I'm interested in the distribution of $\frac{X}{c+X+Y}$, where $X$ and $Y$ follows the exponential distribution with rate parameter $\lambda$, and $c$ is a constant.
I know that without the constant $c$, this ratio has a uniform$(0,1)$ distribution. However, this is not the case when adding the constant $c$.
Trying to follow the same methodology in here. I have the following.
For $\alpha \in [0,1)$ \begin{eqnarray*} \Pr \left(\frac{X}{c+X+Y} \leq \alpha\right) &=& \Pr \left(\frac{c+X+Y}{X} \geq \frac{1}{\alpha}\right)\\ &=& \Pr\left(Y \geq X(1/\alpha -1) -c\right) \end{eqnarray*}
Noting that $X(1/\alpha -1) -c \geq 0$ only for $X \geq \frac{c \alpha}{1-\alpha}$, For simplicity, let's assume that $\lambda =c=1$, thus, we obtain \begin{eqnarray*} \Pr \left(\frac{X}{1+X+Y} \leq \alpha\right) &=& \int_{0}^{\alpha/(1-\alpha)} e^{- x}\cdot 1 \mathrm{d}x + \int_{\alpha/(1-\alpha)}^\infty \lambda e^{-x} e^{- x(\frac{1}{\alpha}-1) -1} \mathrm{d}x &=& 1-e^{-\alpha/(1-\alpha)} + \alpha e^{-\frac{1}{1-\alpha} -1} \end{eqnarray*}
However, simulations I made shows that this is not true.
What have I missed? Is there more general result for ratio of the form $\frac{X}{c + \sum_{i=1}^n Y_i}$?
Any suggestions are much appreciated.
In order to compute the LHS, you should first condition on the event that $Y$ takes the value $y$, and then integrate this with respect to the marginal density of $y$, i.e., \begin{align} \text{Pr}\left(\frac{X}{X+Y+c}\leq \alpha\right)&=\text{Pr}\left(X\leq \frac{\alpha}{1-\alpha}(Y+c)\right)\\ &=\int\limits_{0}^{\infty}\text{Pr}\left(X\leq \frac{\alpha}{1-\alpha}(y+c)\right)f_{Y}(y)~dy\\ &=\int\limits_{0}^{\infty}\text{Pr}\left(X\leq \frac{\alpha}{1-\alpha}(y+c)\right)\lambda e^{-\lambda y}~dy. \end{align}
Under the assumption that $\lambda=c=1$ and $\alpha \in (0,1)$, the above simplifies to \begin{eqnarray} \text{Pr}\left(\frac{X}{X+Y+1}\leq \alpha\right)=\int\limits_{0}^{\infty}\int\limits_{0}^{\frac{\alpha (y+1)}{1-\alpha}}e^{-x} e^{-y}~dx~dy. \end{eqnarray} You may now proceed to do the calculations as usual.