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So I was solving a question which required me to derive the joint distribution function from the following probability density:

$$ f(x,y)= \begin{cases} 2&\text{if } x >0 , y > 0 , x+y<1\\ 0 &\text{otherwise} \end{cases} $$

What I did was for when $x>0$, $y>0$ ,$x+y<1$
$F(x,y)= \int_{0}^{x}\int_{0}^{1-x} 2 dy dx$

$$ F(x,y)= \begin{cases} 2x- x^2&\text{if } x >0 , y > 0 , x+y<1\\ \end{cases} $$

Well I got only this and I'm getting super confused for the other cases (like if x<0) cause I'm not too sure about the limits. Would be very helpful if someone could explain what would be the limits in other cases

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    $\begingroup$ Since $f$ is obviously symmetric in $x$ and $y,$ $F$ needs to be symmetric too, but it is not: evidently you made a mistake in your calculations. Draw a picture. $\endgroup$ – whuber Nov 27 '18 at 17:56
  • $\begingroup$ The answer needs to be symmetric in $x$ and $y$ since the pdf is symmetric. That means your proposed answer cannot be right. You need to give the value of $F(x,y)$ in the following cases: $$ F(x,y) = \begin{cases} \cdots & \text{if } x\ge0,y\ge0, x+y\le1, \\ \cdots & \text{if } x\ge0, 0\le y\le 1, \\ \cdots & \text{if } y\ge 0, 0\le x\le 1, \\ \cdots & \text{if } 0\le x\le1,0\le y\le1, x+y>1, \\ \cdots & \text{if } x\ge1, y\ge1, \\ \cdots & \text{if } x < 0 \text{ or } y < 0. \end{cases} $$ $\endgroup$ – Michael Hardy Nov 27 '18 at 17:58
  • $\begingroup$ The last two above are trivial: $F(x,y) =1$ if $x\ge1 \text{ and } y\ge1,$ and $F(x,y)=0$ if $x<0\text{ or } y<0.\qquad $ $\endgroup$ – Michael Hardy Nov 27 '18 at 17:58
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    $\begingroup$ I suggest posting your work - will make it easier to identify your mistake and point you in the right direction. $\endgroup$ – SecretAgentMan Nov 27 '18 at 19:47

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