Let $X$ and $Y$ be continuous random variables such that the 2dimension random variable $(X,Y)$ is not continuous. why this imply that $$Cov(X,Y)>0$$

I tried to use the definition on expected value so $$Cov(X,Y)=E[XY] - E[X]E[Y]$$ $$E[XY]=\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{\infty}}_{\cssId{lower-bound-mathjax}{0}} P(XY>t)dt\,\cssId{int-var-mathjax}\displaystyle-\int\limits^{\cssId{upper-bound-mathjax}{0}}_{\cssId{lower-bound-mathjax}{- \infty}} P(XY\leq t)dt\,$$

\begin{equation} E[X]E[Y]=\int^{\infty}_{-\infty} xf_x(x)dx \int^{\infty}_{-\infty}yf_y(y)dy \end{equation}

How do I procced from here? may I say that $x=y=t$?

  • 2
    $\begingroup$ May i ask why you think the implication is true? $\endgroup$ – Leander Tilsted Kristensen Feb 23 at 9:31
  • 3
    $\begingroup$ If $X$ and $Y$ are continuous and $(X,Y)$ discontinuous with $\operatorname{Cov}(X,Y) > 0$, then $X$ and $-Y$ are also continuous and $(X,-Y)$ is discontinuous, but $\operatorname{Cov}(X,-Y) < 0$. $\endgroup$ – Leander Tilsted Kristensen Feb 23 at 9:33
  • $\begingroup$ You right I don't know what did I think to my self, this is just imply they not necescery independent. thank you for this note $\endgroup$ – Sagigever Feb 23 at 9:43

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