In case of joint probabiility distribution of $X$ and $Y$, if we define $F(x,y) = \mathbb P(X \le x, Y \le y) = x + y$ for $x, y \in [0,0.5]$, $1$ for $x,y > 0.5$ and $0$ elsewhere, then the joint probability distribution function $f(x,y)$ will be $0$ everywhere (defined as the mixed derivative over both $x$ and $y$). However the integral of $f(x,y)$ can't be 0, it should be equal to 1 over the whole range of $x$ and $y$, especially knowing that the CDF is not zero.
What am I missing here?
PDF and CDF in joint probability distribution
integrationprobabilityprobability distributions
Related Solutions
A joint CDF $F_{X,Y}(x,y)$ gives the probability $$\Pr[(X \le x) \cap (Y \le y)].$$ Geometrically, what this means is that if you have a joint density $f_{X,Y}(x,y)$, then the CDF gives the total volume under the density over the region $(X \le x) \cap (Y \le y)$. That is to say, you are "cutting" the surface along $X = x$ and $Y = y$, and then discarding those pieces for which $X > x$ or $Y > y$. Here is a plot of the density:
Now you can see that if $x < 0$ or $y < 0$, then the point $(x,y)$ is in the L-shaped region to the left of the figure, and there is no volume in that region--the density is zero. That's the first part of the piecewise function in the answer. Now, if you're in the region $y > x > 0$, $(x,y)$ is in the flat triangular area just behind the curved wedge. But the rectangular region $(X \le x) \cap (Y \le y)$ for this point includes part of this wedge, but how much it includes does not depend on $y$ once $y$ is at least as large as $x$. That's the third part of the piecewise function: $$F_{X,Y}(x,y) = 1 - e^{-x} - xe^{-x}, \quad y > x > 0.$$ So $F(3,5) = F(3,10) = F(3,51147034)$. But if you choose a point inside the curved wedge; i.e., $0 < y < x$, then you can see that you're not only cutting away volume to the right, but also some volume to the back. So that's the second part of the piecewise CDF.
I won't go into more mathematical detail since I mainly wanted to give you a visual, intuitive explanation of what's going on. I find that this helps greatly when doing the actual computation.
We always have $X<Y$. So if $x > y$ then
$$F(x,y)-F(y,y)=\int_y^x \int_{-\infty}^y f(u,v) du dv = \int_y^x \int_{-\infty}^y 0 du dv = 0.$$
In other words, if $x>y$ then $F(x,y)=F(y,y)$. So that reduces the problem to case 3. In case 3 you just have to calculate the integral:
$$\int_{-\infty}^y \int_{-\infty}^{\min \{ v,x \}} 2 e^{-u-v} du dv.$$
I got the limits here by rephrasing the region of integration defined as $\{ (u,v) : u \leq x, \, v \leq y, \, u \leq v \}$
Best Answer
Not every distribution function has a density (w.r.t. Lebesgue measure). In this case, suppose $(X,Y)$ had distribution function $F$. Then $P(X= 0)=P(X\leq 0, Y \leq 0.5)=\frac 1 2$. This implies that $X$ has no density function. Neither does $(X,Y)$.