Joint Distribution function and joint density function I am really confused with what the question is asking and where to start. Thanks in advance for help.
Suppose $(a,b) $~uniform$([0,1]\times[0,1])$  (Does that mean we are given the join density function?)


*

*Determine the cdf $F: R^2 \to [0,1]$

*Determine the density $f$ from $F$.


We're given that $(A,B)$ have joint uniform distribution on the unit square $$\mathcal D :=[0,1]\times[0,1] = \{(x,y)\in\mathbb R^2: 0\leqslant x, y\leqslant 1\}. $$
It follows that for any Borel set $S\subset\mathbb R^2$, $$\mathbb P((A,B)\in S) = \lambda(S\cap\mathcal D),$$
where $\lambda$ is $2$-dimensional Lebesgue measure. The joint distribution function is given by 
$$
F_{X,Y}(x,y) = (x\wedge 1)\mathsf 1_{[0,\infty)}(x)(y\wedge 1)\mathsf 1_{[0,\infty)}(y)
$$
where denotes $\wedge$ denotes $\min$ and $\mathsf 1$ the indicator function. Since the joint distribution satisfies $F_{X,Y} = F_XF_Y$ (the product of the marginal distribution functions), $X$ and $Y$ are independent, so the joint density satisfies $f_{X,Y}=f_Xf_Y$. Now, $$f_X(x) = \frac{\mathsf d}{\mathsf dx}F_X(x) = \mathsf 1_{[0,1]}(x)$$ and similarly $f_Y(y)=\mathsf 1_{[0,1])}(y)$, so
$$f_{X,Y}(x,y) = \mathsf 1_{[0,1]}(x)\mathsf 1_{[0,1]}(y). $$
