Joint CDF of dependent random variables: is knowing covariance sufficient? Let $X,Y$ be real-valued random variables, which are dependent.
Want: Calculate $\mathbb{P}[\,\min\{X,Y\}\leqslant0\,]$ (without Monte Carlo)
Know: 


*

*I can compute (numerically) $F_X$ and $F_Y$ (the CDFs of $X$ and $Y$, resp.)

*I know $\operatorname{Var}(X)$, $\operatorname{Var}(Y)$ and $\operatorname{Cov}(X,Y)$.


Thoughts: We have that
$$\mathbb{P}[\,\min\{X,Y\}\leqslant0\,]=F_X(0)+F_Y(0)-F_{XY}(0,0),$$
so my problem reduces to computing $F_{XY}$ (the joint CDF of $X$ and $Y$). I know intuitively that $F_{XY}$ and $\operatorname{Cov}(X,Y)$ both capture information about how $X$ and $Y$ are related, but is knowing the covariance sufficient to evaluate the joint CDF? Since covariance only really measures linear dependence, I'm worried that it's not enough.
Are there any other options? What kind of information is sufficient to evaluate the joint CDF and/or the expression (Want) above?
 A: That covariance is not sufficient can be indicated by counterexample.
If we start with $U$ standard uniform on $[0,1)$ and set $V=U$ when $U> \frac12$ and set $V=\frac12-U$ when $U\leq \frac12$ then $V$ is also standard uniform. 
If we let $X=\Phi^{-1}(U)$ (and similarly for $Y$ in terms of $V$) then $X$ and $Y$ are standard normal and $X$ and $Y$ are dependent. Their covariance is about $-0.66$. In this case $P(\min(X,Y)<0)=\frac12$. 
We can then take a new variable $Z$ such that $(X,Z)$ are bivariate normal (with standard normal margins), with the same covariance as between $X$ and $Y$. 
Then $P(\min(X,Z)<0)$ is considerably larger that $\frac12$ (... it looks to be somewhere in the region of $0.63$).
The joint distribution itself is sufficient to find the probability you seek. To get by with less, you'd usually need sufficient information to calculate the distribution of $\min(X,Y)$; there may be a number of special cases where the probability can be calculated without directly computing that distribution.
