# Distribution of the quotient of two random variables

Let $x$ and $y$ be two random variables with support of $\left[1\hspace{5pt}10\right]$ and $\left[50\hspace{5pt}90\right]$ respectively. The distribution of each of these variables is $p_X(x)$ and $p_Y(y)$ respectively. and their joint distribution is $p_{X,Y}(x,y)$. Now let us define two random variables as follows

$$z_1 = \frac{y}{x} \\ z_2 = x$$

The distribution, $p_{Z_1}(z_1)$, of $z_1$ can be found as follows $$p_{Z_1}(z_1) = \int_{-\infty}^{\infty} z_2p_{X,Y}(z_1z_2,z_2)dz_2$$

My question is:

1. Does the distribution $p_{Z_1}(z_1)$ depend upon the values of $x$, $y$ and $\frac{y}{x}$?
2. Does the distribution $p_{Z_1}(z_1)$ depend only upon $\frac{y}{x}$?

The reason for asking this question is as follows:

Assume $y = 70$ and $x = 7$. Then $Z_1 = 10$. Since these values of $x$ and $y$ are within their support, the joint distribution $p_{X,Y}(x,y)$ will be zon-zero and so will be $p_{Z_1}(z_1)$. On the other hand, assume $y = 150$ and $x = 15$. Both these values are outside their support. But $Z_1$ is still $10$. $p_{X,Y}(x,y)$ will be close to zero. What can we say about $p_{Z_1}(z_1)$? In fact, for given $Z_1 = 10$, we can choose several values of $x$ and $y$ within their support and with different $p_{X,Y}(x,y)$. Does the choice of the values of $x$ and $y$ affect $p_{Z_1}(z_1)$?

• It is stabbing a knife into the heart of a mathematician to say that $p(x)$ and $p(y)$ are different functions. – Matt Samuel Feb 10 '15 at 19:47
• @MattSamuel Actually the whole post is based on abuses of the kind---to the point that the OP themselves finally get lost in these. – Did Feb 10 '15 at 19:51
• Very well pointed out. I changed the notations. – user146290 Feb 10 '15 at 19:56
• Not sure I understand the question either: $p_{Z_1}(z_1)$ is an integral hence it can be positive even though the function it integrates is zero at some arguments. – Did Feb 10 '15 at 20:01
• The support of $z_2$ is $[1\hspace{5pt}10]$. So, the integral is zero beyond this support. – user146290 Feb 10 '15 at 20:05