# Probability exercise

I was doing one of the Papoulis problems and got a bit confused if I got it right, because as you will see in the end, the conditions given in the problem can be relaxed and still be obtained.

The problem is to calculate $P(A\overline{B}+B\overline{A})$ given that $P(A) = P(B) = P(AB)$.

The Papoulis solutions that I have available states:

1. $P(A) = P(AB+A\overline{B}) = P(AB) + P(A\overline{B})$ (given that $AB$ and $A\overline{B}$ are mutually exclusive)
2. $P(B) = P(BA+A\overline{A}) = P(BA) + P(B\overline{A})$ (given that $BA$ and $B\overline{A}$ are mutually exclusive)

therefore, if $P(A) = P(B) = P(AB)$, from eq. (1) we have $P(A\overline{B})=0$ and from eq. (2) $P(\overline{A}B) = 0$, and we conclude that $P(A\overline{B}+B\overline{A})=0$

The issue is that when I developed my answer, I found a way that P(AB) was not needed for this effect to happen. I found that $P(A\overline{B}+B\overline{A})=0$ if only P(A) = P(B), and I would like to know if it is ok what I did:

3.$P(A\overline{B}+B\overline{A})=P(\overline{A}B+A\overline{B})=P(A\overline{B})+P(\overline{A}B)$ (it is quite easy to see that $A\overline{B}$ and $\overline{A}B$ are mutually exclusive, but it is also provable: Let $\overline{A}B=C$ and $A\overline{B}=D$, they are mutually exclusive if, and only if, CD=$\emptyset$. We have $CD = \overline{A}B\overline{B}A = \overline{A}\emptyset A = \emptyset$, thus $\overline{A}B$ and $A\overline{B}$ are mutually exclusive)

We can develop that:

4.$P(A\overline{B})=P(\overline{\overline{A\overline{B}}})=P(\overline{\overline{A}+B})$,

and we know that

5.$P(\overline{K})=1-P(K)$, then we can rewrite (4) as:

6.$P(\overline{\overline{A}+B})=1-P(\overline{A}+B)=1-(P(\overline{A})+P(B)+P(\overline{A}B))$. (edit: as stated, everything below is wrong due to the fact that this equation should be $P(\overline{\overline{A}+B})=1-(P(\overline{A})+P(B)-P(\overline{A}B))$ leading (7) to $P(A) - P(B) + 2\times P(\overline{A}B)$)

Using (6) in (3) we have:

$P(A\overline{B}+B\overline{A})=\overbrace{1-(P(\overline{A})+P(B)+P(\overline{A}B))}^{P(A\overline{B}) \text{ rewritten}}+\overbrace{P(\overline{A}B)}^{P(B\overline{A}) \text{ rewritten}}=1-P(\overline{A})-P(B)$

and we can rewrite that using (5) to obtain:

7.$P(A\overline{B}+B\overline{A})=\overbrace{1-P(\overline{A})}^{P(A)}-P(B)=P(A)-P(B)$.

Please observe that the equation (7) could be obtained as:

8.$P(A\overline{B}+B\overline{A})=P(B)-P(A)$,

if we had used $P(\overline{B}A)$ in (4). The correct equation to be used is the one that obey the first axiom ($P(A\overline{B}+B\overline{A})>0$), that is eq. (7) when $P(A)>P(B)$ and (8) when $P(B)>P(A)$.

So, if $P(A) = P(B)$, we have that both (7) and (8) lead to 0.

Therefore, the condition for $P(A)=P(B)=P(AB)$ can be relaxed to $P(A)=P(B)$. The case when $P(A) = P(B) = P(AB)$ leads to $P(A\overline{B}+B\overline{A})=0$, but also to $P(A\overline{B})=0$ and $P(B\overline{A})=0$, which also means that $A=B$ (that is, A and B are the same subset from $S$)!

• Because this is a homework/self-study question, you should add the self-study tag. See stats.stackexchange.com/tags/self-study/info Commented Jun 14, 2014 at 3:48
• Eq. 6): $P(\bar{A}+B)=P(\bar{A})+P(B)-P(\bar{A}B)$, assuming that you mean unions (i.e. $\cup$) by those nasty $+$ signs! I recommend changing your notation.
– Stat
Commented Jun 14, 2014 at 5:59
• Thanks @Stat and Juho. I knew that there would be something wrong. I used + in place of union because this is what Papoulis uses, and it would be so boring to use /union instead of +. Commented Jun 14, 2014 at 6:54
• @Stat do you want to post the comment about Eq q as an answer? Otherwise I will append it to my answer. Commented Jun 14, 2014 at 7:20

Consider tossing fair coin once, and define the events as $A = \textrm{heads}$, $B = \textrm{tails}$. Now, $P(A)=0.5=P(B)$, so the assumption holds. However, here $A = \overline{B}$, i.e. the events are mutually exclusive and their union is the sample space $S$. Therefore, $$P(A\overline{B} + \overline{A}B) = P(A + B) = 1.$$ That is, almost certainly either the result of the toss is heads and not tails or the result is tails and not heads. Based on this counterexample, we know that your generalized result is false and may proceed to locating the error in the argument.