$X,Y$ are independent exponentially distributed then what is the distribution of $X/(1+Y)$

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}

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]

Probability calculation when the R.V. are the ratio of variables i.i.d. and exponentially distributed

We need to find $P(\frac{X}{1+Y} \leq z)$ for some $z \geq 0$. This resolves to $P(X \leq z(1+Y))$. So now we can set up our integration limits:
\begin{align*} &\int_{y=0}^{\infty} \int_{x=0}^{z(1+y)} \lambda_1 e^{-\lambda_1 y} \lambda_2 e^{-\lambda_2 x} \, dx \, dy \\ &=\lambda_1\lambda_2 \int_{y=0}^\infty e^{-\lambda_1 y}\left(-\frac{1}{\lambda_2} \right) \left. e^{-\lambda_2 x} \right|_{x=0}^{x=z(1+y)} \, dy\\ &=-\lambda_1 \int_{y=0}^\infty e^{-\lambda_1 y} (e^{-\lambda_2 z(1+y)}-1) \, dy \\ &=-\lambda_1 \int_{y=0}^\infty e^{-y(\lambda_1+\lambda_2)-\lambda_2z}-e^{-\lambda_1 y} \, dy \\ &\dots \\ &=1-\frac{\lambda_2 e^{-\lambda_1 z}}{\lambda_2+\lambda_1 z}. \end{align*}
Now differentiate w.r.t. $z$ and you can retrieve your pdf.
• The answer could be more general if the parameters of $X$ and $Y$ where different. (+1) – Math-fun Mar 15 '15 at 21:16