Pairwise independence on the example

[ISTER_REG]

Hello,

I have a question about pairwise independence. First of all, I would like to give my example. I have the three events [MATH]A, B, C[/MATH] and the set [MATH]S = \{1,2,3,4,5,6,7,8\}[/MATH]. Now I have defined my events as follows [MATH]A = \{1,2,3,4\}, B = \{3,4,5,6\}, C = \{5,6,1,2\}[/MATH]. [MATH]P(A) = 0.5, P(B) = 0.5, P(C) = 0.5[/MATH]
To show pairwise independence I have now done the following:

[MATH]P(A \cap B) = P(\{1,2,3,4\} \cap \{3,4,5,6\}) = P(\{3,4\}) = 0.25 = P(A)P(B) =0.5 \cdot 0.5 = 0.25[/MATH]in a similar way for the other ones:
[MATH]P(B \cap C) = P(B)P(C)[/MATH][MATH]P(A \cap C) = P(A)P(C)[/MATH]
Apparently, therefore, pairwise independence is shown here.

My question now is about more events and or changing the basic set. For example, I started with a set of 5 numbers and quickly realized that it is much more difficult to show pairwise independence. Can it be that the pairwise independence can be shown best with power sets [MATH]2^N[/MATH] (I know for example this very simple example to the pairwise independence with the two coins... that is also again [MATH]2^2 = 4[/MATH]).

Related to my example is it also possible to add another event say [MATH]D[/MATH], so that is then still pairwise independent? How would this event have to look like?

 

[Dr.Peterson]

I suspect you may need a composite number; primes may be difficult.

I think of independence of two probabilities in terms of a rectangle, like this, where one event might be the first row, and the other the first column:



Two coins can be represented by a similar table. (How would you do three coins?)

For three events, I would add a third dimension. Your example can be modeled almost that way, but not quite. (My third event would be different from yours). I won't fill in the details, so you can think "independently"; clearly there are different ways to do it.

 

[ISTER_REG]

Hey and thanks four your reply @Dr.Peterson,

So related to their question, but then they also answered I would also introduce a third dimension with three coins, that sounds plausible.

I also think that my example with 5 numbers is more difficult, if not impossible.

But again a step back to my example with the 8 numbers (from my initial post). is it at all possible to have a pairwise independence here, if we now consider the events [MATH]A,B,C,D[/MATH]? I only ever manage to have three events pairwise independent here, the event [MATH]D[/MATH] (no matter how I cast it) destroys my pairwise independence here....

 

[Dr.Peterson]

Hey and thanks four your reply @Dr.Peterson,

So related to their question, but then they also answered I would also introduce a third dimension with three coins, that sounds plausible.

I also think that my example with 5 numbers is more difficult, if not impossible.

But again a step back to my example with the 8 numbers (from my initial post). is it at all possible to have a pairwise independence here, if we now consider the events [MATH]A,B,C,D[/MATH]? I only ever manage to have three events pairwise independent here, the event [MATH]D[/MATH] (no matter how I cast it) destroys my pairwise independence here....

Does your reference to "their question", are you saying that this is an exercise you were given, rather than your own idea? If so, please quote the entire exercise exactly, so I can be sure what you need to do. Otherwise, who is "they"?

I would think that you might need n to have at least 4 factors, so you might need 16 numbers (or 2*3*3*5 numbers, or whatever) to make it work. But I don't think I've ever tried doing this, so I can't be sure. This is just a reasonable guess.

 

[ISTER_REG]

No, So related to their question, but then they also answered should mean related to your question, which you then answered... The Example from my initial post is my own example, here the idea was to extend the "boring/normal" examples from textbox or other sources with a "more complex" one (Because I tried to build up different examples here, I also came to the conclusion that this probably works less well with 5 numbers. ).

 

[Dr.Peterson]

No, So related to their question, but then they also answered should mean related to your question, which you then answered... The Example from my initial post is my own example, here the idea was to extend the "boring/normal" examples from textbox or other sources with a "more complex" one (Because I tried to build up different examples here, I also came to the conclusion that this probably works less well with 5 numbers. ).

I suspected you just used the wrong pronouns, but wanted to be sure I wasn't missing something.