# Thread: Prove the probability inequality

1. ## Prove the probability inequality

If P[A|B] > P[A|B complement], then show that P[A complement] < P[A complement|B complement].

What I've got so far:

0) P[A intersection B]/P[B] > P[A intersection B complement]/P[B complement]

1) P[A intersection B]*P[B complement] > P[B]*P[A intersection B complement]

2) P[A intersection B] - P[A intersection B]*P[B] > P[B]*P[A] - P[B]*P[A intersection B]

3) P[A intersection B] > P[B]*P[A]

4) P[A] < P[A intersection B]/P[B]

5) P[A] < P[A|B]

6) 1 - P[A complement] < P[A|B]

7) P[A complement] > P[A complement|B]

I'm stuck at this part. Any help would be greatly appreciated.

2. What you wrote is correct; after a few lines, you could get the result (replacing P(B) by 1-P(B complement) in the denominator, etc.). However this is not a very direct way to procede... What do you think of the following?:

$\displaystyle P(A|B)>P(A|B^c)$ implies $\displaystyle P(A^c|B)<P(A^c|B^c)$ (taking complement of boths sides), hence $\displaystyle P(A^c)=P(A^c|B)P(B)+P(A^c|B^c)P(B^c)\leq P(A^c|B^c)P(B)+P(A^c|B^c)P(B^c)$ $\displaystyle =P(A^c|B^c)$. qed.

3. Originally Posted by Laurent
hence $\displaystyle P(A^c)=P(A^c|B)P(B)+P(A^c|B^c)P(B^c)\leq P(A^c|B^c)P(B)+P(A^c|B^c)P(B^c)$ $\displaystyle =P(A^c|B^c)$. qed.
Could you show how you arrived at this? I can't figure it out.

4. Originally Posted by My Little Pony
Could you show how you arrived at this? I can't figure it out.
First equality comes from $\displaystyle P(A)=P(A\cap B)+P(A\cap B^c)$ and definition of conditional expectation. This is sometimes called "law of total expectation" or something alike.
The inequality is what I proved right before: $\displaystyle P(A^c|B)\leq P(A^c|B^c)$.
The last equality is $\displaystyle P(B)+P(B^c)=1$.