Question

Let B = {A₁, A2, ..., An) CM, and write

B' = {AT, AT,..., ACM. Show that:
(a) B is independent if and only if B' is
independent.
(b) B spans Mn if and only if B' spans Mam-

Answer 1

B is independent if and only if B' is independent, and B spans Mn if and only if B' spans Mam-.

Step-by-step explanation:

To show that B is independent if and only if B' is independent, we need to prove two statements:

a) If B is independent, then B' is independent:

1. Assume that B is independent. This means that for any two different events Ai and Aj in B, the probability of their intersection is equal to the product of their individual probabilities: P(Ai ∩ Aj) = P(Ai) * P(Aj)

2. Since B' is the set of all complements of the events in B, we can apply De Morgan's laws to the equation above, which gives us: P((Ai ∩ Aj)') = P(Ai') * P(Aj')

3. Since (Ai ∩ Aj)' is the complement of Ai ∩ Aj, we can rewrite the equation as: 1 - P(Ai ∩ Aj) = P(Ai') * P(Aj')

4. This shows that the probability of the complement of the intersection of two events in B' is equal to the product of the individual probabilities of their complements. Therefore, B' is independent.

   b) If B' is independent, then B is independent:

1. Assume that B' is independent. This means that for any two different events Ai' and Aj' in B', the probability of their intersection is equal to the product of their individual probabilities: P(Ai' ∩ Aj') = P(Ai') * P(Aj')

2. Since B is the set of all complements of the events in B', we can apply De Morgan's laws to the equation above, which gives us: P((Ai' ∩ Aj')') = P(Ai) * P(Aj)

3. Since (Ai' ∩ Aj')' is the complement of Ai' ∩ Aj', we can rewrite the equation as: 1 - P(Ai' ∩ Aj') = P(Ai) * P(Aj)

4. This shows that the probability of the complement of the intersection of two events in B is equal to the product of the individual probabilities of the events. Therefore, B is independent.