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suppose we are given $n$ binary random variables, $X_1,\dots,X_n$. A probability distribution $P$ assigns all elementary events a probability; $$ P(X_1=x_1\&\ldots \& X_n=x_n)\in[0,1].$$

A regular marginal $m_I$ is a probability function on a proper(!) subset of variables: Given a proper subset $I\subset \{1,\dots,n\}$, a regular marginal $m_I$ assigns all events described by this set of variables a non-zero probability: $$m_I(X_{i_1}=x_{i_1}\&\dots\&X_{i_{|I|}}=x_{i_{|I|}})>0.$$

Does a given set of regular marginals determine a probability distribution $P$? In other words, are there cases in which there exists only a single distribution $P$ that agrees with all given marginals?

Obviously, the given marginals must agree on joint domains; i.e., they must be consistent.

The regularity condition of the marginals matters here: If $n=2$, $m_1(X_1=x_1)=1$ and $m_2(X_2=x_2)=1$, then there exists a unique $P$ that agrees with these two marginals; $P$ is simply given by $P(X_1=x_1 \& X_2=x_2)=1$.

I think that there are cases for all $n\geq 2$ in which no set of regular marginals determines a unique probability distribution $P$. I would like to be sure! Also, I would like to have a reference.

All help is much appreciated,

Juergen

Here's an example: $X_1$ concerns the outcomes of a coin toss. $X_2$ concerns whether my favourite team wins or looses. $X_3$ concerns whether it rains tomorrow or not. ...

A marginal $m_I$ provides probabilistic information about a subset of these events.

For all fixed $n\geq 3$ (jakobdt solved the case for $n=2$), does there exist a case in which we have specified marginals for all $I$ with $|I|=n-1$ such that there are multiple probability distributions agreeing with all marginals?

Juergen
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  • Try $n=2$ where the marginal probabilities are neither $1$ nor $0$ – Henry Jul 14 '23 at 11:05
  • Yes, for $n=2$ the marginals do not determine the distribution $P$.

    How about the same question for general $n$? I still think that the marginals do not determine a unique $P$.

     – Juergen Jul 14 '23 at 13:50

1 Answers1

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Let $X_1$ be Bernoulli distributed with parameter $p=1/2$. Here are two choices of $X_2$ for which the marginal distribution $m_2$ is the same but the joint distribution of $(X_1,X_2)$ is not:

  1. $X_2=X_1$.
  2. $X_2=1-X_1$.
jakobdt
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  • Thanks for this jakobdt. Unfortunately, I don't quite get this.

    My question rather is (a more general version of the following):

    Suppose $X_1$ is Bernoulli distributed with parameter $p_1\in (0,1)$. Suppose $X_2$ is Bernoulli distributed with parameter $p_2\in (0,1)$.

    Does there exist a unique probability distribution $P$ on $(X_1,X_2)$ that agrees with these two Bernoulli distribution? Note that I do not make any assumptions on $P$. So, it does not matter whether $P$ is Bernoulli distributed or not.

     – Juergen Jul 14 '23 at 12:12
  • If you let $U$ uniformly distributed in $(0,1)$ then you can construct $X_1$ and $X_2$ by taking $X_i=f_i(U)$, where $f_i\colon(0,1)\to{0,1}$ is such that the set ${u\in(0,1)\mid f_i(u)=1}$ has Lebesgue measure $p_i$. Some examples are $f_i=1_{(0,p_i)}$, $f_i=1_{(1-p_i,1)}$ and $f_i=1_{(0,p_i/2)}+1_{(1-p_i/2,1)}$. The point is that $P$ depends on the choice of these functions. For simplicity, just look at how $P(X=Y)$ can vary accordingly. In my example above this probability is either $1$ or $0$. – jakobdt Jul 14 '23 at 12:25
  • I'm beginning to see. If $n=2$, then two marginals (both defined over a single variable) do not determine $P$. Good.

    I'm mostly interested in the case for general $n$. Are there cases in which a number of marginals (defined over at most $n-1$ variables) determine $P$?

     – Juergen Jul 14 '23 at 13:40
  • Let $(X_1,...,X_n)$ be uniformly distributed over the $2^{n-1}$ binary words with an even number of $1$s in them. Then all $n-1$ dimensional marginals are uniform, that is, are the same as those of $(Y_1,...,Y_n)$ where the $Y_i$ are independent uniform 0-1 variables. – kimchi lover Jul 15 '23 at 12:36