Likelihood Ratio for two-sample Exponential distribution Let $X$ and $Y$ be two independent random variables with respective pdfs:
$$f \left(x;\theta_i \right) =\begin{cases} \frac{1}{\theta_i} e^{-x/ {\theta_i}} \quad 0<x<\infty, 0<\theta_i< \infty \\ 0 \quad \text{elsewhere} \end{cases} $$
for $i=1,2$. Two indepedent samples are drawn in order to test $H_0: \theta_1 =\theta_2 $ against $H_1 : \theta_1 \neq \theta_2 $ of sizes $n_1$ and $n_2$ from these distributions. I need to show that the LRT $\Lambda$ can be written as a function of a statistic having $F$ distribution, under $H_0$.

Since the mle of this distribution is $\hat{\theta}=\bar{x} $, the LRT statistic becomes (I am skipping a few tedious steps here):
$$ \Lambda =\frac{\bar{x}^{n_1} \bar{y}^{n_2} \left( n_1+n_2 \right)}{n_1 \bar{x}+n_2 \bar{y}}$$
I know that the $F$ distribution is defined as the quotient of two independent chi-square random variables, each one over their respective degrees of freedom. Additionally, since $X_i,Y_i \sim \Gamma \left( 1,\theta_1 \right)$ under the null, then $\sum X_i \sim \Gamma \left(n_1 ,\theta_1 \right)$ and$\sum Y_i \sim \Gamma \left(n_2, \theta_1 \right) $. 
But how can I proceed from here? Any hints?
Thank you.
 A: If memory serves, it appears you have forgotten something in your LR statistic.  
The likelihood function under the null is
$$L_{H_0} = \theta^{-n_1-n_2}\cdot \exp\left\{-\theta^{-1}\left(\sum x_i+\sum y_i\right)\right\}$$
and the MLE is
$$\hat \theta_0 = \frac {\sum x_i+\sum y_i}{n_1+n_2} = w_1\bar x +w_2 \bar y, \;\; w_1=\frac {n_1}{n_1+n_2},\;w_2=\frac {n_2}{n_1+n_2}$$
So$$ L_{H_0}(\hat \theta_0) = (\hat \theta_0)^{-n_1-n_2}\cdot e^{-n_1-n_2}$$
Under the alternative, the likelihood is 
$$L_{H_1} = \theta_1^{-n_1}\cdot \exp\left\{-\theta_1^{-1}\left(\sum x_i\right)\right\}\cdot \theta_2^{-n_2}\cdot \exp\left\{-\theta_2^{-1}\left(\sum y_i\right)\right\}$$
and the MLE's are
$$\hat \theta_1 = \frac {\sum x_i}{n_1} = \bar x, \qquad \hat \theta_2 = \frac {\sum y_i}{n_2} = \bar y$$
So
$$L_{H_1}(\hat \theta_1,\,\hat \theta_2) = (\hat \theta_1)^{-n_1}(\hat \theta_2)^{-n_2}\cdot e^{-n_1-n_2}$$
Consider the ratio
$$\frac {L_{H_1}(\hat \theta_1,\,\hat \theta_2)}{L_{H_0}(\hat \theta_0)} = \frac {(\hat \theta_0)^{n_1+n_2}}{(\hat \theta_1)^{n_1}(\hat \theta_2)^{n_2}}=\left(\frac {\hat \theta_0}{\hat \theta_1}\right)^{n_1} \cdot \left(\frac {\hat \theta_0}{\hat \theta_2}\right)^{n_2}$$
$$= \left(w_1 + w_2 \frac {\bar y}{\bar x}\right)^{n_1} \cdot \left(w_1\frac {\bar x}{\bar y} + w_2 \right)^{n_2}$$
The sample means are independent -so I believe that you can now finish this. 
A: The likelihood function given the sample $\mathbf x=(x_1,\ldots,x_{n_1},y_1,\ldots,y_{n_2})$ is given by
\begin{align}
L(\theta_1,\theta_2)&=\frac{1}{\theta_1^{n_1}\theta_2^{n_2}}\,\exp\left[-\frac{1}{\theta_1}\sum_{i=1}^{n_1} x_i-\frac{1}{\theta_2}\sum_{i=1}^{n_2}y_i\right]\mathbf1_{\mathbf x>0},\quad\theta_1,\theta_2>0.
\end{align}
The LR test criterion for testing $H_0:\theta_1=\theta_2$ against $H_1:\theta_1\ne \theta_2$ is of the form
\begin{align}
\lambda(\mathbf x)&=\frac{\sup\limits_{\theta_1=\theta_2}L(\theta_1,\theta_2)}{\sup\limits_{\theta_1,\theta_2}L(\theta_1,\theta_2)}
=\frac{L(\hat\theta,\hat\theta)}{L(\hat\theta_1,\hat\theta_2)},
\end{align}
where $\hat\theta$ is the MLE of $\theta_1=\theta_2$ under $H_0$, and $\hat\theta_i$ is the unrestricted MLE of $\theta_i$ for $i=1,2$.
It is easily verified that
$$\left(\hat\theta_1,\hat\theta_2\right)=(\bar x,\bar y)$$
and $$\hat\theta=\frac{n_1\bar x+n_2\bar y}{n_1+n_2}.$$
After some simplification we get this symmetry for the LRT criterion:
\begin{align}
\lambda(\mathbf x)&=\underbrace{\text{constant}}_{>0}\left(\frac{n_1\bar x}{n_1\bar x+n_2\bar y}\right)^{\!\!n_1}\left(\frac{n_2\bar y}{n_1\bar x+n_2\bar y}\right)^{\!\!n_2}
\\&=\text{constant}\cdot\,t^{n_1}(1-t_1)^{n_2},\quad\text{ where }t=\frac{n_1\bar x}{n_1\bar x+n_2\bar y}
\\&=g(t),\,\text{say.}
\end{align}
Studying the nature of the function $g$, we see that $$g'(t)\gtrless 0\;\iff\; t\lessgtr \frac{n_1}{n_1+n_2}.$$
Now since $2n_1\overline X/\theta_1\sim \chi^2_{2n_1}$ and $2n_2\overline Y/\theta_2\sim \chi^2_{2n_2}$ are independently distributed, we have
$$\frac{\overline X}{\overline Y}\stackrel{H_0}{\sim}F_{2n_1,2n_2}.$$
Define $$v=\frac{n_1\overline x}{n_2\overline y},$$
so that $$t=\frac{v}{v+1}\quad\uparrow\, v$$
Therefore,
$$\lambda(\mathbf x)<c \iff v<c_1\quad\text{ or }\quad v>c_2,$$
where $c_1,c_2$ can be found from some size restriction and the fact that, under $H_0$,
$$\frac{n_2}{n_1}\,v\sim F_{2n_1,2n_2}.$$
