Derive the distribution of $X +\frac{a Y}{Y+b}$, with $X$ and $Y$ are exp. r.v Define $X$ and $Y$ to be two exponential random variables. The probability distribution functions are $f_{X}(x)=\lambda e^{-\lambda x}$ and $f_{Y}(y)=\delta e^{-\delta y}$. Note that $X$ and $Y$ are independent.
Define $Z=X +\frac{a Y}{Y+b}$, where $a$ and $b$ are some positive constants. My goal is to derive the CDF of $Z$, i.e. I want to calculate $P(Z <z)$.
My approach consists in deriving the CDF of $R=\frac{a Y}{Y+b}$, and then trying to find the CDF of the sum, which will be a function of the CDF of $R$.
So I have started as follows: $P(\frac{a Y}{Y+b} < r)=P(Y(a-r) <rb)$. It is clear that here $a$ should be $>r$, because otherwise we get something negative.. Thus, following this approach forces us to assume that $a>r$, something that I don't know if I can assume or not.  
Question: is this assumption necessary to be able to derive the CDF? is there any other approach that I can adopt to derive the CDF without making this assumption?
 A: Let $R=\frac{a Y}{ Y + b}$. We have 
$\mathbb{P}\{ R < r \}= \mathbb{P}\{ Y < br (a-r)^{-1} \} = 1- e^{-\delta b r (a-r)^{-1}}$.
Using the remark given by @whuber, it can be noticed that $r < a$ since $R < a$.
Based on the above, we can write
\begin{align}
\nonumber \mathbb{P}\{ X +R < \tau \} & = \int_0^\tau \int_0^{\tau-r}  f_{X}(x)  f_R(r) dx dr  \\ &= \int_0^\tau \int_0^{\tau-r}  \lambda e^{-\lambda x}  f_R(r) dx dr \\ &=  \int_0^\tau (1-e^{-\lambda (\tau-r)})  f_R(r) dr \\ &=F_R(\tau) -e^{-\lambda \tau} \int_0^\tau  e^{\lambda r} f_R(r) dr \\  & = F_R(\tau) - e^{-\lambda \tau} ( F_R(r) e^{\lambda r} |_0^\tau - \lambda \int_0^\tau e^{\lambda r} F_R(r) dr )  \\  &= F_R(\tau)- F_R(\tau)  e^{-\lambda \tau} e^{\lambda \tau} + \lambda e^{-\lambda \tau} \int_0^\tau e^{\lambda r} F_R(r)  dr \\  &= 
 \lambda e^{-\lambda \tau} \int_0^\tau  e^{\lambda r} (1- e^{-\delta b \, r (a-r)^{-1}}) dr \\  &= 1-e^{-\lambda \tau} - \lambda e^{-\lambda \tau} \int_0^\tau  e^{\lambda r-\delta b \, r (a-r)^{-1}} dr  
\end{align}
where the fifth equality is obtained using integration by parts.  
It remains to check if the final integral can be solved or not (?).
A: The answer to this problem should be computable via a technique very similar to the one used in my answer to the OP's previous  similar question CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions. My answer showed how to compute the complementary CDF 
$$P\left\{\frac{XY}{X+Y+a} > v\right\}$$ by converting the problem to
finding the probability that $(X-v)(Y-v)$ exceeds $v^2+av$, that is,
$(X,Y)$ lies above the upper branch of the hyperbola that results
when the graph of the hyperbola $xy = v^2+av$ has been shifted to the right by $v$ and upward by $v$. Here, the
same method applied mutatis mutandis tells us that to compute
$$P\left\{X + \frac{aY}{Y+b} > v\right\},$$ we should find
the probability that $(X+a-v)(Y+b)$ exceeds $ab$. The hyperbola
in question is the result when the graph of the hyperbola
$XY = ab$ has been shifted to the left by $(a-v)$ and downwards
by $b$, and so special cases will need to be considered according as
$v$ is chosen to be smaller than or larger than $a$, and also to
take into account the fact that the hyperbola will cross one or both axes, and so, unlike the single integral in the other problem, the
integral may have to be computed in several parts.

Edit: Actually, after reading the above again, I think that computing
$$P\left\{X + \frac{aY}{Y+b} \leq v\right\}\quad
\text{instead of}\quad
P\left\{X + \frac{aY}{Y+b} > v\right\}$$
as $P\{(X+a-v)(Y+b) \leq ab\}$ might be easier in this problem,
at least in the case when $v < a$ and the hyperbola crosses
both axes.
