Q&A

# Besides guaranteeing 0 ≤ P(A ∩ B) ≤ 1, why can't Independence be defined as P(A ∩ B) = P(A) +,-, or ÷ P(B)?

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I know that $0 \le P(A \cap B) \le 1$ will be violated if $P(A \cap B) = P(A) +,-,\text{or} ÷ P (B)$. I'm not asking about or challenging this reason.

But what are the other reasons against defining independence as $P(A \cap B) = P(A)$

1. $+ P (B)$?

2. or $- P (B)$?

3. or $÷ P (B)$?

### Definition 2.5.1 (Independence of two events).

Events A and B are independent if $P(A \cap B) = P(A)P(B)$.

If P(A) > 0 and P(B) > 0, then this is equivalent to $P(A|B) = P(A)$; and also equivalent to $P(B|A) = P(B)$.

Blitzstein, Introduction to Probability (2019 2 ed), p 63.

We have introduced the conditional probability P(A|B) to capture the partial information that event B provides about event A. An interesting and important special case arises when the occurrence of B provides no such information and does not alter the probability that A has occurred, i.e. , P(A I B) = P(A). When the above equality holds. we say that A is independent of B. Note that by the definition $P(A | B) = P(A \cap B)/P(B)$, this is equivalent to $P(A \cap B) = P(A)P (B).$ We adopt this latter relation as the definition of independence because it can be used even when P(B) = 0, in which case P(A|B) is undefined. The symmetry of this relation also implies that independence is a symmetric property; that is, if A is independent of B, then B is independent of A, and we can unambiguously say that A and B are independent events.

Tsitsiklis, Introduction to Probability (2008 2e), p 34.

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1. Try with different combinations of probabilities: what is their sum, difference, division. Not all of these are probabilities. (Check this.) This would cause a problem.

2. Draw a square; think of it as the unit square $[0, 1] \times [0, 1]$. On $x$-axis mark the (probabilities of) the events $A$ and not $A$. On $y$-axis do the same for $B$ and not $B$. Then try to find $P(A)P(B)$ in the picture; there is a simple geometric interpretation. Next, try drawing a similar picture with dependent probabilities. Playing around with such representations can be helpful in clarifying the concepts.

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