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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 vars..how would one go about finding a joint density or distribution function?

Thanks

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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.2011-10-30

2 Answers 2

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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](http://math.stackexchange.com/a/63572/6179).)2011-12-30
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    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.2011-12-30
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    ...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](http://en.wikipedia.org/wiki/Copula_%28probability_theory%29#Fr.C3.A9chet.E2.80.93Hoeffding_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.2011-12-30
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    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.2011-12-30
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    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](http://en.wikipedia.org/wiki/Conditional_probability_distribution).)2011-12-31
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    Let's see: You have a fixed conditional distribution for $Y$ given X and you have a fixed copula. You also have an arbitrary family of marginal distributions $(p_t^x,p_t^y)$. Now for each $t$, the copula gives you a joint distribution with marginal distributions $p_t^x$ and $p_t^y$. The condtional distribution gives you for each $t$ a joint distribution with marginal distribution $p_t^x$ and a marginal distribution on $Y$ that is usually different from $p_t^y$. So conditional distributions do not give joint distributions with given marginals.2011-12-31
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    Michael: Right. In other words, one wants to decompose the joint distribution of (X,Y) into the marginals of X and Y plus a copula. A move keeping the former fixed while changing the latter is obviously more adapted to this than to the decomposition into the marginal distribution of X plus the conditional distribution of Y conditionally on X. Thanks for your explanations.2012-01-01