I have a question for you and thank you in advance for your answers and ideas.

Let us suppose that we have the marginal distributions of two r.v X and Y, and also the law of X-Y (or any linear function of X and Y).

How could we find the joint distribution f(x,y) (not always unique) of both my r.v?

In case we suppose that (X,Y) is a gaussian vector,the unicity helds. My guess is that same goes for one paramater copulas (Gumbel or Clayton for instance). I am actually looking for the impact of the third marginal on the choice of the copula function and the cases in which the copula to use is unique.



  • 2
    $\begingroup$ I don't have any clever mathematical answer to this, just pointing out a situation where this problem arises. Given three currencies, say JPY, USD and EUR then the distributions (in a risk neutral measure) of the spots of the crosses USD/JPY, EUR/JPY, EUR/USD can each be implied from the options markets. Furthermore, log(Spot(USD/JPY))=log(Spot(EUR/JPY))-log(Spot(EUR/USD)). So, solving your problem is required to imply the joint distributions of these fx spots at any given time from the options markets, so that you can then price derivatives depending on multiple FX crosses. $\endgroup$ Feb 8, 2010 at 20:55
  • $\begingroup$ btw, I'm guessing that you also post on Wilmott (or, at least, someone else asked the same question). Don't know if there's any good answers except for certain special case distributions, such as Gaussians. $\endgroup$ Feb 8, 2010 at 22:25
  • $\begingroup$ Thank you George. I really liked your FX example. Interesting $\endgroup$
    – Averroes
    Feb 9, 2010 at 1:35

1 Answer 1


Let us think about the discrete case; that is let us suppose we are interested in determining a probability distribution $P$ on the discrete set $\Omega ={\mathbb Z}_n^2$.

Such a probability distribution assigns a nonnegative weight to each $(i,j) \in \Omega$. $|\Omega| = n^2$, thus $P$ is determined by $n^2-1$ nonzero variables $p_{i,j}$ whose sum is less than $1$. To fix the marginals of $P$ means to put $2(n-1)$ constraints on $\{p_{i,j}\}$. In addition to these constraints, the present question also imposes a distribution on $X-Y$. These translate into $n-1$ further constraints on $p_{i,j}$. Thus, in general, $P$ will be a function of $n^2-1-3(n-1)$ free variables.

The special cases you mention (i.e, the clayton and gumbel copulas as well as the normal distribution), however, are determined by the marginal distributions and an additional real parameter. In general, if the given data makes sense, $\theta$ can be recovered by first writing an equation that it satisfies and then solving it.

Under any of the above mentioned copulas the joint distribution equals $\Phi(F(x),G(y),\theta)$ where $F$ is the $X$ marginal, $G$ is the $Y$ marginal and $\theta$ is a real number. The only unknown here is $\theta$. Knowing the distribution of $X-Y$, means in particular we know the probability that $X-Y=n-1$[again assuming that we are operating in the discrete setup]. There is only one way this can happen, i.e, if $Y=0$ and $X=n-1$. Thus, we know the weight $p_{(n-1,0)}$ of the point $(n-1,0)$. Then, $\theta$ is the solution of $$\Phi(F(n-1),G(0),\theta) -\Phi(F(n-2),G(0),\theta) = p_{(n-1,0)}.$$

In the case of ${\mathbb R}^2$, one can for example, write the following equation for $\theta$: $$ \int_{\mathbb R}^2 (x-y) d\Phi(F(x),G(y),\theta) = \int z dS(z) $$ where $S$ is the distribution of $X-Y$ given in the problem.

Edit: more importantly, it seems, one has to check if the given distribution of $X-Y$ is compatible with the copula in question. Because, there are many equations that one can write for $\theta$ and all must give the same answer for the solution to be meaningful.

  • $\begingroup$ Thank you has2. Actually you came up with the exact results I found out till the moment. the difference with the use of Kandal's tau of rho here is the fact that we have an infinite set of equations that has to be consistent with a given copula. I have tried to find out a nice formula with the Gumbel's copula but the integral seems to be a bit complicated thus one could consider the integral as a function of the parameter and use an inversion method and compare the values found. I am thinking about finding the distribution of the parameter this way then use Bayes prior/posterior estimator $\endgroup$
    – Averroes
    Feb 9, 2010 at 0:11
  • $\begingroup$ You are welcome AlGoRiS. Another thought, hopefully not totally nonesense is as follows: the density of $X-Y$ equals $f(x,\theta) = \int_{\mathbb R} \Phi_{x,y}(F(y+c),G(y),\theta)dy$. Perhaps one can try to choose $\theta$ to minimize the distance between $f(x,\theta)$ and $S'$. $\endgroup$
    – has2
    Feb 9, 2010 at 0:28
  • $\begingroup$ oh Yes we could use Tykhonov regularisation to obtain a stable value. Again thank you has2 $\endgroup$
    – Averroes
    Feb 9, 2010 at 1:38

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