Let $(X,Y)$ be a random vector. Suppose that the marginal distribution functions of $X$ and $Y$ are known (say $F_1$ and $F_2$). Then the joint law of $(X,Y)$ is given by the following formula:


where C is some copula function. It means, that if the distributions of $X$ and $Y$ are given, we can construct (at least one) joint distribution of $(X,Y)$ conforming with given distributions (e.g. one can set $F_{X,Y}(x,y)=F_1(x)F_2(y)$.

I wonder whether is it possible to extent this property to the case when the distribution of $X+Y$ is known as well (say $F_3$).

My question is: if the distributions of $X$, $Y$ and $X+Y$ are given how can I construct (at least one) joint distribution of $(X,Y)$, conforming with univariate distributions?

Is there any closed-form solution like we have in case when only the marginals are given?

Thank you for the answers!

  • $\begingroup$ I must be missing something in the question. Why not just take $h$ to be anything, and take $D(a,b,c)=C(a,b)$ where $C$ is the function as given before? $\endgroup$ Jul 15, 2010 at 14:38
  • 1
    $\begingroup$ James, thank you for the comment. You can not take $D(a,b,c)=C(a,b)$, because then the law of $X+Y$ would be differ from the given disribution of $X+Y$, (which is $F_3$). I would like to construct the joint distribution of $(X,Y)$, such that univariate distributions of $X$, $Y$ and $X+Y$ are equal to some given distributions $F_1$, $F_2$ and $F_3$ correspondingly. Do you see my point? $\endgroup$
    – Oleg
    Jul 16, 2010 at 10:21
  • $\begingroup$ To rephrase James comment, the choice $D(a,b,c)=ab$ answers your question. This answer may not interest you as it leads to a trivial construction (independence) but then you should reformulate your question! In addition what does a function need to be a copula (definition?)? Anyway, maybe you are rising an interesting idea somewhere but I guess you have to think again about your question. Maybe you want to know the whole class of joint distribution $F_{XY}$ that can be written like in your last equation? also maybe you should work with density or characteristic functions? $\endgroup$ Jul 16, 2010 at 11:49
  • $\begingroup$ Robin, thanks for the comment. However, if we choose $D(a,b,c)=ab$, then the law of $X+Y$ might be differ from given distribution $F_3$, so we can not do it. We would like to construct such distribution of (X,Y) that the law of $X$ is equal to a given law $F_1$, law of $Y$ is equal to a given law $F_2$ and law of $X+Y$ is equal to a given law $F_3$. $\endgroup$
    – Oleg
    Jul 16, 2010 at 12:05
  • $\begingroup$ I just wanted to note that this is equivalent to the previous question mathoverflow.net/questions/12853/… $\endgroup$ Jul 17, 2010 at 9:07

2 Answers 2


You can not prescribe the distribution of the sum. Counterexample: Let X and Y be uniform on [0,1]. Now choose the distribution for X+Y so that P(X+Y < 0.5) = 1. This means P(X > 0.5) = 0 a contradiction to uniform.

A way to visualize this might be looking at mass distributions on the square [0,1]x[0,1]. Prescribing the margins (here uniform) is a restriction on the projections to the axes (i.e. 0x[0,1] and [0,1]x0) and the remaining freedom is distributing the mass in the square.

  • 2
    $\begingroup$ I agree with you, that in general case the solution of the problem might not exist. However, the question is how can we find at least one solution if it exists? $\endgroup$
    – Oleg
    Jul 19, 2010 at 9:29

OK then another answer to another problem:

An easy and practical way might be discretizing all functions. Then your conditions are linear constraints which can be solved for $C$.

Since one can always transform with the inverse cdf we can assume without loss of generality that both marginal distributions are uniform [0,1]. Let $F$ be the distribution of the sum of the marginals.

Now partition the interval [0,1] into n intervals [k-1/n, k/n]. By taking products you arrive at a partition of the square $[0,1]x [0,1]$ into $n^2$ squares.

Let $C_{i,j}$ be the mass of $C$ on each of the squares of the grid. You arrive at 3 different types of constraints for $C_{i,j}$:

(1) $\sum_{i,j} C_{i,j} = 1$ this is the condition that the copula is a probability distribution

(2) $\sum_{i} C_{i,j} = 1/n$, $\sum_{j} C_{i,j} = 1/n$ this is the condition that the marginals are uniform

(3) With $I_d = \{(i,j)| i+j = d\}$ the d-th diagonal with d = 2,...,2n the condition on the distribution of the sum is $\sum_{I_d} C_{i,j} = \sum_{I_d} F(1/i + 1/j)$.

These are 1 + n + n + (2n - 1) linear equations for the $n^2$ unknowns.

I guess this will give a numeric solution, a continous solution could be derived by a proper limit argument.

This should work not only for sums but also other functions of X and Y as well.

I hope this helps

  • $\begingroup$ G g, thanks for this idea. However, you missed one more type of constraints. Namely, since $C_{ij}$ is the mass, then $C_{ij}\geq0$ for all $i$ and $j$. So we have $4n$ equations, $n^2$ inequalities and $n^2$ unknowns. How would you recommend to deal with all this numerically? $\endgroup$
    – Oleg
    Jul 19, 2010 at 17:52
  • $\begingroup$ Well, you are right of course and nothing is perfect. I noted that I was a bit sloppy with (3) as well (F is a distribution not the mass). Dealing with this numerically? Well on the one hand this is simply linear optimization. I would get something like Matlab and just solve the equations. On the other hand, the best approach might depend on your specific requirements. If you have a concrete F in mind, say from measurements, this is would be more a problem of statistics than numerics. $\endgroup$
    – g g
    Jul 19, 2010 at 21:47
  • $\begingroup$ G g, there are two problems if dealing numeric: 1. We have to solve the linear equations and all the inequalities $C_{ij}\geq0$. 2. We have to choose the only one solution among all. We probably can use Lagrange multipliers to deal with this two problems, but I was looking for any kind of closed-form solution (like we have in case where just two marginals are given: $F_{XY}(x,y)=C(F_1(x),F_2(y))$. $\endgroup$
    – Oleg
    Jul 27, 2010 at 10:53
  • $\begingroup$ The real question is, given that X and Y are (marginally) U[0,1], which distributions for X+Y are "realizable". For example, X+Y must be supported on [0,2] and have expectation 1. $\endgroup$ Jul 27, 2010 at 12:42
  • 2
    $\begingroup$ Yuval, the answer to your question is given in this paper: "Best-possible bounds for the distribution of a sum — a problem of Kolmogorov" - Frank, Nelsen and Schweizer, 1987. However, my question is the opposite. Assume, that the distribution of $X+Y$ is realizable. Is there a way to find (at least one) joint distribution of $(X,Y)$. $\endgroup$
    – Oleg
    Jul 27, 2010 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.