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Probability (Part 4/4), Apuntes de Administración de Empresas

Asignatura: Statistics I, Profesor: , Carrera: Administració i Direcció d'Empreses - Anglès, Universidad: UAB

Tipo: Apuntes

2011/2012

Subido el 09/05/2012

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Statistics I Probability Theory
2.4 Probability computation and its properties
The 8 properties below are derived from the 3 axioms by Kolmogorov, and are
very useful to compute the probability of any event.
1. P(/0) = 0
2. P(¯
A) = 1P(A)
3. ABP(A)P(B)
4. If the sample space is composed of ndifferent outcomes (or simple
events), all of them equally likely, then
P(ωi) = 1
n(i=1,2,...,n)
Year 2011-2012 43
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2.4 Probability computation and its properties

The 8 properties below are derived from the 3 axioms by Kolmogorov, and are very useful to compute the probability of any event.

  1. P( 0 /) = 0

2. P( A¯) = 1 − P(A)

3. A ⊆ B ⇒ P(A) ≤ P(B)

  1. If the sample space Ω is composed of n different outcomes (or simple events), all of them equally likely, then

P(ωi) =

n

(i = 1 , 2 ,... , n)

  1. If the sample space Ω is composed of n different outcomes (or simple events), all of them equally likely, and the event A is composed of nA of such outcomes, then

P(A) =

nA n

(i = 1 , 2 ,... , n)

  1. If we are considering m disjoint (incompatible, or mutually exclusive) events, that is, such that Ai ∩ A (^) j = 0 for/ i 6 = j, then

P(

⋃^ m

i= 1

Ai) =

m

i= 1

P(Ai) Ai ∩ A (^) j = 0 for/ i 6 = j

  1. Let A and B be two events. Then:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Exemple 24 A dice is tossed. We are interested in the events A =“The outcome is an even number”, B =“The outcome is less than 3”, C =“The outcome is not 6” and D =“The outcome is 1” (respectively, A = { 2 , 4 , 6 } B = { 1 , 2 } C = { 1 , 2 , 3 , 4 , 5 } D = { 1 } ). We will have:

P(A) =

P(B) =

P(C) =

P(D) =

Furthermore, we might compute other probabilities such as:

P(”The outcome is 3 or more than 3”) = P( B¯) = 1 − P(B) = 1 −

P(”The outcome is an even numner or 1”) = P(A ∪ D) = P(A) + P(D) =

= =

P(”The outcome is an even number or is not 6”) = P(A ∪C) = P(A) + P(C) −

−P(A ∩C) = = P({ 2 , 4 , 6 }) + P({ 1 , 2 , 3 , 4 , 5 }) − −P({ 2 , 4 }) =

=

Definició 25 Let Ω be the sample space associated to a random experiment and let A the corresponding σ -àlgebra of events. Let B ∈ A be an event such that P(B) > 0. Then, for any event A ∈ A , we say that the probability of A conditional on B is: P(A/B) =

P(A ∩ B)

P(B)

  • The interpretation, as seen above, is: “what is the probability of A if we have information saying that the event B has occurred”.
  • This is written in short as “probability of A given B”
  • In many cases, the occurrence of the event B has nothing to do with the probability of A.
  • For instance, the probability of 6 when tossing a dice (event A) given that it’s raining (event B) is, simply, the probability of 6 when tossing a dice (P(A/B) = P(A) in this case).
  • This is the case when A and B are independent events, as we will see later on.

From the conditional probability formula we get:

P(B/A) =

P(A ∩ B)

P(A)

P(A ∩ B) = P(A/B) · P(B) = P(B/A) · P(A)

We can also verify that:

  1. If two events are incompatible (A ∩ B = 0), then if one of them occurs the/ probability of the other is zero:

A ∩ B = 0 / ⇒ P(A/B) = P(B/A) = 0

  1. If the event A occurs if and only if B occurs (A ⊂ B) and B has occurred, then the probability of A conditional on B cannot be lower than the probability of A A ⊂ B ⇒ P(A/B) =

P(A ∩ B)

P(B)

P(A)

P(B)

≥ P(A)

  1. If the event B occurs if and only if A occurs (B ⊂ A) and B has occurred, then A occurs for sure

B ⊂ A ⇒ P(A/B) =

P(A ∩ B)

P(B)

P(B)

P(B)

Notice that when A and B are independent,

P(A ∩ B) = P(A/B) · P(B) = P(A) · P(B)

As a matter of fact, this “multiplication rule”, which is true if and only if A and B are independents, is what is used to define the concept of stochastic independence

Definició 27 Two events A, B ∈ A are stochastically independent or, simply, independent if: P(A ∩ B) = P(A) · P(B)

The concept of independence generalizes to 3 or more events in the following way:

Definició 28 A set of n events A 1 ,... , An ∈ A are stochastically independent or, simply, independent if for any subset of 2 or more events we have that:

P(Ak 1 ∩ Ak 2 ∩... ∩ Akm) = P(Ak 1 ) · P(Ak 2 ) ·... · P(Akm)

Exemple 29 A dice is tossed. Consider the events:

A = { 1 , 3 , 5 }

B = { 2 , 4 , 6 }

C = { 1 , 2 , 3 , 4 }

Intuitively, we have that A=”The outcome is an odd number”, B=”The out- come is an even number” i C=”The outcome is less that 5”.

Clearly A and B ARE NOT independent since the occurrence of any of them implies that the other cannot occur. Thus, the occurrence of one of them yields a lot of information about the probability that the other occurs.

On the other hand, C IS independent both from A and from B. Indeed, if we are told that C has occurred then we know that the outcome is either an even number (2 or 4) or an odd number (1 or 3), with the same probability. Thus we are not getting any relevant information regarding the probability that the outcome is an odd number (event A) or and odd number (event B)

The following properties are important regarding the concept of independence:

  1. The impossible event ( /0) is independent from any other event
  2. If two events are incompatible (or mutually exclusive, A ∩ B = 0), then they/ ARE NOT independent
  3. If one event implies another event (A ⊂ B or B ⊂ A), then they ARE NOT independent
  4. IfA is independent from B, then B is independent from A
  5. A and B independent and B and C independent does NOT imply that A and C are independent

2.6 Total probability and Bayes theorems

We are going to see two important theorems in probability theory. They have important applications in the use and analysis of information.

2.6.1 Total probability theorem

This theorem states that the probability of any event can be computed using the information about other “small” events.

Theorem 30 Let A be a σ -algebra of events and A 1 , A 2 ,... , An a set of events such that:

(i) Ai ∩ A (^) j = 0 / for i 6 = j.

(ii) A 1 ∪ A 2 ∪... ∪ An = Ω

((i) and (ii) mean A 1 , A 2 ,... , An form a partition of Ω)

(iii) P(Ai) > 0 ∀Ai (i = 1 ,... , n)

Then, for any B ∈ A :

P(B) =

n

i= 1

P(B/Ai) · P(Ai)