Pairwise vs Three-way Independence

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Pairwise vs Three-way Independence

Pairwise vs Three-way Independence This is a very classic example, reported in any book on Probability: Example 1 We throw two dice Let A be the event “the sum of the points is 7”, B the event “die #1

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Pairwise vs. Three-way Independence

This is a very classic example, reported in any book on Probability: Example 1.We throw two dice. LetAbe the event "the sum of the points is 7",Bthe event "die #1 came up 3", andCthe event "die #2 came up 4". Now,P[A]=P[B]=P[C]=16 . Also,

P[A∩B]=P[A∩C]=P[B∩C]=1

36
so that all events are pairwise independent. However,

P[A∩B∩C]=P[B∩C]=1

36
while

P[A]P[B]P[C]=1

216
so they are not independent as a triplet. First, note that, indeed,P[A∩B] =P[B∩C] =1 36
, since the fact thatAandBoccurred is the same as the fact thatBandCoccurred.

Example 2.Another example is the case ofΩconsisting of four equally likely points,a1, a2, a3, a4. Let

A={a1,a2},B={a2,a3},C={a3a1}. The three are not independent, but they are pairwise.

However, it is also true that, as long as we consider only specific events (that is, we don"t take into consid-

eration their complements, or, more generally, other members of their algebra), that mutual (3-way) inde-

pendencedoes not imply pairwise independence!

Here is a somewhat trivial example:

Example 3.LetP[A]=p,P[B]=q,P[A∩B]

?pq,P[C]=0then, trivially, butAandBare not pairwise independent.

A less trivial example is the following:

Example 4.Consider the toss of two distinct dice. The sample space is partitioned into equally likely

events of the form(i, j), whereiandjare the points on the first, respectively second die. Obviously,

P[(i, j)]=1

36
. Now, consider the three events A

1=??i=1,2,or3??A2=??i=3,4,or5??A3=i+j=9

We have

A A

1∩A3={(3,6)}

2∩A3={(3,6),(4,5),(5,4)}

1∩A2∩A3={(3,6)}

We have the following probabilities:

P[A1]=P[A2]=12

,P[A3]=19

P[A1∩A2∩A3]=1

36=P[A1]P[A2]P[A3]

but

P[A1∩A2]=16

?14

P[A1∩A3]=1

36?1
18

P[A2∩A3]=1

12?1 18

Note, referring to Example 2, thatP[Cc] = 1, so thatP[Cc∩A∩B] =P[A∩B]?1·p·q, so that consid-

ering the complement of one of the sets makes the new triplet dependent. Similarly, referring to Example

3,P[A3c]=89

, and A which has probability 5 36
?12·12·89=29. Note that this fact does not apply to pairs of events: Fact:IfAis independent ofB, then so are, pairwise,AcandB,AandBc, andAcandBc. That"s

because, for example,P[A∩Bc] =P[A∩(Ω\B)]andP[A∩(Ω\B)] +P[A∩B] =P[A], soP[A∩Bc] +

P[A]P[B] =P[A], henceP[A∩Bc] =P[A]-P[A]P[B] =P[A](1-P[B]) =P[A]P[Bc]. Similarly for the other cases. This points to a better definition of independence of multiple events:

Theorem:Suppose eventsA,B,Csatisfy the conditions

P[X∩Y∩Z]=P[X]P[Y]P[Z]

whereX, Y , Zare, respectively,A, orAc,B, orBc, andC, orCc.Then they are also pairwise indepen- dent. The result extends to any finite collection of events, in an obvious way.

Proof:We can writeP[A∩B]=P[(A∩B∩C)?(A∩B∩Cc)]=P[A]P[B]P[C]+P[A]P[B]P[Cc], because

the two parts are disjoint. This is equal toP[A]P[B](P[C] +P[Cc]) =P[A]P[B]. All other cases are treated in the same way. Remark:Checking all intersections of the sets and their complementcan be seen as checking indepen-

dence of all couples built from theminimal algebra generated by each of the events, which, for an eventA,

is the collection{A,Ac,Ω,∅}. Of course, trivially,Ω, and∅are independent of any event.

While some scholars have looked at the taxonomy of events that arek-independent, but noth-inde-

pendent forh < k, this is not a very exciting subject, since, in considering independence and, more gener-

ally, conditional probabilities, it is much more significant to look at all events in the algebras the events

belong to naturally - at the very least the ones generated by each event and its complement. 2quotesdbs_dbs8.pdfusesText_14

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