Assume we have 3 random variables $X_1,X_2,X_3$, and we know the pairwise marginal distribution $P(X_1,X_2), P(X_2,X_3), P(X_3,X_1)$, but we don't know anything else (such as conditional independence). Can we get the joint distribution $P(X_1,X_2,X_3)$?



Consider a trivariate distribution with bivariate (standard, independent) normal margins, but with half the octants having 0 probability and half having double probability. Specifically, consider octants ---, -++, +-+, ++- have double probability.

Then the bivariate margins are indistinguishable from the one you'd get with three iid standard normal variates. Indeed, there's an infinity of trivariate distributions which would produce the same bivariate margins

As Dilip Sawarte points out in comments he has discussed essentially the same example in an answer (but reversing the octants which are doubled and zeroed), and defines it in a more formal way. Whuber mentions an example involving Bernoulli variates that (in the trivariate case) looks like this:

  X3=0      X1                  X3=1      X1
          0    1                        0    1

    0    1/4   0                  0     0   1/4 
 X2                         X2
    1     0   1/4                 1    1/4   0

... where every bivariate margin would be

          0    1       

    0    1/4  1/4      
    1    1/4  1/4    

and so would be equivalent to the case of three independent variates (or indeed to three with exactly the reverse form of dependence).

A closely related example I initially started to write about involved a trivariate uniform with alternating "slices" in a checkerboard pattern of greater and lower probability (generalizing the usual zero and double).

So you can't compute the trivariate from bivariate margins in general.

  • 5
    $\begingroup$ +1. Another standard example--the simplest possible, and closely related to yours--is to let the $X_i$ be independent Bernoulli$(1/2)$ variables. The full distribution can be tabulated since there are only eight equiprobable outcomes. Their marginals and pairwise marginals are the same after conditioning the $X_i$ to have an even number of zeros (just cross out the other rows in the table and double all their probabilities), but the two joint distributions obviously differ. $\endgroup$ – whuber Sep 26 '17 at 17:29
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    $\begingroup$ +1 The trivariate distribution is written out in detail in this answer of mine except that I used the octants $+++, +--, -+-, --+$ instead. It is, of course, related to the Bernoulli random variables mentioned by @whuber which example goes back to Bernstein, I believe. $\endgroup$ – Dilip Sarwate Sep 26 '17 at 17:45
  • $\begingroup$ But, in less artificial cases, maybe some bounds could be made? $\endgroup$ – kjetil b halvorsen Sep 26 '17 at 18:27
  • $\begingroup$ there's got to be a copula solution here. Sklar's theorem has extension to n-dimensional case, and there you have only marginals, not the bivariate marginals which have more information $\endgroup$ – Aksakal Sep 27 '17 at 4:36
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    $\begingroup$ Aksakal The copula itself completely specifies the dependence structure, not the marginals. The fact that you can keep the marginals but change the copula is a simpler version of the same issue here. $\endgroup$ – Glen_b Sep 27 '17 at 8:00

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