3
$\begingroup$

A textbook I'm currently reading defines independence as follows:

Two events $A$ and $B$ are independent if $P(A|B)=P(A)$ (provided that the probability of the events are positive)

Then derives the following as a theorem:

Two events $A$ and $B$ are independent iff $P(A \cap B)=P(A)P(B)$

My question is this: can we give this theorem as a definition, that is, if we don't have the concept of conditional probability, can we still define independence as follows:

Two events $A$ and $B$ are independent if $P(A \cap B)=P(A)P(B)$

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Please cite the textbook that you are reading. All the textbooks on my bookshelf either give the theorem as the definition of independence and then point out the interpretation in terms of conditional probabilities as intuition as to why the definition makes sense, or develop the notion in terms of conditional probabilities, point out the deficiency when the conditioning event has probability $0$ and then give the formal definition of independence as $P(A\cap B) = P(A)P(B)$. $\endgroup$
    – Dilip Sarwate
    Commented Sep 14, 2017 at 3:52
  • $\begingroup$ @DilipSarwate: Below are two sources dartmouth.edu/~chance/teaching_aids/books_articles/… . Also here onlinecourses.science.psu.edu/stat414/node/38 $\endgroup$
    – Sanyo Mn
    Commented Sep 14, 2017 at 5:42
  • $\begingroup$ Applied Statistics and Probability for Engineers, Third Edition, Douglas C. Montgomery, George C. Runger $\endgroup$
    – Sanyo Mn
    Commented Sep 14, 2017 at 6:00
  • $\begingroup$ A First Course in Probability, Sheldon Ross $\endgroup$
    – Sanyo Mn
    Commented Sep 14, 2017 at 6:01
  • $\begingroup$ @DilipSarwate: Can you cite a textbook where the above theorem is introduced before mentioning conditional probability. $\endgroup$
    – Sanyo Mn
    Commented Sep 14, 2017 at 6:02

2 Answers 2

Reset to default
8
$\begingroup$

As a counterpoint to Matthew Drury's answer, I prefer the second definition $$\text{Events}~A ~\text{and}~ B~\text{are said to be independent if and only if}~ P(A\cap B) = P(A)P(B)$$ over the first because it avoids the asymmetry in the first definition where $B$ can be "independent of" $A$ because $P(B\mid A) = P(B)$ holds while $A$ cannot be said to be "independent of" $B$ because $P(B) = 0$ and so $P(A\mid B)$ is undefined. Yes, the definition in terms of conditional probabilities is more intuitive (when it works) but to my mind, independence is a fundamental concept that should not (and does not) need the notion of conditional probability to define.

Improve this answer
$\endgroup$
3
  • $\begingroup$ (+1) Yah, agree to disagree. In my mind, conditional probability is a much more fundamental concept than independence : ) $\endgroup$
    – Matthew Drury
    Commented Sep 13, 2017 at 22:03
  • $\begingroup$ The question was "Can independence be defined without using conditional probability" and your opinions about what is the best way to define it is not at issue. You are free to have differing opinions on the best definition. $\endgroup$
    – Michael R. Chernick
    Commented Sep 14, 2017 at 2:24
  • 2
    $\begingroup$ "independence is a fundamental concept that should not need the notion of conditional probability to define" is the answer I'm looking for. $\endgroup$
    – Sanyo Mn
    Commented Sep 14, 2017 at 5:47
6
$\begingroup$

You can, as the two equations are equivalent (at least in the cases where $P(B)$ is not zero).

On the other hand, I prefer the first as a definition, as it communicates the intent of introducing the concept. The equations

$$ P(A \mid B) = P(A) $$

directly expresses that knowledge of $B$ does not influence our state of knowledge about $A$, which, in a sense, is what independence actually means.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.