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!


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.

  • $\begingroup$ Aha. Is there any criteria to avoid such examples? I mean, $f_X,f_Y$ continuous + something else implies $f_{(X,Y)}$ is jointly continuous. $\endgroup$ – Martingalo Jun 4 '15 at 12:52

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