How to tell if a bivariate random vector is of continuous or discrete type from its given distribution function?

Question

Let $ X = (X_1, X_2) $ be a bivariate random vector have distribution function

$$F(x,y) = \begin{cases} 0 && x<0 \;\;\text{or}\;\; y<0\\\\ \frac{1+xy}{2} && 0\leq x < 1,\; 0\leq y < 1 \\\\\frac{1+x}{2} && 0\leq x < 1, y\geq 1 \\\\ \frac{1+y}{2} && x\geq 1, \;0\leq y < 1 \\\\ 1 && x\geq 1, \;y\geq 1 \end{cases}$$

How can I tell if $X$ is continuous or discrete or neither?

My Attempt:

Clearly, $$ F(0,0) = \frac12$$

But if we approach $(0,0)$ using the path $y = x$ where $ x\to 0^-$, then

$$\lim_{x\to 0^-}F(x,x) = 0 \neq \frac12$$

So $F$ is not continuous at $x=0$ $\Rightarrow X$ can't be of continuous type.

How do I prove / disprove for the discrete case?

Answer 1

Hint: By definition, the distribution of $\ (X,Y)\ $ is discrete if and only if there is a countable set $\ S\subseteq\mathbb{R}^2\ $ such that $$ \sum_{(a,b)\in S}P(X=a,Y=b)=1\ . $$ But the only value of $\ (a,b)\ $ for which $\ P(X=a,Y=b)\ne0\ $ is $\ (0,0)\ $, for which you have $\ P(X=0,Y=0)=\frac{1}{2}\ $.