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So I know one can go from a joint density function $f(x,y)$ to marginal density functions, like $f_x(x)$ by integrating against the other variables as in $f_x(x) = \int f(x,y) dy$...but given $f_x(x)$ and $f_y(y)$ as densities for dependent random would one go about finding a joint density or distribution function?


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In general, you can't recover the joint density from the marginals. You need more information. In a few cases (eg joint gaussian variables, or two bernoullis) the correlation would suffice. – leonbloy Oct 30 '11 at 22:27
up vote 5 down vote accepted

For example, suppose the marginal densities for $X$ and $Y$ are both 1 on the interval $[0,1]$, 0 otherwise. One family of possibilities for the joint density is $f(x,y) = 1 + g(x) h(y)$ for $0 < x < 1$, $0 < y < 1$, 0 otherwise, for functions $g$ and $h$ such that $\int_0^1 g(x)\, dx = \int_0^1 h(y)\, dy = 0$, $-1 \le g(x) \le 1$ and $-1 \le h(y) \le 1$. And there are infinitely many other possibilities.

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There will be many different distributions with the same marginal distributions, so one needs to select a specific way to aggregate the marginal distributions into joint distributions. Assuming they are independent is essentially making one of these possible choices. The most common way to make the choice, is by working with a copula.

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Michael: Since you mention copulas, you might wish to explain which aspects of theory and/or applications they provide, that the usual conditioning notion does not. (Compare with this.) – Did Dec 30 '11 at 9:32
I'm not sure what usual conditioning is in this context. If you mean generating the joint distribution by amploying conditional distributions, then the advantage is that copulas always work out- for all marginal distributions, you get a joint distribution. – Michael Greinecker Dec 30 '11 at 16:52
...Just like one obtains the joint distribution of every random couple (X,Y) from the marginal distribution of X and the conditional distribution of Y conditionally on X. I see no advantage here, rather a reformulation of some well-known characterization. Likewise, the WP page mentions Fréchet-Hoeffding copula bounds, which are (i) unsourced and (ii) a trivial translation of $P(\cap A_i)\le\min P(A_i)$ and $P(\cup A_i)\le\sum P(A_i)$. It seems that my previous question remains. – Did Dec 30 '11 at 18:29
If you have a marginal distribution for $X$ and $Y$ and a conditional distribution on $Y$ given $X$, you might get two different marginal distributions on $Y$. Copulas give you function that maps a family of marginal distributions into a joint distribution. Conditional distributions don't d that. – Michael Greinecker Dec 30 '11 at 20:26
Michael: Hmmm... conditional distributions (plus a marginal distribution) do exactly that. In fact, the marginal distribution of X plus the conditional distribution of Y given X determine uniquely the joint distribution of (X,Y), and in particular determine uniquely the marginal distribution of Y. (On conditional distribution probabilities, see here.) – Did Dec 31 '11 at 0:34

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