# Extension of copulas

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:

$$F_{X,Y}(x,y)=C(F_1(x),F_2(y)),$$

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?

• 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? – James Martin Jul 15 '10 at 14:38
• 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? – Oleg Jul 16 '10 at 10:21
• 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? – robin girard Jul 16 '10 at 11:49
• 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 $F_3$. – Oleg Jul 16 '10 at 12:05
• I just wanted to note that this is equivalent to the previous question mathoverflow.net/questions/12853/… – Gjergji Zaimi Jul 17 '10 at 9:07

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.

• 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? – Oleg Jul 19 '10 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

• 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? – Oleg Jul 19 '10 at 17:52
• 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. – g g Jul 19 '10 at 21:47
• 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: $F_{XY}(x,y)=C(F_1(x),F_2(y))$. – Oleg Jul 27 '10 at 10:53
• 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. – Yuval Filmus Jul 27 '10 at 12:42
• 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 $(X,Y)$. – Oleg Jul 27 '10 at 15:59