Is there a definition of positively correlated that involves $P(X,Y) > P(X)P(Y)$? Suppose $X,Y$ are random variables. Then $X \perp Y$ if 
$$P(X,Y)=P(X)P(Y)$$ 
Now, suppose 
$$P(X,Y) > P (X)P(Y)$$ 
Is there a notion of correlation where $P(X,Y) > P(X)P(Y)$ is called "positively correlated", $P(X,Y) < P(X)P(Y)$ is "negatively correlated" and $P(X,Y) = P(X)P(Y)$ is "uncorrelated"? 
I encountered this reading 
https://plato.stanford.edu/entries/causation-probabilistic/

But Pearson correlation was defined as 
$$\rho = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y}$$ 
and I thought you could have no correlation (i.e. $\rho = 0$) while $P(X,Y)\ne P(X)P(Y)$
 A: The common cause principle uses event notation and $A,B,C$ denotes specific events, such that their complements, i.e. $\tilde C$, are different events. But, you define $X,Y$ as random variables, not specific events. This way, when you say $P(X,Y)>P(X)P(Y)$, which is a vague notation, it means
$$P_{X,Y}(x,y)>P_X(x)P_Y(y) \ \ \forall x,y$$ 
So, it should hold for all values of $x,y$. But in CCP, when we have $P(A,B)>P(A)P(B)$, it doesn't mean the following need to hold: $$P(\tilde A,\tilde B)>P(\tilde A)P(\tilde B)\\P(A,\tilde B)>P(A)P(\tilde B)\\P(\tilde A, B)>P(\tilde A)P(B)$$
Actually, they can't all be true together. So, there is a notational confusion in your assertion. 
Additionally, if you  integrate/sum both sides of your inequality, both will give $1$ and you'll end up $1>1$.
A: 
Is there a notion of correlation where $P(X,Y) > P(X)P(Y)$ is called "positively correlated", $P(X,Y) < P(X)P(Y)$ is "negatively correlated" and $P(X,Y) = P(X)P(Y)$ is "uncorrelated"? 

No, there is no such notion. As gunes and user20160 have pointed out, $P(X,Y) > P(X)P(Y)$ cannot possibly hold for all choices of $X$ and $Y$. Very broadly speaking with accompanying vigorous hand-waving and more as an intuitive feeling about things than a formal notion, two centered (zero-mean) random variables are positively correlated (Pearson correlation) if the total probability mass in the first and third quadrants exceeds the total probability mass in the second and fourth quadrants (For concentered random variables, define the quadrants with respect to the mean point $(\mu_X, \mu_Y)$ instead of with respect to the origin $(0,0)$).

I thought you could have no correlation (i.e. $\rho = 0$) while $P(X,Y)\ne P(X)P(Y)$

Yes indeed. Consider discrete random variables taking on values $(0,1)$, $((1,0)$, $(-1,0)$ and $(0,-1)$ with equal probability  $\frac 14$. $P(X,Y)$ does not equal  $(P(X)P(Y)$ for any choice of $X$ and $Y$ and yet the random variable are uncorrelated. If the four points were $(\pm 1, \pm 1)$ instead, $P(X.Y)$ would equal $P(X)P(Y)$ everywhere, that is, $X$ and $Y$ would be independent random variables and hence uncorrelated random variables. Note that in both cases, my broad notion of correlation (actually, lack thereof) holds: the total probability mass in the first and third quadrants equals the total probability mass in the second and fourth quadrants.  If you would like continuous random variables, consider $(X,Y)$ being uniformly distributed on the unit disc. $X$ and $Y$ are dependent random variables but they are uncorrelated random variables.
