Probability Proof

**dwsmith** · February 3rd 2011, 04:46 PM

Let A and B be events in a sample space with positive probabilities. Prove that $P(B|A)>P(B)\iff P(A|B)>P(A)$

The seems relatively easy.

1st direction: $P\Rightarrow Q$

$\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}\cdot\frac{P(A)}{P(A)}>P(B)\cdot\frac{P(B )}{P(B)}$

$\displaystyle\Rightarrow\frac{P(A\cap B)}{P(A)}\cdot\frac{P(A)}{P(B)}>P(B)\cdot\frac{P(A )}{P(B)}$

$\displaystyle\frac{P(A\cap B)}{P(B)}>P(A)\Rightarrow P(A|B)>P(A)$

The second direction is identical so I am not going to post it.

Is this how it is proven though?

**Random Variable** · February 3rd 2011, 04:58 PM

I think it's even simpler than that.

$\displaystyle P(B|A) = \frac{P(A \cap B)}{P(A)} > P(B)$

then $\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} > \frac{P(A)P(B)}{P(B)} = P(A)$

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