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I have to compute CDF of random vector (X,Y) using the following probability density function

       c   x>=0, y>=0 and x+y<=2 

f(x,y) = 0 otherwise

The solution has several cases (x>=0,y>=0, x+y<=2) (x>=2,0<=y<2), (y>=2,0<=x<2), (0<=x<2, 0<=y<2,x+y>=2). Why I should be interested to find these cases? Could someone help me to understand this exercise? thank you in advance.

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  • $\begingroup$ The joint CDF is the probability $F(x,y)=P(X\le x,Y\le y)=\int_{-\infty}^y\int_{-\infty}^x f(u,v)\,du\,dv$ for all $(x,y)$. We should consider the different cases you mention because we need them to specify the CDF completely. It is only natural that $F(x,y)$ turns out to be a piecewise function, completely defined for all possible values that $x$ and $y$ can take. Note that $F:\mathbb R^2\to \mathbb R$. $\endgroup$ – StubbornAtom Nov 19 '18 at 19:52
  • $\begingroup$ For example if I want to compute F(x,y) for this case (0<=x<2, 0<=y<2,x+y>=2), what are the endpoints of the integrals? $\endgroup$ – Francisco Nov 19 '18 at 19:56

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