MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I saw in a paper an argument, that seems simple. (Let $\triangle$ be the triangle operator of symmetric set difference between two sets) It states that if $P(A \triangle B) < \epsilon$, for some small $\epsilon>0$, then if $x \in A$ then it is very likely to be in $B$ and vice versa.

I am trying to figure out how we would formalize this notion. More specifically, I am trying to formalize the idea that if $P(A \triangle B) < \epsilon$ then $P(A) \approx P(B)$, or that we can use $A$ and $B$ interchangeably in a statement. Is it true, for example, that if $P(A \triangle B) < \epsilon$, then $|P(A)-P(B)|<\epsilon$?

I know that $P(A \triangle B)$ defines a pseudometric, so the question is really whether this pseudometric is dominated by the divergence measure of $|P(A)-P(B)|$. Any other ideas how to formalize this idea that set difference is small means sets are interchangeable would be great.

share|cite|improve this question

closed as off-topic by Carlo Beenakker, Chris Godsil, Andrey Rekalo, Ricardo Andrade, David White Oct 4 '13 at 13:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Carlo Beenakker, Chris Godsil, Andrey Rekalo, Ricardo Andrade, David White
If this question can be reworded to fit the rules in the help center, please edit the question.

This question may be more appropriate for Math StackExchange.

Nevertheless,

some hints:

For the argument from the paper, consider what would happen if $\mathbb P(A)$ and $\mathbb P(B)$ are close to $\epsilon$ as well.

For the other question: $$ A\triangle B = (A\setminus B)\cup (B\setminus A),$$ $$ A= (A\setminus B) \cup (A\cap B),\quad\text{and}$$ $$\mathbb P(X\cup Y)=\mathbb P(X)+\mathbb P(Y)\quad\text{if }X\cap Y=\emptyset.$$

share|cite|improve this answer

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