We have the iid random variables $(X,Y)$ where $f_x(x)=\lambda e^{-\lambda x}$, $x>0$.
We are given $Z=\frac{X}{Y}$ and asked to find the cdf and the density function.
Here's my attempt.
$Z=\frac{X}{Y}$, $W=Y$.
I think this is a one-to-one transformation because we can express $X,Y$ in terms of $Z,W$ as $X=ZW$, $Y=W$.
Then,
$f_{ZW}(z,w)=\frac{f_{XY}(x,y)}{|\partial(z,w) / \partial(x,y)|_{x=zw,y=w}}$.
With:
$\frac{\partial(z,w)}{\partial(x,y)}=\left[ \begin{array}{ c c } \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} \\ \end{array} \right]$
$=\left[ \begin{array}{ c c } \frac{1}{y} & x \\ 0 & 1 \\ \end{array} \right]$
Therefore,
$f_{ZW}(z,w)=\frac{f_X(x) f_Y(y)}{w} = \frac{f_X(zw)f_Y(w)}{w}$.
From there, though, I'm not quite sure how to finish.
I know that it follows:
$f_Z(z)=\int_{-\infty}^{\infty}f_{ZW}(z,w)dw$ $=\int_{-\infty}^{\infty}\frac{f_{X}(zw)f_Y(w)dw}{w}$.
But I don't know how to get my PDF and density function from this result. Do I just use the exponential function and plug in my new variables? Any help appreciated.
$${\partial(x,y)\over\partial(z,w)}=\begin{bmatrix}w&z\\0&1\end{bmatrix}$$
$${f_Z(z)}=\int_{0}^\infty f_X(x)f_Y(y)\Big{|}{\partial(x,y)\over\partial(z,w)}\Big{|}dw=\lambda^2\int_{0}^\infty we^{-(z+1)\lambda w}dw={1\over(z+1)^2}$$