Conditional probability greater than unconditional probability

Question

I am trying to derive conditions for where $P(A|B) > P(A)$,

By Bayes' theorem, I get that:

$P(B|A) P(A) / P(B) > P(A)$,

which means that I am left with: $P(B|A) > P(B)$.

What would the interperetation of this result be?

Thanks.

Answer 1

As Graham Kemp commented, just use $$P(A\mid B) = \frac{P(A\cap B)}{P(B)} $$ to get $P(A\cap B) > P(A)P(B)$. There is no need (or benefit) to appealing to Bayes' theorem here.

Answer 2

Well, the correlation coefficient is $\rho_{\lower{0.5ex}{A,B}}=\dfrac{\mathsf P(A\cap B)-\mathsf P(A)\,\mathsf P(B)}{\sqrt{\bbox[0.5ex]{\mathsf P(A)\,(1-\mathsf P(A))\,\mathsf P(B)\,(1-\mathsf P(B))}}}$ .

So the condition is met when there is a positive correlation between the random variables.

Answer 3

Here is a simple condition directly derived from the definition of conditional probability. You want this: $$P(A|B)=\frac{P(A\cap B)}{P(B)} \stackrel{!}{>} P(A)$$

But fact is that $A\cap B \subseteq A$. So, $$P(A|B)=\frac{P(A\cap B)}{P(B)} \leq \frac{P(A)}{P(B)} \mbox{ and } \frac{P(A)}{P(B)} \gt P(A) \mbox{ for }0 \lt P(B) \lt 1$$

So, a working condition follows easily:

$$A \subseteq B \mbox{ and } P(B) <1$$

Answer 4

Suppose that after observing Alice for an entire year, including both rainy days and dry days, we observe that she carried an umbrella on $15\%$ of the occasions when she left the house. But we also observe that Alice took an umbrella $90\%$ of the time when it was raining. We estimate that if $B$ is the event that Alice carries an umbrella the next time she leaves her house, and $A$ is the event that it is raining then, then $P(B) = 0.15$ and $P(B\mid A) = 0.8.$

Alice just left the house carrying her umbrella. Is it more likely than usual for it to be raining right now?