Let $X$, $Y$ be exponentially distributed with means $\lambda$ and $\mu$ ($\lambda < \mu$), respectively (Note that $X$ and $Y$ do not need to be independent). Let $Z$ be a random variable, which with probability $\lambda/\mu$ is equal to $X$ and with probability $1 - \lambda/\mu$ is equal to $X+Y$. Show that $Z$ is exponentially distributed with parameter $\lambda$.
My solution: I used the Laplace-Stiltjes transform, but ran into trouble while finding the joint density function of $X+Y$. I have $$\mathrm{LST}(Z)=\mathbb{E}(e^Z) = \mathbb{E}(e^{\lambda/\mu X + (1-\lambda/\mu)(X+Y)})\\ = \mathbb{E}(e^{X+(1-\lambda/\mu)Y}) = \int_{\mathbb{R}} e^{x+(1-\lambda/\mu)y} f_{X+Y}(x,y)\,\mathrm dx(?)$$ I don't know how to continue further. I tried calculating the coint density of $X+Y$, but got something like $$f_Z(z)=\lambda/\mu(1-e^{-\lambda z})(1-\lambda/\mu)(\lambda/\mu^2 + 1/\mu)^z.$$
If $X$ and $Y$ are independent, \begin{align} Z^\star(s) &= Z^\star(s|Z=X)P(Z=X) + Z^\star(s|Z=X+Y)P(Z=X+Y) \\ &= Z^\star(s|Z=X) \frac{\lambda}{\mu} + Z^\star(s|Z=X+Y) \left( 1- \frac{\lambda}{\mu} \right) \\ &= \frac{1/\lambda}{1/\lambda+s} \frac{\lambda}{\mu} + \frac{1/\lambda}{1/\lambda+s} \frac{1/\mu}{1/\mu+s} \left( 1- \frac{\lambda}{\mu} \right) \end{align} If $\lambda=\mu$ your result holds. Otherwise, I do not think so.
Now, consider the following different problem. Let $G$ be a geometric random variable, with probability of success equal $\lambda/(\lambda+\mu)$. Let $W$ be defined as the waiting time until a success, assuming each slot has duration given by the minimum of $X$ and $Y$, $$U=\min(X,Y)$$
Then,
\begin{align} W^\star(s) &= G(U(s)) \\ &= \frac{pz}{1-(1-p)z} \end{align} evaluated at point $z=(\lambda+\mu)/(\lambda+\mu+s)$ and $p=\lambda/(\lambda+\mu)$
Then, the answer is the one you desired
where
$$W^\star(s)=\lambda/(\lambda+s)$$
however, the problem is very different