I am trying to find joint PDF function from a CDF, which is defined as $F(x,y)=\frac{(x+y)}{2}$ for $x,y \in [0,1]$.
I guess just taking derivatives w.r.t to $x$ and $y$ will not help me here. Could you please provide me with a hint on how to approach this?
Let $Z\sim U[0,1]$ be a standard uniform random variable, and let $B$ be an independent random bit with $P(B=0)=P(B=1)=1/2$. Then the random variables $(X,Y) = (BU,(1-B)U)$ have the given cdf. That is, the probability mass for $X,Y$ is $1/2$ of one-dimensional Lebesgue measure spread on the line segment connecting $(0,0)$ with $(0,1)$ plus $1/2$ of one-dimensional Lebesgue measure on the line segment connecting $(0,0)$ with $(1,0)$.
A hint making this plausible: the function $F$ is not continuous on the line segments connecting $(0,0)$ with $(0,1)$ and $(1,0)$.
This joint distribution is not absolutely continuous with respect to 2-dimensional Lebesgue measure, and hence does not have a joint probability density function. Or, if one's love of formulas wants one to fib a little bit, one can say the density "function" is $(\delta(x)+\delta(y))/2$ on $[0,1]^2$, where $\delta$ is the usual delta "function".