Transference of properties from marginals to joint density functions

Question

Let $(X,Y)$ be an absolutely continuous random vector and denote by $f_{(X,Y)}(x,y)$ its joint density function and $f_X(x)$, resp. $f_Y(y)$ the marginal density functions.

If $f_X$ and $f_Y$ are continuous and $X$ and $Y$ are independent then $f_{(X,Y)}(x,y)=f_X(x)f_Y(y)$ and hence $f_{(X,Y)}$ is continuous as well.

So, let us assume $X$ and $Y$ are not independent. How can one obtain regularity of the joint density having only properties from the marginal densities? Is it possible? Is there any criteria?

Maybe an illustrative example of a random vector with marginal continuous densities but discontinuous joint density would help. Is there a toy example of this?

Thanks a lot!

Answer 1

Let $f_{X,Y} \colon \def\R{\mathbf R}\R^2 \to \R$ by given by $$ f_{X,Y}(x,y) = \begin{cases} \frac 12 & \def\abs#1{\left|#1\right|}\abs x + \abs y \le 1\\ 0 & \text{otherwise} \end{cases} $$ Then $f_{X,Y}$ is discontinuous, but we have for $\abs x \le 1$ \begin{align*} f_X(x) &= \int_{\R} f_{X,Y}(x,y)\, dy\\ &= \int_{-1+\abs x}^{1- \abs x} \frac 12 \, dy\\ &= 2 - \abs x \end{align*} Hence $$ f_X(x) = \begin{cases} 2 - \abs x & x \in [-1,1]\\ 0 & \text{otherwise} \end{cases} $$ is continuous. Due to symmetry, $f_Y = f_X$ is also continuous.

Answer 2

For the first question, "So, let us assume X and Y are not independent. How can one obtain regularity of the joint density having only properties from the marginal densities? Is it possible? Is there any criteria?", Perhaps Copula could help you.

lets start with a example. In this example I want to generate distribution $F_{(X,Y)}(x,y)$ such that their marginals be $F_1$ and $F_2$ and $F_{(X,Y)}(t_1,t_2) \neq F_{1}(t_1) F_{2}(t_2) $ (so they are not independent).

lets $ F_1$ and $ F_2$ are continues and known.

let $\Phi$ is cumulative distribution of uni-variate standard normal . suppose $$(e_1,e_2) \sim \Phi_2$$ where $\Phi_2$ is CDF of standard bi-variate normal distribution, that is,

\begin{eqnarray} (e_1,e_2)\sim Normal \bigg( \left( \begin{array}{c} 0 \\ 0 \end{array} \right) , \left( \begin{array}{cc} 1 & \rho \\ \rho & 1 \end{array} \right) \bigg) \end{eqnarray}

define

$$X=F^{-1}_1(\Phi(e_1))$$

$$Y=F^{-1}_2(\Phi(e_2))$$

so

$$F_X(t_1)=P(X\leq t_1)=P(F^{-1}_1(\Phi(e_1))\leq t_1)$$

$$=P\bigg(e_1 \leq \Phi^{-1}\left( F_1( t_1) \right) \bigg)= \Phi \bigg( \Phi^{-1}\left( F_1( t_1) \right) \bigg)=F_1(t_1)$$

so $X\sim F_1$. similarity $Y\sim F_2$. In next step I show they are not independent.

$$F_{(X,Y)}(t_1,t_2)=P(X\leq t_1 , Y\leq t_2)$$ $$=P(F^{-1}_1(\Phi(e_1))\leq t_1 , F^{-1}_2(\Phi(e_2))\leq t_2)$$ $$=P\bigg(e_1 \leq \Phi^{-1}\left( F_1( t_1) \right) , e_2 \leq \Phi^{-1}\left( F_2( t_2) \right) \bigg)$$ $$=\Phi_2\bigg(\Phi^{-1}\left( F_1( t_1) \right) , \Phi^{-1}\left( F_2( t_2) \right) \bigg)$$

so

$$F_{(X,Y)}(t_1,t_2)=\Phi_2\bigg(\Phi^{-1}\left( F_1( t_1) \right) , \Phi^{-1}\left( F_2( t_2) \right) \bigg)$$

now if $\rho \neq 0 $

$$F_{(X,Y)}(t_1,t_2)=\Phi_2\bigg(\Phi^{-1}\left( F_1( t_1) \right) , \Phi^{-1}\left( F_2( t_2) \right) \bigg)$$

$$\neq F_1( t_1) \times F_2( t_2) $$

but if $\rho = 0 $ $$F_{(X,Y)}(t_1,t_2)=\Phi_2\bigg(\Phi^{-1}\left( F_1( t_1) \right) , \Phi^{-1}\left( F_2( t_2) \right) \bigg)$$ $$=\Phi \bigg(\Phi^{-1}\left( F_1( t_1) \right) \bigg) \times \Phi \bigg(\Phi^{-1}\left( F_2( t_2) \right) \bigg)$$ $$= F_1( t_1) \times F_2( t_2) $$

This method known as "Copula".

In general case let $F_1$ and $F_2$ are known.

$$(X,Y)\sim F_{(X,Y)}(t_1,t_2)=c\left(F_1(t_1),F_2(t_2)\right)$$

Now If $$c(a,b)=a\times b$$ so $X$ and $Y$ are independent. if $$c(a,b)\neq a\times b$$ they are dependent. you can have a different way by choosing different $c(a,b)$ . In this paper you can see three different choise for $c\left(F_1(t_1),F_2(t_2)\right)$.

For the second question, find example with such condition ,you may find a $c\left(F_1(t_1),F_2(t_2)\right)$ such that satisfy your condition.