Does UMVUE of $\frac{\theta_{x}}{\theta_{y}}$ exist? X $\sim$ exp($\theta_{x}$), Y $\sim$ exp($\theta_{y}$) I have got the following variables. 
$$
X\sim exp(\theta_{x_{i}}), \ Y \sim exp(\theta_{y_{i}})
$$
and want to find the UMVUE of 
$$ 
\frac{\theta_{x}}{\theta_{y}}.
$$
As the complete statistics are $\bar{X}$ and $\bar{Y}$ the UMVUE should be found through $E(\frac{\bar{X}}{\bar{Y}})$, but as the expected value of an inverse exponential distribution does not exist, does the UMVUE exist? 
(I am sure I have understood something wrong here) 
 A: If $\bar X$ and $\bar Y$ are independent, since $\bar Y$ is an unbiased estimator of $\theta_y^{-1}$, the question boils down to whether or not there exists an UMVUE of $_x$  which calls first for the resolution of the question as to whether or not there exists an unbiased estimator of $\theta_x$. While $X\sim\mathcal{E}(\theta_x)$ does not allow¹ for an unbiased estimator of $\theta_x$, $X_1,X_2\stackrel{\text{iid}}{\sim}\mathcal{E}(\theta_x)$ does: since$$Z=X_1+X_2\sim\mathcal{Ga}(2,\theta_x)$$it satisfies$$\mathbb{E}[Z^{-1}]=\int_0^\infty z^{-1}\theta_x^2ze^{-\theta_x z}\,\text{d}z=\theta_x \int_0^\infty e^{-u}\,\text{d}u=\theta_x$$ UMVU-ness then follows by Lehmann-Scheffé.

¹A quick proof that there is no such estimator is to consider the contrapositive: if there exists an unbiased estimator $h(Y)$ of $\theta$, then$$\int_0^\infty h(y) \theta \exp\{-\theta y\}\,\text{d}y=\theta$$is equivalent to$$\int_0^\infty h(y) \exp\{-\theta y\}\,\text{d}y=1$$Differentiating in $\theta$,
$$\int_0^\infty h(y)y  \exp\{-\theta y\}\,\text{d}y=0$$
or$$\int_0^\infty h(y)y \,\theta \exp\{-\theta y\}\,\text{d}y=0$$which implies that $h(y)y=0$ or $h\equiv 0$ by the completeness theorem for exponential families (itself following from the uniqueness of the Laplace transform). Hence there is no solution.
