Deriving independence from conditional probability

**My Little Pony** · January 25th 2010, 04:43 PM

If [A|B] = [A|B complement], show that A and B are independent.

Here's what I've got so far:

Using the definition of conditional probability and the multiplicative law, I arrived at this:

P[A|B] = (P[B complement|A]*P[A])/P[B complement]

I'm not sure how I'm supposed to arrive at the definition of independence (P[A|B] = P[A]) from this.

**courteous** · January 26th 2010, 12:43 AM

We use the known equality: $P(A|B)=\frac{P(AB)}{P(B)}=\frac{P(AB')}{P(B')}$ .
We also know that $P(B')=1-P(B)$ and use that on last fraction: $P(AB')=P(A|B)-P(A|B)P(B)=P(A|B)-P(AB)$ $\Rightarrow P(A|B)=P(AB')+P(AB)=P(A)$

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