## Introduction

This article is about the relationship between two random variables, x and y, that have a joint density function. We will explore how their individual densities are related to one another and how this affects the overall shape of the joint density function.

## Theorem

Let X and Y have a joint density function f(x,y) > 0 for all x in (-infinity, infinity), y in (-infinity, infinity). If f(x,y)=f(y,x) for all x and y then X and Y are independent.

## Proof

We will prove this by showing that the joint density function of x and y is equal to 1y – (if x and y have joint density function 1y).

First, let’s recall the definition of a joint density function:

Let x and y be two random variables with joint probability distribution P(x,y). The set of all possible values of (x,y) is called the support of the joint distribution. The joint density function of x and y is a function f(x,y) defined on the support of the joint distribution such that for any Borel set B in the plane,

Pr(x ∈ B, y ∈ B) = ∫∫ B f(x,y)dydx.

Now, let’s consider the left-hand side of our equation: Pr(x ∈ B, y ∈ B). This is equal to the probability that x is in the set B and y is in the set B. Thus, we have:

Pr(x ∈ B, y ∈ B) = Pr(x ∈ B)*Pr(y ∈ B).

Now, let’s consider the right-hand side of our equation: ∫∫B f(x,y)dydx. This is equal to the integral over all possible values of x and y in the set B. Thus, we have:

∫∫B f(x,y)dydx = ∫B∫Bf(x,y)*1*dy*dx.

By substituting in our expressions for Pr(x ∈ B)*Pr(y ∈ B), we have:

Pr(x∩B=1)*Pr(Y^-b)=∫BfX^-b)(Y>dy*dx

We now know that 1 – (if x and y have joint density function 1 – (1 – x))=1/xy .

## Conclusion

We have seen that if x and y have a joint density function f(x,y) then the marginal density function of x is given by: f_x(x)=int_{-infty}^{\infty} f(x,y) dy and the marginal density function of y is given by: f_y(y)=int_{-infty}^{\infty} f(x,y) dx.