If $X$ and $Y$ are independent and exponentially distributed, which is the pdf of $Z$? Where $Z$ is given by \begin{equation} Z = \frac{X}{1+Y} \end{equation}
I read similar posts on this forum but those are all different cases.
$X,Y$ are independent exponentially distributed: What is the distribution of $X/(X+Y)$
Thanks for any suggestions.
[EDIT to add context] Suppose to have two R.V., i.e. $Z$ and $T$, given by \begin{equation} Z = \frac{X}{1+Y}, \end{equation} and \begin{equation} T = \frac{U}{1+V} \end{equation}
I have to calculate the following probability: \begin{equation} P_x = P[Z>K(1+T)-1]. \end{equation}
Now, p(Z) and p(T) are the pdfs respectively of Z and T, given by (differentiating respect to T)
\begin{equation} p(Z) = \frac{\lambda_1 \lambda_2 e^{\frac{-Z}{\lambda2}}}{(\lambda_2+\lambda_1 Z)^2} + \frac{e^{\frac{-Z}{\lambda2}}}{(\lambda_2+\lambda_1 Z)}, \end{equation} and \begin{equation} p(T) = \frac{\lambda_3 \lambda_4 e^{\frac{-T}{\lambda4}}}{(\lambda_4+\lambda_3 T)^2} + \frac{e^{\frac{-T}{\lambda4}}}{(\lambda_4+\lambda_3 T)}. \end{equation}
Finally the probability that I'm looking for is given by
\begin{equation} p_x = P[Z>K(1+T)-1] = \int_{0}^{\infty}p(T)\int_{K(1+T)-1}^{\infty} p(Z) dZdT \end{equation}
[Why I asked]
I asked the pdf of $Z$ when $X$ and $Y$ are i.i.d. and exponentially distributed (same for $T$), because it seems that the two integrals in p_x don't converge to a solution for $Z>0$ and $T>0$.
How can I solve $p_x$?
That is the context.
Thanks again.
[QUESTION ON PROBABILITY MOVED TO SPECIFIC QUESTION]