So I've started by saying that since $P(A | B^c) = P(A | B)$ we know that $\frac{P(A \cap B^c)}{P(B^c)} = \frac{P(A \cap B)}{P(B)}$. However I'm not sure where to go from there. Any help would be great!
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Hint: $P(A) = P(A\mid B^c)P(B^c)+P(A\mid B)P(B)$ – Graham Kemp Sep 17 '15 at 23:54
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How do you define "the events $A$ and $B$ are independent" ? – Henry Sep 18 '15 at 00:02
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I define it as $P(A \cap B) = P(A)P(B)$. – mconn7 Sep 18 '15 at 00:09
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@mconn7 Henry's comment was a Socratic question. – Did Sep 18 '15 at 08:48
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$$\frac{P(A \cap B^c)}{P(B^c)} =\frac{P(A) -P(A\cap B)}{1-P(B)}= \frac{P(A \cap B)} {P(B)}$$ now $$(P(A) -P(A\cap B))P(B)=P(A \cap B)(1-P(B))$$ thus $P(A)P(B)-\color{red}{P(A\cap B)P(B)}=P(A \cap B)-\color{red}{P(A\cap B)P(B)}$ so $P(A)P(B)=P(A \cap B) $ hence $A$ and $B$ are independent.
R.N
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So I understand everything here except how you got $P(A-B)$. Where does that come from? – mconn7 Sep 18 '15 at 00:12
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@mconn7 $P(A)=P(A\cap B)+P(A\cap B^c)$ so $P(A\cap B^c)=P(A)-P(A\cap B)$. Here $A-B$ is only a different notation of $A\cap B^c$ and it is not really functional in this proof. – drhab Sep 18 '15 at 08:57
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Hint: By the Total Probability Theorem: $\mathsf P(A) = \mathsf P(A\mid B^c)\,\mathsf P(B^c)+\mathsf P(A\mid B)\,\mathsf P(B)$
Then since: $\mathsf P(A\mid B)=\mathsf P(A\mid B^c)$ ...
Graham Kemp
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The defining relation is: $$P(A\cap B)=P(A|B)P(B)$$
You could also write: $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$ but that asks for an apart treatment of the special case $P(B)=0$.
Let's say that $P(A\mid B)=c=P(A\mid B^c)$.
Then:
$$P(A)=P(A\mid B)P(B)+P(A\mid B^c)P(B^c)=cP(B)+cP(B^c)=c=P(A\mid B)$$
So: $$P(A\cap B)=P(A|B)P(B)=P(A)P(B)$$
drhab
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