what is the CDF of (X + Y)/Z Can you please help me with this exercise?
Given $X\sim \exp(\lambda_X)$, $Y\sim \exp(\lambda_Y)$ and $Z\sim \exp(\lambda_Z)$ are non-negative independent non-identically distributed random variables. 
Find the CDF $F_T(t)$ where is given by
$$ T = \frac{X+Y}{Z}$$
My comment: as can be seen, T is also a non-negative random variables.
This is what I have tried so far:
The CDF of T is expressed as
$$F_T(t)=\int_0^\infty\int_0^{\gamma z} (1 - exp(-\frac{tz}{\lambda_X} + \frac{y}{\lambda_X}))p_Y(y)p_Z(z)dydz$$
$$F_T(t)=\int_0^\infty\int_0^{\gamma z} (1 - exp(-\frac{tz}{\lambda_X} + \frac{y}{\lambda_X}))exp(-\frac{y}{\lambda_Y})exp(-\frac{z}{\lambda_Z})dydz$$
Can you please help me with this exercise?
Given $X\sim \exp(\lambda_X)$, $Y\sim \exp(\lambda_Y)$ and $Z\sim \exp(\lambda_Z)$ are non-negative independent non-identically distributed random variables. 
Find the CDF $F_T(t)$ where is given by
$$ T = \frac{X+Y}{Z}$$
My comment: as can be seen, T is also a non-negative random variables.
This is what I have tried so far:
The CDF of T is expressed as
$$F_T(t)=\int_0^\infty\int_0^{\gamma z} (1 - exp(-\frac{tz}{\lambda_X} + \frac{y}{\lambda_X}))p_Y(y)p_Z(z)dydz$$
$$F_T(t)=\int_0^\infty\int_0^{\gamma z} (1 - exp(-\frac{tz}{\lambda_X} + \frac{y}{\lambda_X}))exp(-\frac{y}{\lambda_Y})exp(-\frac{z}{\lambda_Z})dydz$$
$$F_T(t)=\int_0^\infty\int_0^{\gamma z}exp(-\frac{y}{\lambda_Y}-\frac{z}{\lambda_X})dydz$$
$$-\int_0^\infty\int_0^{\gamma z} exp[-(\frac{t}{\lambda_X}+\frac{1}{\lambda_X})z]exp[-(\frac{1}{\lambda_Y}-\frac{1}{\lambda_X})y]dydz$$
Is this correct?? If not, do you have any other approach?? :((
 A: For positive $x,y,t$ we have:
$$\Pr\left(T\leq t\mid X=x,Y=y\right)=\Pr\left(Z\geq t^{-1}\left(x+y\right)\right)=e^{-\lambda_{Z}t^{-1}\left(x+y\right)}$$
Applying: $$\Pr\left(T\leq t\right)=\int_{0}^{\infty}\int_{0}^{\infty}\Pr\left(T\leq t\mid X=x,Y=y\right)f_{X}\left(x\right)f_{Y}\left(y\right)dxdy$$
we find:
$$\Pr\left(T\leq t\right)=\lambda_{X}\lambda_{Y}\int_{0}^{\infty}\int_{0}^{\infty}e^{-\lambda_{Z}t^{-1}\left(x+y\right)}e^{-\lambda_{X}x-\lambda_{Y}y}dxdy$$
Setting: $$\rho_{X}:=\lambda_{Z}t^{-1}+\lambda_{X}\text{ and }\rho_{Y}:=\lambda_{Z}t^{-1}+\lambda_{X}$$
we find:
$$\int_{0}^{\infty}\int_{0}^{\infty}e^{-\lambda_{Z}t^{-1}\left(x+y\right)}e^{-\lambda_{X}x-\lambda_{Y}y}dxdy=\int_{0}^{\infty}\int_{0}^{\infty}e^{-\rho_{X}x-\rho_{Y}y}dxdy=$$$$\int_{0}^{\infty}e^{-\rho_{X}x}dx\int_{0}^{\infty}e^{-\rho_{Y}y}dxdy=\frac{1}{\rho_{X}\rho_{Y}}$$
So we end up with: $$F_T(t)=\frac{\lambda_{X}\lambda_{Y}}{\rho_{X}\rho_{Y}}=\frac{t^{2}}{\left(\frac{\lambda_{Z}}{\lambda_{X}}+t\right)\left(\frac{\lambda_{Z}}{\lambda_{Y}}+t\right)}$$
Don't forget to check me on mistakes though. I have not much trust in myself by things like this.
