right)} \tag{1}$$
where $\omega$ is the set of values for the parameter under the null hypothesis and $\Omega$ the respective set under the alternative hypothesis. Note that $\omega$ here is a singleton, since only one value is allowed, namely $\lambda = \frac{1}{2}$. On the other hand the set $\Omega$ is defined as 
$$\Omega = \left\{\lambda: \lambda >0 \right\}$$
as the parameter of the exponential distribution is positive, regardless if it is rate or scale. To obtain the LRT we have to maximize over the two sets, as shown in $(1)$. How do we do that? By maximum likelihood of course. 
You have already computed the mle for the unrestricted $ \Omega $ set while there is zero freedom for the set $\omega$: $\lambda$ has to be equal to $\frac{1}{2}$. All you have to do then is plug in the estimate and the value in the ratio to obtain
$$L = \frac{  \left( \frac{1}{2} \right)^n \exp\left\{ -\frac{n}{2} \bar{X} \right\} } { \left( \frac{1}{ \bar{X} } \right)^n   \exp \left\{ -n \right\} } $$
and we reject the null hypothesis of $\lambda = \frac{1}{2}$ when $L$ assumes a low value, i.e. when
$$L = \frac{  \left( \frac{1}{2} \right)^n \exp\left\{ -\frac{n}{2} \bar{X} \right\} } { \left( \frac{1}{ \bar{X} } \right)^n   \exp \left\{ -n \right\} }  \leq c $$
Merging constants, this is equivalent to rejecting the null hypothesis when
$$ \left( \frac{\bar{X}}{2} \right)^n \exp\left\{-\frac{\bar{X}}{2} n \right\}  \leq k $$
for some constant $k>0$. This is clearly a function of $\frac{\bar{X}}{2}$ and indeed it is easy to show that that the null hypothesis is then rejected for small or large values of $\frac{\bar{X}}{2}$. You can show this by studying the function
$$ g(t) = t^n \exp\left\{ - nt \right\}$$ 
noting its critical values etc. All that is left for us to do now, is determine the appropriate critical values for a level $\alpha$ test. That is, determine $k_1$ and $k_2$, such that we reject the null hypothesis when
$$\frac{\bar{X}}{2} \leq k_1 \quad \text{or} \quad \frac{\bar{X}}{2} \geq k_2$$
and this is done with probability $\alpha$. This can be accomplished by considering some properties of the gamma distribution, of which the exponential is a special case. Some transformation might be required here, I leave it to you to decide. Remember, though, this must be done under the null hypothesis.
