Let's define two independent random variables, $x$ and $y$ as:
$x$ ~ $D_A$ = $\exp ( \alpha)$
$y$ ~ $D_B$ = $\exp ( \beta)$
where $x$ gives the time differences between successive events of type $X$, and $y$ gives time differences between successive events of type $Y$.
What is the mixture of the two distributions $D_M = F(\alpha, \beta)$, that gives random variables that correspond to time differences between successive events of any type ($X$ or $Y$)?
For example, here are the histograms for waiting times drawn from exponential distributions with parameters $\alpha$ = 0.1 and $\beta$ = 0.02, and the histogram of the mixture (all in log scale). I want to know the symbolic distribution $D_M$ that generates the third histogram (which is clearly not of exponential distribution).
NOTE: The third histogram is erroneous in the sense that it is the mixture of identical sample sizes drawn from the two exponential distributions $D_A$ and $D_B$, which does not comply with the idea that in any given timeframe, random Poisson processes with intensities $\alpha$ ≠ $\beta$ yield different event counts.
How does this relate to the convolution of the two distributions?
Let $N_A=(N_{A,t})_{t \geq 0}$ be the Poisson process counting the occurrences of the type $X$ events, with intensity $\alpha$. Let $N_B=(N_{B,t})_{t \geq 0}$ be the Poisson process counting the occurrences of the type $Y$ events, with intensity $\beta$.
Suppose these are independent.
The process $N_A+N_B$ counts occurrences of both types, and is a Poisson process with intensity $\alpha+\beta$, and thus the waiting times between occurrences are IID exponential random variables with parameter $\alpha+\beta$ and thus mean $1/(\alpha+\beta)$.
Suppose $E\sim \textrm{Exp}(\alpha),F\sim \textrm{Exp}(\beta)$, independent. Then their sum is distributed with a pdf given by the convolution of their pdfs; its solution is not exponential with parameter $\alpha +\beta$. So the convolution of the two distributions does not relate to the distribution of the waiting times of $N_A+N_B$. The distribution of the waiting times of $N_A+N_B$ can be recovered with $\min(E,F)$. Indeed: $$P(\min(E,F)> x)=P(E>x)P(F>x)=e^{-(\alpha+\beta)x}$$ so $\min(E,F)\sim \textrm{Exp}(\alpha+\beta)$.