CALCU-Consider an abstract probability space (Ω, F

Consider an abstract probability space (Ω, F, P). Answer the following questions. (a) Suppose that A1 and A2 form a partition Ω, and that B is another subset of Ω. Write down the smallest σ-algebra F containing A1, A2, and B. How many elements are in F? [1] (b) Extending (a), suppose now A1, A2, . . . , An form a partition of Ω for n ∈ {3, 4, 5, . . . }, and B is another subset of Ω. What is the cardinality of the smallest σ-algebra, Fn, containing A1, . . . , An and B?

Solution by Steps

step 1

Identify the elements of the smallest σ-algebra containing

A_1, A_2,

and

B

step 2

The elements are

A_1, A_2, B, A_1 \cap B, A_2 \cap B, A_1 \cup B, A_2 \cup B, \Omega,

and

\emptyset

step 3

Remove redundant elements, noting that

A_1 \cup A_2 = \Omega

and

A_1 \cap A_2 = \emptyset

step 4

The final σ-algebra

F

contains

A_1, A_2, B, A_1 \cap B, A_2 \cap B, A_1^c \cap B, A_2^c \cap B, \Omega,

and

\emptyset

step 5

Count the number of elements in

F

[1] Answer

There are 9 elements in

F

.

step 6

For the general case with

n

partitions, identify the elements of the smallest σ-algebra containing

A_1, \ldots, A_n,

and

B

step 7

The elements include all

A_i

, their intersections with

B

, their complements intersected with

B

,

\Omega

, and

\emptyset

step 8

Calculate the number of elements generated by the partitions and

B

step 9

The total number of elements is

2n + 2^n

, accounting for all possible intersections and unions with

B

[2] Answer

The cardinality of the smallest σ-algebra

F_n

is

2n + 2^n

.

Key Concept

Construction of a σ-algebra from a partition and an additional set

Explanation

The smallest σ-algebra containing a partition of

\Omega

and another set

B

includes the partition sets, their intersections with

B

, their complements intersected with

B

,

\Omega

, and

\emptyset

. For

n

partitions, the cardinality includes these elements plus all possible unions of the partition sets with

B

, resulting in

2n + 2^n

elements.

aUsing only the three Kolmogorov axioms of a probability measure, show that if A, B and C are arbitrary events then P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).

Solution by Steps

step 1

Use the first Kolmogorov axiom which states that for any event

E

,

P(E) \geq 0

step 2

Use the second Kolmogorov axiom which states that

P(\Omega) = 1

, where

\Omega

is the sample space

step 3

Use the third Kolmogorov axiom which states that for any sequence of mutually exclusive events

E_1, E_2, \ldots

, the probability of their union is the sum of their probabilities:

P\left(\bigcup_{i=1}^{\infty} E_i\right) = \sum_{i=1}^{\infty} P(E_i)

step 4

Apply the third axiom to the union of two events to get

P(A \cup B) = P(A) + P(B) - P(A \cap B)

step 5

Apply the third axiom to the union of three events

A \cup B \cup C

by considering

A \cup B

as one event and

C

as another, and then use the result from step 4

step 6

Calculate

P(A \cup B \cup C)

by breaking it down into

P(A \cup B) + P(C) - P((A \cup B) \cap C)

step 7

Expand

P((A \cup B) \cap C)

using the distributive property of sets to get

P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)

step 8

Combine the results from steps 6 and 7 to get

P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)

Answer

P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)

Key Concept

Inclusion-Exclusion Principle for Three Events

Explanation

The Inclusion-Exclusion Principle is used to find the probability of the union of three events by adding the probabilities of the individual events and subtracting the probabilities of their pairwise intersections, then adding back the probability of their triple intersection.

a).Let (Ω, F, P) be a probability space and A, B ∈ F. Suppose there exist constants α, β > 0 such that P(A) = α, P(B) = β, and P(A ∪ B) = α + β − αβ. Are A and B independent events? b).Let (Ω, F, P) be a probability space, and B ∈ F an event for which P(B) > 0. Define F|B := F ∩ B and, for each C ∈ F|B, set Q(C) := P(C)/P(B). Show that (B, F|B, Q) is a probability space.

Solution by Steps

step 1

To determine if events A and B are independent, we need to check if the product of their individual probabilities equals the probability of their intersection

step 2

The definition of independence for two events A and B is P(A ∩ B) = P(A)P(B)

step 3

Given P(A) = α, P(B) = β, and P(A ∪ B) = α + β − αβ, we use the formula P(A ∪ B) = P(A) + P(B) - P(A ∩ B) to find P(A ∩ B)

step 4

Substituting the given values into the formula, we get P(A ∩ B) = α + β − αβ

step 5

To check for independence, we calculate P(A)P(B) = αβ

step 6

Comparing P(A ∩ B) with P(A)P(B), we see that P(A ∩ B) = αβ, which is equal to P(A)P(B)

a) Answer

Yes, A and B are independent events because P(A ∩ B) = P(A)P(B).

Key Concept

Independence of Events

Explanation

Two events A and B are independent if and only if P(A ∩ B) = P(A)P(B). In this case, the given probabilities satisfy this condition.

step 1

To show that (B, F|B, Q) is a probability space, we need to verify that Q satisfies the axioms of probability

step 2

First, we check that for any event C in F|B, Q(C) is non-negative. Since P(C) ≥ 0 and P(B) > 0, Q(C) = P(C)/P(B) ≥ 0

step 3

Next, we check that Q(B) = 1. Since B ∈ F|B and B ∩ B = B, Q(B) = P(B)/P(B) = 1

step 4

Finally, we verify that Q is countably additive. For any countable collection of mutually exclusive events {C_i} in F|B, we have Q(∪C_i) = P(∪C_i)/P(B)

step 5

Since P is countably additive, P(∪C_i) = ΣP(C_i), and thus Q(∪C_i) = ΣP(C_i)/P(B) = ΣQ(C_i)

b) Answer

(B, F|B, Q) is a probability space because Q satisfies all the axioms of probability.

Key Concept

Probability Space Axioms

Explanation

The function Q defined as Q(C) := P(C)/P(B) for each C ∈ F|B satisfies the axioms of probability: non-negativity, normalization, and countable additivity.

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