Question

Let b be an event with p(b)>0 and define measure p 0 b(a|b). show that p 0[a||g]

Answer 1

The conditional probability of A given B under the measure P0[B] is equal to the probability of A under the measure P0.

Step-by-step explanation:

Let's assume there are two events A and B, with P(B) > 0. We need to show that P[A|B] = P(A) for the measure P0[B](A|B). In other words, we want to prove that the conditional probability of A given B under the measure P0[B] is equal to the probability of A under the measure P0.

Using the definition of conditional probability, we have:

P[A|B] = P[A AND B] / P[B]

But since P0[B](A|B) is defined as P[A AND B] / P0[B](B), we can rewrite the equation as:

P[A|B] = P0[B](A|B) / P0[B](B)

Since B is the sample space under P0[B], the probability of B under the measure P0[B] is equal to 1. Therefore, we can simplify the equation to:

P[A|B] = P0[B](A|B) / 1 = P0[B](A|B)

Thus, we have shown that P[A|B] = P0[B](A|B), which means that the conditional probability of A given B under the measure P0[B] is equal to the probability of A under the measure P0.