What is the distribution of a exponential distribution, whose expected values is drawn from the expontial distribution
$$X\sim\mathrm{Exp}(\text{mean}=\alpha) f(x\mid α) = (1/α) \exp(-x/α) $$
$$Y\sim\mathrm{Exp}(\text{mean}=X) f(y\mid x) = (1/x) \exp(-y/x) $$
how is $Y$ distributed?
I have found this question answered when $X$ and $Y$ are parameterized in terms of rates (Pareto), but I can't figure it out when $X$ and $Y$ are parameterized in terms of expected values.
If I understand correctly, you're trying to find the distribution of $Y$ where $Y\mid X=x\sim \mathrm{Exp}\left(\frac1x\right)$. To find the marginal density of $Y$ given the marginal density of $X$ and conditional density of $Y$ given $X$, we would compute
\begin{align} f_Y(y) &= \int_0^\infty f_{X,Y}(x,y)\mathsf dx\\ &=\int_0^\infty f_{Y\mid X}(y\mid x)f_X(x)\mathsf dx\\ &= \int_0^\infty \frac1xe^{-\frac 1xy}\frac1\alpha e^{-\frac1\alpha x}\mathsf dx\\ &= \int_0^\infty \frac1{\alpha x} e^{-\left(\frac yx+\frac x\alpha\right)}\mathsf dx, \end{align} but this integral can't be expressed in terms of elementary functions (please correct me if I'm wrong). On the other hand, if we have $X\sim\mathrm{Exp}(\alpha)$ and $Y\mid X\sim\mathrm{Exp}(X)$ then with a straightforward application of integration by parts we obtain $$ f_Y(y) = \int_0^\infty xe^{-xy}\alpha e^{-\alpha x}\mathsf dx = \int_0^\infty \alpha xe^{-x(\alpha+y)}\mathsf dx = \frac\alpha{(\alpha + y)^2}.$$
(This doesn't exactly answer the question, but it is too long for a comment.)
