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$X$ 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 $\mu$.

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    $\begingroup$ With a uniform random variable, this is trivial by picking odd and even binary digits ($g$ and $h$ are not continuous though). It's not hard to transform exponential to uniform. $\endgroup$
    – Anthony Quas
    Commented Sep 29, 2015 at 20:48
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    $\begingroup$ It's a bit weird because your example uses two independent variables to produce another two... but if we can we use two independent exponentials $X,Y$, then I think the minimum $M = \min\{X,Y\}$ and the absolute difference $Z = |X-Y|$ should be independent by the memoryless property (and both are exponential). $\endgroup$
    – usul
    Commented Sep 29, 2015 at 21:14

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One curious property of Gamma distributions is their relation with the Dirichlet distributions. 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.

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Yes. For example, if we take $Y=\lfloor X \rfloor$ and $Z=X-Y$.

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