SLIDE 1 ECE 730 Lectures 2 and 3
John A. Gubner
UW-Madison ECE Dept.
SLIDE 2
Outline
1.4 Axioms and Properties of Probability Axioms Consequences of the Axioms 1.5 Conditional Probability The Law of Total Probability and Bayes’ Rule
SLIDE 3 1.4 Axioms and Properties of Probability
Axioms
(i) The empty set ∅ is called the impossible event. P(∅) = 0. (ii) Probabilities are nonnegative; i.e., for any event A, P(A) ≥ 0. (iii) If A1, A2, . . . are events that are pairwise disjoint, then P ∞
An
∞
P(An). The technical term for this property is countable
- additivity. In other words, “the probabilities of disjoint
events add.” (iv) The entire sample space Ω is called the sure event or the certain event, and its probability is one; i.e., P(Ω) = 1. If an event A = Ω satisfies P(A) = 1, we say that A is an almost-sure event.
SLIDE 4 Consequences of the Axioms
Basic Results
◮ Finite Disjoint Unions.
P
An
N
P(An), An pairwise disjoint.
◮ Probability of a Complement (not compliment).
P(Ac) = 1 − P(A).
◮ Monotonicity. A ⊂ B
implies P(A) ≤ P(B). B A
SLIDE 5
Consequences of the Axioms
Basic Results – continued
◮ Inclusion–Exclusion. P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
B A
SLIDE 6 Consequences of the Axioms
Limit Results
◮ P
∞
An
N→∞ P
An
◮ P
∞
An
N→∞ P
An
SLIDE 7 Consequences of the Axioms
Limit Results
◮ P
∞
An
N→∞ P
An
◮ P
∞
An
N→∞ P
An
( a )
A1 A2 A3
SLIDE 8 Consequences of the Axioms
Limit Results
◮ P
∞
An
N→∞ P
An
◮ P
∞
An
N→∞ P
An
( a )
A1 A2 A3 A1 A2 A3
( b )
SLIDE 9 Consequences of the Axioms
Limit Results – continued
( a )
A1 A2 A3 A1 A2 A3
( b )
◮ P
∞
An
N→∞ P(AN),
if An ⊂ An+1.
◮ P
∞
An
N→∞ P(AN),
if An+1 ⊂ An. These last two properties are called sequential continuity.
SLIDE 10 Consequences of the Axioms
Limit Results – continued
◮ Union Bound or Countable Subadditivity.
P ∞
An
∞
P(An).
SLIDE 11
1.5 Conditional Probability
Given two events A and B, P(A|B) := P(A ∩ B) P(B) . (1) This is equivalent to P(A ∩ B) = P(A|B)P(B). Interchanging the roles of A and B in (1) yields P(B|A) = P(A ∩ B) P(A) . It follows that P(B|A) = P(A|B)P(B) P(A) . In many cases, we are given P(B), P(A|B), P(Bc), and P(A|Bc), but we have to find P(A).
SLIDE 12
The Law of Total Probability and Bayes’ Rule
Write A as the disjoint union A = (A ∩ B) ∪ (A ∩ Bc). Then P(A) = P(A ∩ B) + P(A ∩ Bc). We then have the Law of Total Probability, P(A) = P(A|B)P(B) + P(A|Bc)P(Bc). Substituting this last formula into P(B|A) = P(A|B)P(B) P(A) yields Bayes’ Rule, P(B|A) = P(A|B)P(B) P(A|B)P(B) + P(A|Bc)P(Bc).