Find the distribution of $R=\bar X / \bar Y$ given that exponential distribution with mean parameter 2 is equivalent to Chi squared with 2 dof.
I'm given 2 samples $x_i$ and $y_i$ of sizes $n$ and $m$ respectevely such that $$f(x,\theta)=\frac{1}{\theta}e^{-x/\theta}$$ and $$f(y,\theta)=\frac{1}{\theta}e^{-y/\theta}$$
Construct confidence interval of $R$?
I think it should also me exponentially distributed, but I'm struggling even starting this question. In particular, the different sizes of samples put me off doing $f(R,\theta)=f(x)/f(y)$
I though about about finding the mean $E(X)=\int^{\infty}_0 x f(x,\theta)dx$ and finding $E(R)=E(X)/E(Y)$