4

Given two independent events $A$ and $B$, with given conditions:
$0 \lt P(A) , P(B) <1 $.
Which one of the following options is/are false?

  1. $A$ and $B’$ are independent.
  2. $A’$ and $B’$ are independent.
  3. $P(A|B) = P(A|B’)$
  4. For any event c, with $0 \lt P(c) \lt 1$, $P(AB|c)= P(A|c)\cdot P(B|c)$

Here is what I tried:

  1. A and B are independent iff: $P(A \cap B)$ $=$ $P(A)\cdot P(B)$
    Now, we have : $P(A) = P(A \cap B) + P(A \cap B')$
    So,
    $ P(A \cap B')$ $=$ $P(A) - P(A \cap B)$
    $=$ $P(A) - P(A)\cdot P(B)$
    $=$ $[1-P(B)]\cdot P(A)$
    $=$ $P(A)\cdot P(B')$
    Thus, 1 is true.

  2. We know that, $P(A’ \cap B’) =P(A \cup B )’$
    $ =1 - P(A \cup B)$
    $ =1 - P(A) - P(B) + P( A \cap B)$
    $ =1 - P(A) - P(B) + P(A)\cdot P(B)$
    $ = [1-P(A)] \cdot [1-P(B)]$
    $ =P(A’)P(B’) $
    Thus, 2 is also true.

  3. By conditional probability, $P(A | B)$ $=$ $\frac{P(A \cap B)} {P(B)}$
    $=$ $\frac{P(A)\cdot P(B)}{P(B)}$
    $=$ $P(A) $
    And
    $P(A | B') $ $ =$ $\frac{P(A \cap B')}{P(B')}$
    $=$ $\frac{P(A)\cdot P(B')}{P(B')}$
    $=$ $P(A) $

The problem is with 4. I tried to disprove it, by finding a counter-example, and I couldn't.

What is the correct answer?

Tom Zych
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ABcDexter
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1 Answers1

10

Is option 4. true?

No : Assume that $C = A\cup B$, then $P(C)\geqslant P(A)>0$ and $$P(AB \mid C)=\frac{P((A \cap B) \cap C)}{P(C)}=\frac{P(A \cap B)}{P(C)} = \frac{P(A)P(B)}{P(C)}$$ while $$P(A\mid C)P(B\mid C) = \frac{P(A \cap C)}{P(C)}\frac{P(B \cap C)}{P(C)} = \frac{P(A)P(B)}{P(C)^2}$$ hence the equality in option 4. holds if and only if $P(C)=1$. Now, $A$ and $B$ are independent hence $$P(C)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$$ Thus, $P(C)=1$ would mean that $$0=1-(P(A)+P(B)-P(A)P(B))=(1-P(A))(1-P(B))$$ which does not hold since $P(A)\ne1$ and $P(B)\ne1$. Thus, for $C=A\cup B$, $0<P(C)<1$ as required for a counterexample to option 4.

To sum up, option 4. never holds.

Did
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Sarvesh Ravichandran Iyer
  • 76,105
  • Wow, looks good, much thanks :) – ABcDexter Mar 26 '17 at 10:18
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    I think when we have counterexamples to show with independent events, we usually take dice rolls as a starting point. You are welcome! Also, the early down vote seems to have been cancelled out. – Sarvesh Ravichandran Iyer Mar 26 '17 at 10:29
  • Could you please give a non-trivial example in which the C doesn't rely on the mutual-exclusiveness of events (A,B)? – ABcDexter Mar 26 '17 at 10:33
  • @ABcDexter First of all, thank you for editing my answer. Second, are you looking for a counterexample in which $A$ and $B$ are not independent? Because if $C$ is independent of $A$ and $B$, then you can check that actually equality holds above. – Sarvesh Ravichandran Iyer Mar 26 '17 at 10:44
  • $A$ and $B$ are given independent, it's about choosing a proper C (which you just did). I just wanted another counter-example. Also, can we disprove it symbolically. – ABcDexter Mar 26 '17 at 10:46
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    This answer gives an example of a probability space $\Omega$ and events $(A,B,C)$ such that $A$ and $B$ are independent, $0<P(A)<1$, $0<P(B)<1$ and $P(A\cap B\mid C)\neq P(A\mid C)P(B\mid C)$. A more ambitious (and more interesting) result is that, for every probability space $\Omega$ and every independent events $A$ and $B$ such that $0<P(A)<1$ and $0<P(B)<1$, there exists some event $C$ such that $P(A\cap B\mid C)\neq P(A\mid C)P(B\mid C)$. Actually, it seems that this is what a full answer to the question as it is formulated, requires. – Did Mar 26 '17 at 10:47
  • @Did How would you do this? Specifically, abstract construction of an event is involved here, while in the counterexample I took a concrete example. Will we have to transfer this question into a measure-theoretic context and deal with it, maybe? (I don't know how to, I'm just trying to see if I can start from there). – Sarvesh Ravichandran Iyer Mar 26 '17 at 10:50
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    There is nothing to transfer to anything, the question is clearly stated and my last comment explains why, in my opinion, a specific example does not answer it (after all, there might exist some Omega and independent (A,B) such that 4 holds). You can probably get an idea of the abstract construction that allows to prove that 4 never holds if you read carefully the thread of comments on main, where I rather transparently points at the solution... – Did Mar 26 '17 at 10:53
  • OH, all right. I'll look for an event depending on $A$ and $B$ itself, that will give a result. All right, certainly the answer is incomplete in that case. Thank you for the point out, @Did. – Sarvesh Ravichandran Iyer Mar 26 '17 at 10:55
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    @ABcDexter See the improved answer. Turns out the question wasn't too difficult! – Sarvesh Ravichandran Iyer Mar 26 '17 at 11:08
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    Yet another step to add and the answer will be complete: why can we be sure that C=A∪B is such that 0<P(C)<1? – Did Mar 26 '17 at 11:12
  • @Did I shall add that too in. – Sarvesh Ravichandran Iyer Mar 26 '17 at 11:14
  • @Did Added. Also, thank you for helping me improve this answer. – Sarvesh Ravichandran Iyer Mar 26 '17 at 11:21
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    Revamped your post, trying to focus on the relevant points. If this does not suit your taste, please revert to a previous version. (+1) – Did Mar 26 '17 at 11:30
  • Oh, but of course, it was certainly on the cards. Once again, thank you, this answer is effectively your suggestion! – Sarvesh Ravichandran Iyer Mar 26 '17 at 11:33
  • The last part with $P(C)=1$ is essential (+1). – msm Mar 26 '17 at 11:35
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    @астонвіллаолофмэллбэрг Thanks for the bounty on the other answer. 'twas not required, and is much appreciated all the same... – Did Mar 28 '17 at 18:57
  • @Did I must give back for this answer, which I am very happy with. You are welcome. – Sarvesh Ravichandran Iyer Mar 29 '17 at 09:24