# Rewriting expressions involving Bernoulli random variables

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?

Thanks.

• 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 $p_2+p_3-p_2p_3$ Mar 16 at 12:02
• 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) Mar 16 at 12:04
• 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 Mar 16 at 12:06
• 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)?
– whuber
Mar 16 at 14:21

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.

• Since it is not possible, is there a way to approximate the required behaviour? Mar 16 at 13:39
• @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.
– Tim
Mar 16 at 14:58