If $X$ and $Y$ are independent variables from each other, then are $(X,Y)$ and $X$ (necessarily) independent?

Assume one is trying to analyze the effect of peer education (education from patient to another patient) and nurse education on acceptance of a certain operation.

Nurses and peer patients are independent people.

The comparison of two independent samples can be performed via Wilcoxon-Mann-Whitney test.

Assume the candidate patients for operation are two groups:
1. Those that receive education only from nurses. ($X$ in my question)
2. Those that receive education from nurses AND from peer patients. ($(X,Y)$ in my question).

Then, are these two groups independent from each other? Why or why not?

Edit:(the suggestion by Dave offers the following tabular structure)

Educated by (some) nurses only  X=0    
Educated by the very same (some) nurses AND by peers (some other patients) X=1     Acceptance of Operation (0/1 variable)
-------------------------------------------------------------------------------    ---------------------------------------
1                                                                                          0
1                                                                                          1
0                                                                                          1
0                                                                                          0
:                                                                                          :

Multiple observations that take only the above 4 possible values.

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    $\begingroup$ I don't think that your second group is the vector variable $(X,Y)$. It seems more like the intersection $X \cap Y$, which is a different question. $\endgroup$ – Dave Jun 21 '19 at 13:55
  • $\begingroup$ @Dave I added your logic (as far as I understand) to solve the problem. So, in principle, you seem to suggest X and Y to be taken differently; Y for acceptance, for example. $\endgroup$ – Erdogan CEVHER Jun 21 '19 at 15:23
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    $\begingroup$ Your question (and title) need to be corrected. However given that you have defined $X$ to be "those that receive education only from nurses", as it stands that would exclude combining with any other education. The only way that "Those that receive education from nurses AND from peer patients" becomes "X$\cap$Y" is if you drop the "only" from the definition of X. $\endgroup$ – Glen_b Jun 21 '19 at 16:05
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    $\begingroup$ @Glen_b I'm wondering if (X,Y) might be the concatenation of the X observations and Y observations ($X \cup Y$, or maybe more like $X_1,\dots ,X_{n_1},Y_1,\dots , Y_{n_2}$). That would allow for "only" to appear, thought I do wonder if its inclusion is a mistake because we're looser with regular language than we are with language in mathematics. To Erdogan CEVHER, I don't follow that chart at all. $\endgroup$ – Dave Jun 21 '19 at 16:16
  • $\begingroup$ @Dave "AND" there is the mathematical AND; i.e., there is NO patient who took education only from the peer educator (other patient). Second, you are right that (X,Y) notation used NOT to indicate a vector variable; but, for the sake of solution, if there is a way to solve the question with that meaning, (X,Y) could be qualified as vector variable, as well. $\endgroup$ – Erdogan CEVHER Jun 21 '19 at 18:28

Based on the discussion in the comments, I can post an answer. Ronald and Brad are the control group. This is your baseline. Then you examine the effect of education by other patients and check how George and John do. Or maybe George and John form your treatment group and you test the effect of removing education from patients. Either way is fine.

This is a archetypal two-sample comparison, very similar to seeing how patients in a trial for a new drug perform compared to patients in a control/placebo group do. Both groups have the human immune system (akin to your X--education by nurses), but one group also gets the experimental drug (akin to your (X,Y)--education by both nurses and patients).

It looks like your measurement is a binary yes/no, in which case, you would be comparing probabilities of getting 1 in each group (proportion test). Use whatever proportion test you desire after putting the yes/no into two groups based on what kind of education each patient received.

Finally, your notation is extremely confusing. When all is said and done, your $X$ distribution is education by nurses only, while your $Y$ distribution is education a different way (by both nurses and other patients). The notation $(X,Y)$ means a bivariate random variable, which is not what's going on here. It might even be better to call them $X_0$ and $X_1$ or $X_{nurse}$ and $X_{both}$ instead of $X$ and $Y$, and $X_0$ and $X_1$ would be the yes/no outcome of whether the patient decided to go for surgery.

(Yes, I mean Ronald Fisher, Brad Efron, George Box, and John Tukey.)

(Also, there is something going on here that's bivariate, but that is because each patient has two distinct observations: education and treatment. This fact will not be especially helpful to you at this point, and you can solve your problem without thinking about it.)

  • $\begingroup$ Many thanks, Dave. You revealed all the complexities of the problem, and even more. $\endgroup$ – Erdogan CEVHER Jun 21 '19 at 20:22

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