Let $X,Y$ be real-valued random variables, which are dependent.

Want: Calculate $\mathbb{P}[\,\min\{X,Y\}\leqslant0\,]$ (without Monte Carlo)


  1. I can compute (numerically) $F_X$ and $F_Y$ (the CDFs of $X$ and $Y$, resp.)
  2. 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?

  • 1
    $\begingroup$ No, covariance is not sufficient in general.. The joint distribution is always sufficient. I'll post an example of why the covariance doesn't do it $\endgroup$ – Glen_b -Reinstate Monica Jul 27 '18 at 0:04
  • $\begingroup$ It's not quite clear what you mean by "What kind of information is sufficient to evaluate the joint CDF". Knowing the joint distribution is clearly sufficient, as is knowing the distribution of $\min(X,Y)$ for this specific problem. But generally speaking other kinds of information will not be sufficient unless they effectively give us enough information to get that. Can you give more of an indication in what form you have information? How do you come to know the population correlation but not the joint distribution? Are you really talking about estimation of various quantities from a sample? $\endgroup$ – Glen_b -Reinstate Monica Jul 27 '18 at 2:40
  • $\begingroup$ I admit the "what kind of information" is a vague question. And no, I'm not talking about samples in this case. The variables $X$ and $Y$ in my application are a little long-winded to write down. I may post a new question with more info if I can't figure it out. In the meantime, your counterexample succinctly answered the question above. Thank you $\endgroup$ – David M. Jul 27 '18 at 20:52
  • $\begingroup$ @Glen_b I have posted a question related to this with more information here, if you’re interested $\endgroup$ – David M. Aug 3 '18 at 20:40

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.


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