# Is the merging of DEPENDENT Bernoulli processes still a Bernoulli process?

This is the question:

Suppose that the two Bernoulli processes $X_n$ and $Y_n$ are dependent. We still assume, however, that the pairs ($X_n, Y_n$) are independent. E.g., ($X_2, Y_2$) is independent from ($X_1, Y_1$), etc.

Is $Z_n$, the process formed by recording an arrival in a given time slot if and only if both of the original processes record an arrival in that same time slot, guaranteed to be a Bernoulli process?

The text answer is no, but I have difficulties to find a case where, given the assumptions, the merged process would not be a Bernoulli one.

For example, I may think that the dependency of $Y$ to $X$ changes with time, but if so $Y_n$ would have a different probability across time slots, and it would not have been a Bernoulli process to start with..

EDIT:

I am still thinking to the merging of two DEPENDENT Bernoulli processes. For example $p_{X}(x=1) = 0.7$ and $p_{Y|X}(y=1|x= 0) = 0.2, p_{Y|X}(y=1|x= 1) = 0.8$.

The second stream and the merged one are, on my thought, Bernoulli processes a priori (with $p_Z(z=1) = 0.7*0.8$), but not a posteriori conditional to the realisation of the X stream, as the time homogeneity property would then be lost.

Am I right ?

Everywhere* you read about Bernoulli processes merging, it is referred to independent ones (with the independent word often highlighted), while for me this is irrelevant (obviously $p_Z$ would be different in the two cases).

*for example:

• I'm not exactly sure what you are asking, but if you have a sequence of mutually independent vectors $\{(X_i,Y_i)\}_{i=1}^{\infty}$ and you define $Z_i = f(X_i,Y_i)$ for some deterministic and measurable function $f$, then $\{Z_i\}_{i=1}^{\infty}$ is a sequence of mutually independent random variables. In your case I think $X_i , Y_i$ are binary-valued and $f(X_i,Y_i) = \min[X_i,Y_i]$, so $Z_i$ is also binary-valued. – Michael Mar 14 '18 at 14:23
• You could of course have $\{X_i\}$ iid, $\{Y_i\}$ iid, and the vectors $\{(X_i,Y_i)\}$ iid, but play around with dependencies between $X_i$ and $Y_i$ (such as alternating $X_i=Y_i$ for odd $i$ and $X_i$ indep of $Y_i$ on even $i$) to make the $Z_i$ variables have distribution that depends on $i$. – Michael Mar 14 '18 at 14:30
• To provide a concrete example of @Michael's comment: flip two numbered coins. $X_i$ records "Heads on coin 1"; $Y_i$ records "Heads on coin 1 if $i$ is odd, or on coin 2 if $i$ is even." Then $Z_i$ alternates probabilities $1/2$ and $1/4$. – Connor Harris Mar 14 '18 at 15:26
• @Michael , ConnorHarris: Yes, but in your examples the dependent process would not be a Bernoulli process to start with, because of time variability. What the text claims (and that I suspect to be wrong) is that the merging with rule AND of two dependent Bernoulli processes may not be a Bernoulli process. – Antonello Mar 14 '18 at 15:49
• Antonello : Why do you say that "the dependent process would not be a Bernoulli process to start with"? I give my example again in more detail below. – Michael Mar 15 '18 at 1:05

The example again, in a bit more detail, is this. I will use $X_i$ and $W_i$ for the 0/1 outcome for Connor's two coins on flip $i$. Define:

• Coin 1: $\{X_i\}_{i=1}^{\infty}$ i.i.d. Bernoulli, $P[X_i=1]=P[X_i=0]=1/2$.

• Coin 2: $\{W_i\}_{i=1}^{\infty}$ i.i.d. Bernoulli, $P[W_i=1]=P[W_i=0]=1/2$.

• Assume $\{X_i\}$ and $\{W_i\}$ are independent.

• For $i \in \{1, 2, 3, ...\}$ define $Y_i = \left\{ \begin{array}{ll} X_i &\mbox{ if$i$is odd} \\ W_i & \mbox{ if$i$is even} \end{array} \right.$

• For $i \in \{1, 2, 3, ...\}$ define $Z_i = \min[X_i, Y_i]$.

Thus, $\{X_i\}_{i=1}^{\infty}$ is an i.i.d. Bernoulli process with $P[X_i=1]=1/2$ for all $i$. Likewise, $\{Y_i\}_{i=1}^{\infty}$ is an i.i.d. Bernoulli process with $P[Y_i=1]=1/2$ for all $i$. Also, $\{Z_i\}_{i=1}^{\infty}$ is a sequence of mutually independent random variables, but is not i.i.d. since $P[Z_i=1]$ depends on $i$.