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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.

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    $\begingroup$ Also $\mathsf P(A\cap B)>\mathsf P(A)\,\mathsf P(B)$, so there is a positive correlation between the events. $\endgroup$ – Graham Kemp Aug 6 '17 at 3:45
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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.

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  • $\begingroup$ Interesting! Would you have a reference for this definition of the correlation coefficient? I've never seen it before! Also, if one replaces A by say $C \cap D$, does this relationship still hold? Thanks! $\endgroup$ – Thomas Moore Aug 6 '17 at 3:56
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    $\begingroup$ It uses the expectation definition: $\mathsf P(A)=\mathsf E(\mathbf 1_A)$ where $\mathbf 1_A$ is the indicator random variable that event $A$ occurs. Then $$\dfrac{\mathsf E(\mathbf 1_A\mathbf 1_B)-\mathsf E(\mathbf 1_A)~\mathsf E(\mathbf 1_B)}{\sqrt{\bbox[0.5ex]{\mathsf {Var}(\mathbf 1_A)~\mathsf {Var}(\mathbf 1_B))}}}~=~\dfrac{\mathsf P(A\cap B)-\mathsf P(A)\,\mathsf P(B)}{\sqrt{\bbox[0.5ex]{\mathsf P(A)\,(1-\mathsf P(A))\,\mathsf P(B)\,(1-\mathsf P(B))}}}$$ $\endgroup$ – Graham Kemp Aug 6 '17 at 4:04
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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.

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  • $\begingroup$ Hi. Okay, that's great. Thanks. Could you please explain why there's a positive correlation between events as is stated by Graham Kempf above? $\endgroup$ – Thomas Moore Aug 6 '17 at 3:52
  • $\begingroup$ @ThomasMoore In the formula for correlation, the numerator is $P(A\cap B) - P(A)P(B),$ which is a positive number (as shown by the answer above), and the denominator also is positive. Hence the correlation is positive. $\endgroup$ – David K Aug 6 '17 at 6:45
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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?

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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:

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

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