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In studying joint distribution and density functions I have become confused by this exponential distrobution:

$$F_{X,Y}(x,y)=(1-e^{-(x+y)})u(x)u(y)$$

It seems this should be a proper distribution as it is:

  1. monotonically increasing
  2. $F_{X,Y}(0,0)=F_{X,Y}(-\infty,y)=F_{X,Y}(x,-\infty)=F_{X,Y}(-\infty,-\infty)=0$
  3. $F_{X,Y}(\infty,\infty)=1$
  4. $0\le F_{X,Y}(x,y)\le1$

However if I take the partial derivatives with respect to X and Y to derive the density I get a function that is not a valid density:

$$\frac{\partial}{\partial x \partial y} F_{X,Y}(x,y)=\delta(x) \delta(y)-e^{-(x+y)}u(x)u(y)+e^{-(x+y)}u(x)\delta(y)+e^{-(x+y)}\delta(x)u(y)-e^{-(x+y)}\delta(x)\delta(y)$$

Which basically reduces to $-e^{-(x+y)}u(x)u(y)$ for non-zero values of X and Y. But this is not a valid density function. Can anyone offer insight on how I am going astray?

amerigo
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    Well, the reasonable conclusion is that $F$ is not a proper distribution function. Indeed $F(0,\infty)=F(\infty,0)=1$ leaving no place $for X,Y$ other than $(0,0)$ which is impossible. Distribution of a joint exponential is $F=(1-e^{-x})(1-e^{-y})$. – A.S. Oct 28 '15 at 05:31
  • Thanks for the prompt reply. I was leaning toward this answer, but the marginals seemed ok since they were 1 at $\infty$ and 0 at $-\infty$ but your comment about X,Y not having any place but the origin makes sense. thank you. – amerigo Oct 28 '15 at 07:32

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