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I was wondering if marginal densities always define the joint density function or if this is only true for statistically independent variables?

PS: I can imagine this is a duplicate but I haven't found anything on the topic via search. If so, please excuse me

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    $\begingroup$ It is not true in general, except, as you mention, for independent variables, for which the joint density is just the product of the marginal densities. The conditional densities, however, do, jointly, define the joint density function: $f(x|y,z), f(y|x,z), f(z|x,y)$ define $f(x,y,z)$ etc. $\endgroup$ – jbowman Mar 14 '18 at 19:39
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    $\begingroup$ Search for "copula." $\endgroup$ – whuber Mar 14 '18 at 19:55
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A simple counterexample is taking the interval $I=[0,1]$ with the Lebesgue measure on it and take $(X, Y)(z) = (z, z)$. Now $P_{(X,Y)}(\{(x,y) | x = y\}) = 1$, but as this is a null set on the unit cube, the integral of any integrable function over it would be 0. The marginal densities of $X$ and $Y$ are simply the indicator function on $I$.

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  • $\begingroup$ +1. More generally, when the marginal distribution functions are $F_1$ and $F_2$ and $F_2$ is invertible (as is always the case when it has a density), let $X\sim F_1$ and define a bivariate random variable as $(X, F_2^{-1}(F_1(X))).$ $\endgroup$ – whuber Apr 30 at 18:02

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