The distribution of $Y/X$

Question

Let $X$ and $Y$ be r.v.'s whose joint distribution is the uniform distribution on the triangle $A = \{(x,y) \in \mathbb{R} : 0\leq y \leq x \leq 1\}$. The density function is then given by $f_{X,Y}(x,y) = 2 $ whenever $(x,y) \in A$ and $f_{X,Y} = 0$ otherwise. NOTE: $X$ and $Y$ does not have to be independent.

My question is: how can I compute the distribution of $Y/X$?

There are several methods that I've tried:

1. The hints that I've found so far on this website is by computing $F(k) = P(Y \leq kX)$. I am not sure what this is, to be honest. Is it the following integral? $$P(Y \leq kX) = \int_0^{kx}\frac{1}{x}\ dy$$

2. I thought maybe using the Jacobian transformation might help. Let $g(x,y) = (y/x, x)$. The problems with this are the following: First of all, I have no idea what domain should I choose such that $g$ is injective. My idea is to remove the boundaries of $A$, that is, $g: A\backslash\partial A \to \mathbb{R}$. However, I am not sure if that is a good approach or not. Secondly, does it matter if I define $g$ to be $g(x,y) = (y/x, y)$ or $g(x,y) = (y/x, x)$? If I can find a suitable domain for $g$, I can surely compute the joint distribution and afterward, compute its marginal distribution, right?

3. I thought maybe I can compute it like this: Let $Z = Y/X$, then $$P(Z \leq z) = E[1_{\{Z \leq z\}}(Z)]= E[1_{\{Z \leq z\}}(Y/X)]$$ So, $$P(Z \leq z) = \iint 1_{\{Z \leq z\}}(y/x)f(x,y)\ dxdy = 2\iint 1_{\{Z \leq z\}}(y/x)\ dxdy.$$ Is that even correct? Even if it is correct, I have no idea what to do afterward.

4. On Wikipedia, the ratio distribution is a thing apparently. The page says something about the distribution of $Z$ (here $Z = X/Y$). But $X$ and $Y$ are independent on that Wikipedia page.

As you can see, I am stuck. Maybe I think too hard about this, but I just don't see how I can compute the distribution of $Y/X$. I hope someone can help me with this.

Answer 1

let $k\leq 1$ $$P(Y\leq k X)=\int_{0}^{1} \int_{0}^{x} 1_{\{y\leq k x\}}2 \, dy \, dx$$

$$=\int_{0}^{1} \int_{0}^{kx} 2 \, dy \, dx = \int_{0}^{1} 2 kx \, dx=k$$

for $k>1$

$$P(Y\leq k X)=\int_{0}^{1} \int_{0}^{x} 1_{\{y\leq k x\}}2 \, dy \, dx$$ $$=\int_{0}^{1} \int_{0}^{x} 2 \, dy \, dx=1$$

Answer 2

Wikipedia's treatment of ratio distributions actually covers more than just the ratio of independent variables. The PDF of $Z:=Y/X\in[0,\,1]$ is $\int_{\Bbb R}|x|f_{X,\,Y}(x,\,zx)dx=\int_0^1|x|2dx=1$, i.e. the ratio is $U(0,\,1)$-distributed. This is geometrically obvious, because the marginal distribution of $Z$ is $U(0,\,1)$ for any fixed value of $X$. This agrees with @Masoud's answer, which derives the CDF instead of the PDF.