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If the ratio is greater than 1, then the posterior probability of either A or B will be higher than the prior probability of either A or B.

I suppose it's as if A correlates with B.

However, if A correlates with B, then A and B are random variables instead of "events".

So I wonder if we have something like this in probability. Basically, 2 variables, indicate one another.

I wonder if

P(A&B)/P(A)P(B) = P(A|B)/P(B) = P(B|A)/P(A) has a name?

That name, let's call it L, (I think it should be called evidence strength or something) has an interesting property.

It shows how much increase of probability an evidence give and it is always symmetrical.

Imagine a drug test with 99% sensitivity and 99% specificity. Imagine .5% drug users. So out of 100000 people tested we have 500 users. 495 of which is tested positive. We also have 99500 non users. 995 of which is tested positive.

Here, the probability of user is .5% Probability of those tested positive is around 1.5%

The ratio I am talking about is around 66.

If we know a guy is a user, we know how likely he is tested positive. Just multiply the probability of being tested positive by 66 and we get around 33%.

If we know a guy is tested positive, we know the probability that he is a user. Just multiply the probability that he is a user, which is .5% by 66 and we get 33%

Wow.

That means the strength of evidence of one path of reasoning is exactly the same both way. A->B doesn't always mean B->A. However, they increase the probability of the conclusion by the same amount.

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  • $\begingroup$ You could say that $A,B$ have positive correlation iff the rv's $\mathbf1_A,\mathbf1_B$ have positive correlation of course. But let me add that I never encountered that terminology. $\endgroup$ – drhab Sep 22 '19 at 10:40
  • $\begingroup$ Correlation is only defined for random variable with scalar value. In fact, I wonder if P(A&B)/P(A)P(B) = P(A|B)/P(B) = P(B|A)/P(A) has a name? $\endgroup$ – user4951 Sep 22 '19 at 10:43
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    $\begingroup$ But that is what indicator functions do, so I cannot recognize any obstacle in what you're saying in your comment. Actually we could go for:$$\mathsf{Cov}(A,B):=\mathsf{Cov}\left(\mathbf{1}_{A},\mathbf{1}_{B}\right)=P\left(A\cap B\right)-P\left(A\right)P\left(B\right)$$and:$$\mathsf{Corr}\left(A,B\right):=\mathsf{Corr}\left(\mathbf{1}_{A},\mathbf{1}_{B}\right)=\frac{P\left(A\cap B\right)-P\left(A\right)P\left(B\right)}{\sqrt{P\left(A\right)\left(1-P\left(A\right)\right)P\left(B\right)\left(1-P\left(B\right)\right)}}$$Unfortunately I do not know any name for the expression in your comment. $\endgroup$ – drhab Sep 22 '19 at 10:58
  • $\begingroup$ I noticed that the ratio is the ratio between the posterior and prior probability. And it's symmetric. So I bet it must be very interesting. If we know the probability of A, we can know the probability of A given B by just multiplying by that number. And that is the exact same number to know the probability of B given A if we know the probability of B. $\endgroup$ – user4951 Sep 22 '19 at 12:26
  • $\begingroup$ Ah I see. Yea. I prefer to see the ratio not the difference. $\endgroup$ – user4951 Sep 22 '19 at 12:41
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I can't find a name for the ratio itself, but the $\log$ of the ratio has a name: pointwise mutual information

The average of the pointwise mutual information, i.e. the average of $\log {P(A\cap B) \over P(A)P(B)}$, averaged over the joint distribution, is more well-known: it is mutual information

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