Division of Two exponential Random variables Division of Two exponential Random variables
I have two exponential random variables $X$ and $Y$ with rate parameter $\lambda_1$ and $\lambda_2$. I have to find distribution function of $Z$ where $Z$ is given by:
$$Z=\frac{X}{Y+1}$$
and $f_{X}(x)=\lambda_1.\exp(-\lambda_1 x)$ and $f_{Y}(y)=\lambda_2.\exp(-\lambda_2 y)$.
How to find pdf of $Z$ and How to check its validity?
I followed this procedure and got an answer as:
\begin{eqnarray}
F(Z)=p{(Z \leq z)}\\
&=&P\left\{\frac{X}{Y+1} \leq z \right\}\\
&=&P\left\{X \leq z.(y+1) \right\}\\
&=&\int_{y=0}^{+\infty} \int_{x=0}^{z(y+1)}f(x,y) dx\,dy$\\
&=&\int_{y=0}^{+\infty} \int_{x=0}^{z(y+1)}\lambda_1\lambda_2\exp(-\lambda_1x)\exp(-\lambda_2y) dx\,dy\\
&=&\lambda_1\lambda_2\int_{y=0}^{+\infty} \exp(-\lambda_2y)\int_{x=0}^{z(y+1)}\exp(-\lambda_1x)dx\,dy\\
&=&\lambda_2\int_{y=0}^{+\infty}\exp(-\lambda_2y).(1-\exp(-\lambda_1(1+y).z))dy\\
&=&\lambda_2\left(\frac{1}{\lambda_2}-\frac{exp(-\lambda_1z)}{\lambda_2+\lambda_1.z}\right)
\end{eqnarray}
which is CDF of $F(Z)$. To find pdf $F_{Z}(z)$ I derivated $F(Z)$ and got the result as:
\begin{eqnarray}
f_{Z}(z)&=&\frac{dF(Z)}{dz}\\
&=&\lambda_1\lambda_2\exp(-\lambda_1z).\left[\frac{1}{\lambda_1 z+\lambda_2}+\frac{1}{(\lambda_1z+\lambda_2)^2}\right]
\end{eqnarray}
Now, How do I check it's Validity?
 A: Let $f$ be a continuous real-valued bounded function defined on $[0,+\infty[$. 
We want to compute $\mathbb{E}\big[ f(Z) \big]$. Given that $X$ and $Y$ are independent (and both follow exponential distributions), we have:
$$ \mathbb{E}\big[ f(Z) \big] = \int_{0}^{+\infty} \int_{0}^{+\infty} f\Big(\frac{x}{y+1}\Big) \lambda_1 e^{-\lambda_{1}x} \lambda_{2}e^{-\lambda_{2}y} \; dxdy. $$
Consider the change of variables $\Phi \, : \, [0,+\infty[^2 \, \rightarrow \, [0,+\infty[^2$ defined by:
$$ \forall (x,y) \in [0,+\infty[^2, \; \Phi\big( (x,y) \big) = \Big(\frac{x}{y+1}, y\Big) = (u,v) \in [0,+\infty[^2. $$
$\Phi$ is a $\mathcal{C}^{1}$ diffeomorphism from $[0,+\infty[^2$ to itself and:
$$ \Phi^{-1}\big( (u,v) \big) = \big( u(v+1), v \big) $$
$$ \mathrm{Jac}\big( \Phi^{-1}, (u,v) \big) = \begin{bmatrix} v+1 & u \\ 0 & 1 \end{bmatrix}. $$
It follows from the change of variable theorem that:
$$ 
\begin{align*} 
\mathbb{E}\big[ f(Z) \big] & = {} \int_{0}^{+\infty} \int_{0}^{+\infty} f(u) \lambda_1\lambda_2 e^{-\lambda_{1}u(v+1)}e^{-\lambda_{2}v} (v+1) \, du dv. \\[2mm]
 & = \int_{0}^{+\infty} f(u) \Bigg( \lambda_1 \lambda_2 e^{-\lambda_1 u} \int_{0}^{+\infty} (v+1) e^{-(\lambda_{1}u + \lambda_{2})v} \, dv \Bigg).
\end{align*}$$
Also, note that:
$$ 
\begin{align*}
\int_{0}^{+\infty} (v+1)e^{-(\lambda_{1}u + \lambda_{2})v} dv & = {} \Bigg[-\frac{e^{-av}(av + a + 1)}{a^2} \Bigg]_{v=0}^{v=+\infty} \\
 & = \frac{a+1}{a^2}.
\end{align*}$$
with $a = \lambda_{1}u + \lambda_{2}$.
Therefore, the probability density function of $Z$ (with respect to the Lebesgue measure on $[0,+\infty[$) is given by:

$$ u \in [0,+\infty[ \, \mapsto \; \lambda_{1}\lambda_{2}e^{-\lambda_{1}u} \frac{\lambda_{1}u + \lambda_{2} + 1}{(\lambda_{1}u + \lambda_{2})^2}. $$

