my problem is the following. Consider only Bernoulli random variables $X_1,\dots, X_n$ where $P(X_i = T) = p_i$ ($T$ stands for true, success). All the r.v.s are independent. Starting from the simplest case, consider 3 r.v.s $X_1$, $X_2$, and $X_3$. Define $Z$ as follows: $Z = X_1 \cdot X_2 + X_1 \cdot X_3$. Then, I can compute $P(Z = T)$ .

My question is: is there a general way to remove the dependency from $X_1$ from both $X_1\cdot X_2$ and $X_1 \cdot X_3$ by introducing two more variables $X_1^{'}$ and $X_1^{''}$ and re-writing $Z$ as $Z = X_1^{'} \cdot X_2 + X_1^{''} \cdot X_3$? What should be the distribution of $X_1^{'}$ and $X_1^{''}$? The new variables will have (i think) a Bernoulli distribution, but it is possible to apply this approach in the general case? How can I compute $P(X_1^{'} = T)$ and $P(X_1^{''} = T)$? I'm interested in the computation of the success probability of the newly inserted variables, and not in the computation of the success probability of the random variable $Z$. The goal is to have no shared variables in the terms invoved in the summations. I consider only expression composed by summations of products of random variables with Bernoulli distribution.

This is a simple example, but is there a general formula that accounts also longer expressions with more than two terms and more than a single shared variable? Any pointer to existing scientific literature? Do other solutions already exist?


  • $\begingroup$ The easier approach is to say $Z=X_1\cdot X_{23}^{'}$ where $X_{23}^{'}= X_2+X_3$ and is independent of $X_1$ and is $T$ or $1$ with probability $p_2+p_3-p_2p_3$ $\endgroup$
    – Henry
    Commented Mar 16, 2022 at 12:02
  • $\begingroup$ A random variable either depends on another one or it doesn't. You can't "remove the dependency" by rewriting it differently. (Or you didn't describe your actual goal clearly enough) $\endgroup$
    – J. Delaney
    Commented Mar 16, 2022 at 12:04
  • $\begingroup$ Alternatively write $Z=X_1\cdot X_2\cdot X_3 + X_1\cdot X_2\cdot (1-X_3)+X_1\cdot (1-X_2)\cdot X_3$ where the three terms being added are mutually exclusive rather than independent $\endgroup$
    – Henry
    Commented Mar 16, 2022 at 12:06
  • $\begingroup$ Since $Z$ can take on the values $0,$ $1,$ and $2,$ I wonder what you might mean by the event "$Z=T$" or, more generally, what you mean by "success probability." You are trying to equate numbers to a logical value ($T$), which makes no sense. Would you perhaps have intended to define $Z = \min(1, X_1X_2+X_1X_3)$ (and thereby equate $1$ with $T$ and $0$ with false)? $\endgroup$
    – whuber
    Commented Mar 16, 2022 at 14:21

1 Answer 1


If the variables are binary than

$$ \Pr(X_1 \land X_2) = \Pr(X_1 X_2 = 1) $$

by binary arithmetic (for $x y = 1$ to hold, both need to be ones). The result of multiplying zeroes and ones is also zeroes and ones, it's still Bernoulli distributed. You know that all the variables are independent, so

$$ \Pr(X_1 \land X_2) = \Pr(X_1) \Pr(X_2) $$

As noticed by J. Delaney in the comment, you cannot "remove the dependency" from the variables. You cannot replace $X_1$ with two (different) variables for it to hold. To convince yourself, write down all the combinations of results for

$$ z = a b + a c $$

where $a, b, c \in \{0, 1\}$. You cannot replace $a$ with $a'$ and $a''$ other than $a = a' = a''$ for the equation to always have the same result as initially. For example, for $z=2$ there is only one possibility $a = a' = a''= 1$, so the new variables would depend on $a$, it would not work.

  • $\begingroup$ Since it is not possible, is there a way to approximate the required behaviour? $\endgroup$ Commented Mar 16, 2022 at 13:39
  • $\begingroup$ @damianodamiano you can always approximate, it is just a question of how bad you want the approximation to be. You could assume that $X_1', X_2''$ are i.i.d. and distributed as $X_1$, but then you are getting the probability distribution and the variability of the result pretty wrong. $\endgroup$
    – Tim
    Commented Mar 16, 2022 at 14:58

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