On page 48, the book says:
“Let Ω = [0,1], A = [0,1/4] as before; then A is independant of B = [1/8,5/8] and of C = [1/8,3/8] U [3/4,1]. In addidtion, B and C are independant. However, P(A∩B∩C) ≠ P(A)P(B)P(C). Thus, given three events, the pairwise independance of each of the three possible pairs does not sufice for the extension of ‘independance’ to all three events.”
When I first read this I was like :dubious: so I put together a spreadsheet using the RAND function to get over 100000 rows of random numbers in columns A,B and C. In column D I put a big IF statment that returns a 1 if event A above occurs in column A and event B occurs in column B etc. It returns 0 if even one of the events do not occur.
On their own, I calculate the probability of event A = 0.25, P(B) = 0.5, P(C) = 0.5
Then P(A)P(B)P(C) = 0.0625
If I do a sum of column D on my spreadsheet and divide that by a count then it gets quite close to 0.0625. Obviously never exacly the same but never too far.
The problem is that A∩B∩C = A∩B = A∩C – it’s the interval [1/8,1/4], and so has probability 0.125. You don’t need a spreadsheet with random functions for that: you just need to draw a diagram, and see on that where sets A, B and C intersect.
I don’t think that approach is appropriate for independant events.
What you are saying is as if I only had random numbers in one column and was testing for each condition (A,B,C) on this column.
As these are all independant, each condition is tested on a sepparate column of random numbers. So the result is different. No?
If you are taking random numbers in the interval [0,1], then:
A is satisfied if the number is between 0 and 1/4
B is satisfied if the number is between 1/8 and 5/8
C is satisfied if the number is between 1/8 and 3/8, or between 3/4 and 1
So all three of A, B and C are satisfied if the number is between 1/8 and 1/4 – but each time you should be checking the same number.
A is satisfied if the number1 is between 0 and 1/4
B is satisfied if the number2 is between 1/8 and 5/8
C is satisfied if the number3 is between 1/8 and 3/8, or between 3/4 and 1
Also, I have just realised the point the author was making:
The probability of the intersection does not = the product of the probabilities for independant events. So you calculaiton of 0.125 does not = 0.0625
By themselves knowing I’m in region A tells me nothing about whether I’m in region B (I have a 1/4 chance whether B is true or not) and vis-versa P(B)=P(B|A)=P(B|!A)=1/2
Similarly A and C provide no information about each other, and B and C provide no information about each other.
However in combination, if I know that B and C are both true, then that tells me that A must be true. So even though every pair is independent, the whole set of events isn’t.
Here is simpler example: I flip 2 coins and define the following three events.
A = first coin is heads
B = second coin is heads
C = both coins show the same face.
It is clear that A and B are independent, since there is no relationship. Also I just know A that the first coin is a head, I still don’t have any information about whether the 2 coins have the same face, so A and C are independent. Similarly B and C are independent. However knowledge of both A and B in combination gives me complete knowledge of C. So the combined three events aren’t independent.