Is it possible to simulate pairs of random variables with a given marginal distribution and population correlation where one random variable is larger than the other?

More formally, I need to simulate pairs of random variables $(X_1, Y_1), \dots(X_n, Y_n),$ where $X_1, \dots X_n \sim f(\cdot;\Theta)$, $Y_1, \dots Y_n \sim g(; \Psi)$, $f$ and $g$ are continuous probability distributions, and $Cor(X_i, Y_i) = \rho$. Is it possible to simulate these pairs of random variables such that $X_i \leq Y_i$ for all $1 \leq i \leq n$?

I can weaken the conditions slightly: a strict inequality, $X_i < Y_i$, would be fine, the population or sample correlation can close to $\rho$, $X$ and $Y$ are from the same family of distributions but with different parameters.

Without the inequality constraint, this is easy to simulate using a copula: Generate $(W_{i1} ,W_{i2}) \sim \mathcal{N}_2(0, \Sigma)$, use probability-integral-transform to get correlated $U[0,1]$ random variables, then plug them into the marginals. I don't know how to do this with the constraint. My instinct is to keep working with a copula of some kind, but I am not tied to this approach. Is this kind of random variable generation possible?

  • $\begingroup$ You probably do not want perfect correlation, right? Else, something like $X\sim U[a,b]$ and $Y=b+\epsilon+X$ would do the trick. $\endgroup$ – Christoph Hanck Jun 22 '20 at 14:01
  • $\begingroup$ Writing $Y=X+\epsilon$, the question boils down to finding the conditional distribution of $\epsilon>0$ such that $Y$ is marginally distributed from $g$ and the correlation is $\rho$. Assuming this is possible. $\endgroup$ – Xi'an Jun 22 '20 at 14:29
  • $\begingroup$ A counter example is when $X$ and $Y$ are both marginally $\mathcal N(0,1)$ because, unless $X=Y$ with probability one, it is impossible that $Y>X$ with probability one. $\endgroup$ – Xi'an Jun 22 '20 at 14:32
  • $\begingroup$ @ChristophHanck, what do you mean by perfect correlation? Cor(X, Y) = 1? That wouldn't work $\endgroup$ – Eli Jun 22 '20 at 14:44
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    $\begingroup$ Inverting the question sheds considerable light on it: if you can guarantee that $X_i\le Y_i$ for all $i,$ then the support of the distribution a fortiori must lie in the half plane $x \le y.$ Thus, when the marginals and $\rho$ are consistent with a bivariate distribution supported in that region you can succeed; otherwise you cannot. $\endgroup$ – whuber Jun 22 '20 at 18:07

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