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I've asked a question about getting the joint probability distribution for $N$ Bernoulli random variables, given the expected value for each one ($E[X_i]=p_i)$ and it's correlations ($\rho_{12},\rho_{13},\dots$). Someone kindly directed me to this paper, and it's been really helpful, but I still can't figure one last point out:

Given that I know the correlation matrix for the $N$ Bernoulli random variables: How can I get the expected value for the product of all of them: $$ E\left[\prod_{j=1}^N X_j \right]=? $$

For two Bernoulli random variables, I know that expected value is: $$ E\left[X_1·X_2\right]=\sigma_{12}+p_1·p_2 $$ where $\sigma_{12}$ is the covariance of $X_1,X_2$ and $p_j=E[X_j]$. However, when working with three random variables, I can't find a way. My gut tells me that it can be a function of the covariances, but I don't know for sure.

Can you point me in the right direction?

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Ordinarily, bivariate relationships do not determine multivariate ones, so we ought to expect that you cannot compute this expectation in terms of just the first two moments. The following describes a method to produce loads of counterexamples and exhibits an explicit one for four correlated Bernoulli variables.


Generally, consider $k$ Bernoulli variables $X_i, i=1,\ldots, k$. Arranging the $2^k$ possible values of $(X_1, X_2, \ldots, X_k)$ in rows produces a $2^k \times k$ matrix with columns I will call $x_1, \ldots, x_k$. Let the corresponding probabilities of those $2^k$ $k$-tuples be given by the column vector $p=(p_1, p_2, \ldots, p_{2^k})$. Then the expectations of the $X_i$ are computed in the usual way as a sum of values times probabilities; viz.,

$$\mathbb{E}(X_i) = p^\prime \cdot x_i.$$

Similarly, the second (non-central) moments are found for $i\ne j$ as

$$\mathbb{E}(X_iX_j) = p^\prime \cdot (x_i x_j).$$

The adjunction of two vectors within parentheses denotes their term-by-term product (which is another vector of the same length). Of course when $i=j$, $(x_ix_j) = (x_ix_i) = x_i$ implies that the second moments $\mathbb{E}(X_i^2) = \mathbb{E}(X_i)$ are already determined.

Because $p$ represents a probability distribution, it must be the case that

$$p \ge 0$$

(meaning all components of $p$ are non-negative) and

$$p^\prime \cdot \mathbf{1} = 1$$

(where $\mathbf{1}$ is a $2^k$-vector of ones).

We can collect all the foregoing information by forming a $2^k \times (1 + k + \binom{k}{2})$ matrix $\mathbb{X}$ whose columns are $\mathbf{1}$, the $x_i$, and the $(x_ix_j)$ for $1 \le i \lt j \le k$. Corresponding to these columns are the numbers $1$, $\mathbb{E}(X_i)$, and $\mathbb{E}(X_iX_j)$. Putting these numbers into a vector $\mu$ gives the linear relationships

$$p^\prime \mathbb{X} = \mu^\prime.$$

The problem of finding such a vector $p$ subject to the linear constraints $p \ge 0$ is the first step of linear programming: the solutions are the feasible ones. In general either there is no solution or, when there is, there will be an entire manifold of them of dimension at least $2^k - (1 + k + \binom{k}{2})$. When $k\ge 3$, we can therefore expect there to be infinitely many distributions on $(X_1,\ldots, X_k)$ that reproduce the specified moment vector $\mu$. Now the expectation $\mathbb{E}(X_1X_2\cdots X_k)$ is simply the probability of $(1,1,\ldots, 1)$. So if we can find two of them that differ on the outcome $(1,1,\ldots, 1)$, we will have a counterexample.

I am unable to construct a counterexample for $k=3$, but they are abundant for $k=4$. For example, suppose all the $X_i$ are Bernoulli$(1/2)$ variables and $\mathbb{E}(X_iX_j) = 3/14$ for $i\ne j$. (If you prefer, $\rho_{ij} = 6/7$.) Arrange the $2^k=16$ values in the usual binary order from $(0,0,0,0), (0,0,0,1), \ldots, (1,1,1,1)$. Then (for example) the distributions

$$p^{(1)} = (1,0,0,2,0,2,2,0,0,2,2,0,2,0,0,1)/14$$

and

$$p^{(2)} = (1,0,0,2,0,2,2,0,1,1,1,1,1,1,1,0)/14$$

reproduce all moments through order $2$ but give different expectations for the product: $1/14$ for $p^{(1)}$ and $0$ for $p^{(2)}$.

Here is the array $\mathbb{X}$ adjoined with the two probability distributions, shown as a table:

$$\begin{array}{cccccccccccccc} & 1 & x_1 & x_2 & x_3 & x_4 & x_1x_2 & x_1x_3 & x_2x_3 & x_1x_4 & x_2x_4 & x_3x_4 & p^{(1)} & p^{(2)} \\ & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{14} & \frac{1}{14} \\ & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & \frac{1}{7} & \frac{1}{7} \\ & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & \frac{1}{7} & \frac{1}{7} \\ & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & \frac{1}{7} & \frac{1}{7} \\ & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{14} \\ & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & \frac{1}{7} & \frac{1}{14} \\ & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & \frac{1}{7} & \frac{1}{14} \\ & 1 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & \frac{1}{14} \\ & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & \frac{1}{7} & \frac{1}{14} \\ & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & \frac{1}{14} \\ & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & \frac{1}{14} \\ & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \frac{1}{14} & 0 \\ \end{array}$$


The best you can do with the information given is to find a range for the expectation. Since the feasible solutions are a closed convex set, the possible values of $\mathbb{E}(X_1\cdots X_k)$ will be a closed interval (possibly an empty one if you specified mathematically impossible correlations or expectations). Find its endpoints by first maximizing $p_{2^k}$ and then minimizing it using any linear programming algorithm.

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