Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs

Dashboard
Notifications
Mark all as read
Q&A

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

+0
−3

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.

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

0 comment threads

1 answer

+1
−0
  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.

Why does this post require moderator attention?
You might want to add some details to your flag.

1 comment thread

1. Thanks for your answer. I agree that P(A) + P(B) can $>1$, but then why can't we change the defini... (4 comments)

Sign up to answer this question »

This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!

Like what we're doing? Support us! Donate