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I have $N$ Bernoulli random variables $X_1, ..., X_{N}$ with known parameters $p_1, ..., p_{N}$. I want generate a joint distribution in which these random variables are not independent as I know that joint distribution would just be the product of their marginals.

How can I create this joint distribution that can be updated as I add more Bernoulli random variables?

Dependent Bernoulli trials

^ This seems to be my exact question but I am looking for a much simpler way of defining the distribution than selecting $2^n - 1$ parameters

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  • $\begingroup$ Your title says dependent while your questions says not dependent $\endgroup$
    – Henry
    Jul 2, 2016 at 23:46
  • $\begingroup$ Editing in 'not $in$dependent' doesn't clarify much; if the joint distribution is a product of marginals, doesn't that make them independent? Hypergeometric trials can be S vs F, or B vs W, if you have blue and white balls in an urn sampled without replacement. But updating would be a problem. For updating, you could have a Markov chain with state space $\{0, 1\}$ in which each value depends on the one just before? Might help if you said what kind of dependence you want, and why. $\endgroup$
    – BruceET
    Jul 3, 2016 at 0:43
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    $\begingroup$ @BruceET In the original model, independence of N Bernoulli random variables was assumed. My goal is to generate a joint distribution without independence and see how things change. I haven't thought about what kind of dependence I want yet. $\endgroup$
    – user265634
    Jul 3, 2016 at 1:14

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Based on your most recent comment, I think you should consider a 2-state Markov chain to produce a sequence of random variables $X_i,$ taking values in $\{0, 1\},$ roughly as follows:

Start with a deterministic or random $X_1.$ Then

(i) $P\{X_{i+1} = 1|X_i = 0\} = \alpha,$ and (ii) $P\{X_{i+1} = 0|X_i = 1\} = \beta.$

The parameters $\alpha$ and $\beta$ are the respective probabilities of 'changing state' from one $X_i$ to the next. To avoid certain kinds of deterministic sequences, you may want to use $0 < \alpha, \beta < 1.$ If $\alpha = 1 - \beta,$ then the sequence is independent.

By induction, one can show that $$P\{X_{1+r} = 0|X_1 = 0\} = \frac{\beta}{\alpha+\beta} + \frac{\alpha(1-\alpha - \beta)^r}{\alpha+\beta}.$$ If $|1-\alpha - \beta| < 1$, then in the long run $P\{X_n = 0\} \approx \beta/(\alpha+\beta),$ regardless of the value of $X_1.$

Moreover, there are similar formulas for the '$r$-step transitions' from 0 to 1, 1 to 0, and 1 to 1. Of course, I am skipping over a lot of detail here.

Perhaps there is a rich enough variety of models here to satisfy your curiosity as to what happens when independence fails in this way.

Later chapters in many probability books have a complete development of the theory of 2-state Markov chains. Also there are several good elemeentary books just on Markov chains. [Google '2-state Markov Chain'. One reference among many is Chapter 6 of Suess and Trumbo (2010), Springer, in which I have a personal interest.]

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