Is the following function

$F(x,y) = 1 - e^{-xy}$ for $x,y \ge 0$

$F(x,y)= 0 $ otherwise

a joint distribution function for two random variables X and Y ? Provide reasons.

This is my homework problem. I'm not sure if he means CDF or PDF. I integrated the function $\int_0^\infty \int_0^\infty F(x,y)\space dx \space dy$ in Wolfram Alpha and it doesn't converge (not surprising because the constant term 1 blows up). So it's not a PDF.

If it's a CDF then taking $\frac{\partial}{\partial x} \frac{\partial F}{\partial y} = e^{-xy}(1-xy)$ gives the alleged PDF, but integrating it $\int_0^\infty \int_0^\infty e^{-xy}(1-xy) \space dx \space dy$ in Wolfram Alpha yields 0. So either way that's not a PDF.

So either way the original F is not a joing distribution function.

Is this accurate? Are these valid tests? Is there a different sense of "joint distribution function" other than CDF and PDF to test for?

  • $\begingroup$ You do not have to derive it to verify wether or not it is a CDF. Use the definition : $F(x, y) = P(X \le x, Y \le y)$ for some r.v.s $X$ and $Y$. $\endgroup$ – Olivier Mar 4 '15 at 19:03
  • $\begingroup$ @Olivier Thanks for your time. I'm not sure what you mean. Is my argument invalid? $\endgroup$ – Leo 254 Mar 4 '15 at 19:08
  • $\begingroup$ For the CDF test, yes. You simply have to verify that $F$ is a probability (some conditions here) that increases with $x$ and $y$. $\endgroup$ – Olivier Mar 4 '15 at 19:14
  • $\begingroup$ this could help. $\endgroup$ – drhab Mar 4 '15 at 19:18
  • $\begingroup$ @Olivier I think you missed a link in your comment on "here" $\endgroup$ – Leo 254 Mar 4 '15 at 19:32

If they say "distribution function" that means CDF. If they meant PDF they would say "density function".

The given function is NOT a valid distribution function.

First check that $\lim_{x,y \to -\infty}{F(x,y)} = 0$ and that $\lim_{x,y \to \infty}{F(x,y)} = 1$. That's OK in this case.

$F$ is a continuous function so it must have a density function given by $f(x,y) := \dfrac{\partial F(x,y)}{\partial x\partial y}$.

Check that $f(x,y) \geq 0$ for all $x,y$.

If this holds then $F$ is a joint distribution function.

You found $f(x,y) = e^{-xy}(1-xy)$. This is negative when $xy\gt 1$. Therefore, $f$ is not a density function and so $F$ is not a distribution function.


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