0
$\begingroup$

I am trying to solve a probability theory question which asks to prove the following for any events $A$ and $B$

$-P[(A-B)\cup (B-A)] \le P[A]-P[B] \le P[(A-B)\cup (B-A)]$

The following is my attempt:
$P[(A-B)\cup (B-A)] = P[(A\cap B^c)\cup(B\cap A^c)]$
$= P[A\cap B^c] + P[B\cap A^c]$, since $(A\cap B^c)$ and $(B\cap A^c)$ are mutually disjoint.
$=P[A-A\cap B] + P[B-A\cap B] = P[A] + P[B] - 2P[A\cap B]$

So I have simplified the left-most and right-most expression to $P[A] + P[B] - 2P[A\cap B]$. But I am unable to prove that the inequality is correct. I may have done the decomposition wrong and if so, I would appreciate if someone could show me the alternative decomposition which will make it possible to prove the inequality.

2 Answers 2

0

You're basically there, then. You have that the RHS is $P[A]+P[B]-2P[A\cap B]$, but the middle is $P[A]+P[B]-2P[B]$, and $P[A\cap B]\leq P[B]$.

  • 0
    Oh, I see it now. But how do you know that $P[A\cup B] \le P[B]$? – 2017-02-08
  • 0
    It's $P[A\cap B]\leq P[B]$. This is true simply because every time that A and B both happen, B certainly happens, so B is at least as likely to happen as (A and B). – 2017-02-08
  • 0
    But if $P[A\cap B] \le P[B]$, then $-2P[B]\ge -2P[A\cap B]$. But according to the inequality, $-2P[B]\le -2P[A\cap B]$. Am I missing something here? Sorry, just can't get my head around this. I would appreciate an explanation – 2017-02-08
  • 0
    No, $P[A\cap B]\leq P[B]$ implies $-2P[A\cap B]\geq-2P[B]$ - you need to reverse the inequality when you change the sign. – 2017-02-09
0

We have that $$ P(A) \leq P(A \cup B) = P((A-B)\cup(B-A)) + P(A\cap B)$$ since disjoint events. Now, $A\cap B \subseteq B$ and so we have that $$P(A) \leq P((A-B)\cup(B-A)) + P(B).$$ This gives us $P(A)-P(B)\leq P((A-B)\cup(B-A))$. Swapping $A$ and $B$ then gives us $$|P(A)-P(B)|\leq P((A-B)\cup(B-A)) $$ and the result follows.