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In the Frechet Bounds inequality equation,

max(0, P(A) + P(B) − 1) ≤ P(A & B) ≤ min(P(A), P(B))

I can totally understand the intuition behind the minimum value of P(A & B) to be "0" and maximum value to be minimum of "P(A) and P(B)" but in what possible case will the minimum value be P(A)+ P(B)-1?

According to my understanding, the maximum value of P(A)+P(B) can be 1 or less than 1. So P(A)+P(B) -1 will always be "0" or negative. In what possible way, P(A) +P(B)-1 will be greater than "0"?

Sandipan Dey
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AKSHAYAA VAIDYANATHAN
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1 Answers1

3

P(AUB)=P(A)+P(B)-P(A&B), by the union rule.

=> P(A&B)=P(A)+P(B)-P(AUB) >= P(A)+P(B)-1, since P(AUB) <= 1, being a probability measure. (1)

Also, P(A&B) >= 0, being a probability measure. (2)

Combining (1) & (2), P(A&B) >= max(0,P(A)+P(B)-1)

Consider the following 2 cases in the following figure: enter image description here

Sandipan Dey
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