$X$ follows Exponential $(\lambda)$. Can we split $X$ into two independent r.v.'s, i.e.,


do there exist functions $g$ and $h$ such that $g(X)$ and $h(X)$ are independent for any fixed $\lambda$? $g(X)$ and $h(X)$ can have non-exponential distribution.


The context. The above can be thought as the parallel of the normal case:
$X \sim N(\mu,1)$ and adding independent Gaussian noise $Z \sim N(0,1)$ we get 
$X+Z$ and $X-Z$ are independent for any fixed $\mu$. 



splitting exponential random variable into independent components
One curious property of [Gamma distributions](https://en.wikipedia.org/wiki/Gamma_distribution) is their relation with the [Dirichlet distributions](https://en.wikipedia.org/wiki/Dirichlet_distribution). If a random variable $Y \sim \mathrm{Gamma}(\alpha, \gamma)$ is independent of a random vector $(\pi\_1, \dots, \pi\_k) \sim \mathrm{Dirichlet}(\alpha\_1, \dots, \alpha\_k)$ where $\alpha\_1+\dots+\alpha\_k = \alpha$, then the vector $(\pi\_1 Y, \dots, \pi\_k Y)$ consists of $k$ mutually independent components, where $Y\_j \sim \mathrm{Gamma}(\alpha\_k, \gamma)$, $j=1,\dots,k$.


In particular, for example, if $Y$ follows Exponential distribution with parameter $\gamma$, and $b$ is a Beta-distributed random variable with parameter $r \in (0,1)$, then $bY$ and $(1-b)Y$ are independent, Gamma distributed with parameters $(r, \gamma)$ and $(1-r, \gamma)$, respectively.


