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:
- https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-436j-fundamentals-of-probability-fall-2008/lecture-notes/MIT6_436JF08_lec20.pdf
- http://ocw.metu.edu.tr/pluginfile.php/8197/mod_resource/content/1/EE230_lecture27-print.pdf
- http://slpl.cse.nsysu.edu.tw/cpchen/courses/probability/arrivalProcess.pdf
- http://web.it.nctu.edu.tw/~chungliu/courses/StochasticProcesses/slides/Chapter5.pdf
- http://www.telecom.otago.ac.nz/tele302/ref/Bertsekas_ch5.pdf
- https://www.scribd.com/document/324508062/Bernoulli-Process-pdf
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 $i$.