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I have a dataset composed of 3 random variables X, Y and Z. However, at each sample one of the random variable remains hidden. As an arbitrary example, the observation 1 is $(x_1,y_1)$, the observation 2 is $(x_2,z_2)$, the observation 3 is $(y_3,z_3)$ and so on.

The task I need to solve is to predict the most likely value of the missing random variable for each sample.

Is it possible to estimate the joint probability distribution $P(X,Y,Z)$ of the described dataset to employ it in the inference of the missing variable? The variables are highly coupled so I don't think it is possible to make any conditional independence assumption.

Rodrigo CIlla
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  • Do your variables take discrete or continuous values? – Salmonstrikes Feb 08 '16 at 22:50
  • This might be relevant, if you look at the Kernel density estimates from your samples as pairwise joint distributions: http://mathoverflow.net/questions/33145/recovering-joint-distribution-from-marginals – Salmonstrikes Feb 08 '16 at 23:39

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I don't think this can be done in general without additional assumptions.

You're observing samples from the bivariate margins, and trying to infer the trivariate distribution.

That this can't really be done in general can be seen from the fact that the bivariate distributions don't determine the trivariate distribution.

For example, it's possible to have a situation where each of the margins is bivariate normal, but the sign of the third variable is determined by the signs of the two variables you observe (where alternating octants are empty or occupied). Since you don't observe all three together you'd never realize whether you should be treating it as one sign (as at the link) or the other (constructing the opposite situation from the one at the link, with the signs flipped), or both at random (effectively trivariate iid normal), or indeed some other arrangement (some other trivariate copula not considered yet that has uniform bivariate margins).

With additional assumptions or some trivariate observations you could perhaps get somewhere.

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Glen_b
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