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Most of the problems in my textbook have numeric solutions in the back of the book except the proofs. Is this proof correct?

Prove that if $A \cup B$ and $A \cap B$ are independent events, then either $P(A \cap B)=0$ or $P(A \cup B) = 0)$.

  1. $P((A \cup B) \cap (A \cap B)) = P(A \cup B)P(A \cap B)$ (definition of independence)
  2. $P(A \cap B)=P(A \cup B)P(A \cap B)$ (properties of sets and intersections)
  3. Either $P(AB)=0$ or $P(A\cup B)=1$. In the first case we already proved the proposition. In the second case $P(A\cup B)=1\implies P(\bar{A} \cap \bar{B})=0$. $\square$
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What does $P(AB)$ mean? Do you mean $P(A \cap B)$? –  Fly by Night Jan 19 '13 at 22:29
    
If $A$ and $B$ are each equal to the whole sample space, then $A\cup B$ and $A\cap B$ are independent and each has probability $1$. –  Michael Hardy Jan 19 '13 at 22:29
    
Yes. Some older probability texts write $AB$ instead of $A\cap B$. –  ncmathsadist Jan 19 '13 at 22:32
    
@ncmathsadist The OP used both $AB$ and $A\cap B$. –  Fly by Night Jan 19 '13 at 22:42
    
I noticed that. That is indeed a gaffe Mr. Dong has perpetrated. He should use the $A\cap B$ notation, this being the early 21st century. –  ncmathsadist Jan 19 '13 at 23:24

2 Answers 2

up vote 1 down vote accepted

The second line should read

$$(A \cap B) = (A\cap B) \cap (A\cup B) \Rightarrow P(A \cap B) = P[(A\cap B) \cap (A\cup B)]$$

This is the properties of sets mentioned, and we do not apply the probability aspect as yet. Your second line currently isn't explained, but you'd see that it is true by taking probabilities on the sets above.

The third line should then read

$$ P(A\cap B) = P(A\cup B) P(A\cap B) \Rightarrow P(A\cap B) = 0 \mbox{ or } P(A \cup B) = 1$$

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Let $X$ and $Y$ be any events such that $X\subseteq Y$. If $X$ and $Y$ are independent, then $P(X\cap Y)=\Pr(X)\Pr(Y)$. But clearly $\Pr(X\cap Y)=\Pr(X)$, since $X\cap Y=X$. It follows that $$\Pr(X)=\Pr(X)\Pr(Y).$$ If $\Pr(X)\ne 0$, by cancellation we obtain $\Pr(Y)=1$.

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