How to find the PDF of a function of two random variables

Question

Y is a uniform continuous random variable between [0,L] and X is a uniform continuous random variable given Y=y between [0,y]. What is the PDF of Z = X/Y.

Answer 1

First of all you have to find the joint density $f_{XY}(x,y)$

You know:

1. Marginal $$f_Y(y)=\frac{1}{L}\mathbb{1}_{[0;L]}(y)$$

2. Conditional $$f_{X|Y}(x|y)=\frac{1}{y}\mathbb{1}_{[0;y]}(x)$$

So your joint density is

$$f_{XY}(x,y)=\frac{1}{yL}\mathbb{1}_{[0;L]}(y)\mathbb{1}_{[0;y]}(x)$$

....now you can proceed as you explained...please post you efforts

EDIT:

First observe that $z \in [0;1]$

Theb, using the definition of CDF you get

$$F_Z(z)=\mathbb{P}[Z \leq z]=\mathbb{P}[Y >\frac{X}{z}]=$$

$$=\int_0^{Lz}dx\int_{\frac{X}{z}}^L \frac{1}{Ly}dy...$$

Some intermediate results:

$$\frac{1}{L}\int_0^{Lz}dx[ln y]_{\frac{x}{z}}^L=\frac{1}{L}\int_0^{Lz}[lnL-ln(x/z)]dx=...=zln L-(lnL-1)z=z$$

Note: the integral $\int_0^{Lz}ln(x/z)dx$ is solved by parts.