probability-part-1 - ARE 210 BASIC CONCEPTS FOR THE ALGEBRA...

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ARE 210 Course Notes page 2 Date of last Update: 8/29/2003 page 2 B ASIC C ONCEPTS FOR THE A LGEBRA OF S ETS Many types of study may be characterized by repeated experimentation under the same set of conditions. Each experiment has an outcome , but prior to the experiment the outcome cannot be predicted with certainty. If every possible outcome can be described a priori , and the experiment can be repeated, we have a random experiment . The collection of every possible outcome is the outcome set , or sample space . Generically, space means the totality of all elements, or outcomes. A sample space is a set C , a collection of members or elements, e , called events. Let C be a subset of C . That is, C is a collection of some of the possible out- comes. This is denoted C C . Then C is an event . The outcomes, e , are elements (members) of the set of all possible outcomes, C . This is denoted e C . The theory of probability is defined and developed axiomatically in terms of sets and an algebra system defined over sets. To develop an algebra of sets, we must have notions of order, equality, zero, addition, and multiplication similar to the algebra of numbers. Definitions: 1. Subset - A set C 1 is a subset of the set C 2 , denoted CC 12 , if each element of C 1 is also an element of C 2 . C C eC eC 1 2 ⊂⇔ lq 2. Equality - If and 21 , then = . Two sets are equal if and only if they have exactly identical elements. 3. Null set - If C has no elements it is empty , the null set . This is de- noted C =∅ . The null set is a subset of every other set. Logical note: If the hypothesis of a statement is false, then the conclusion always follows (the statement is vacuous). If e ∈∅ (there are no such e 's), then eC for all C C . If 11 0 + = , then the world is flat. (Same type of statement.) 4. Union - The union of the sets C 1 and C 2 , denoted , is the set of all elements in at least one of C 1 or C 2 . 5. Intersection - The intersection of two sets C 1 and C 2 is the set of all elements belonging to both of the sets, CC e 1 2 ≡∈ : and Note: Intersections of sets satisfy commutative and associative laws. CC C C CC C CC 3 1 23 2 13 ∩∩ bg bgbg == . Similarly unions are commutative and associative. 2 ∪∪ Theorem: If , then CC C 2 = and 1 = . Proof: Suppose . Then eC C 1 2 & . Hence C 2 . 2 1 2 since . Hence 2 . Therefore, 2 = . ∈⇒ 1 2 & . Hence CCC 112 . eC C eC 1 (by definition). Hence 1 . Therefore, 1 = . Q.E.D.
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ARE 210 Course Notes page 3 Date of last Update: 8/29/2003 page 3 D ISTRIBUTIVE L AW FOR S ETS C CCC C C C 12 3 13 23 ∪∩ ∩∪∩ bgbg bg = C C C C 3 ∩∪ ∪∩∪ = eCC C eCC eC ∈⇒ 3 3 & 1 or 2 & 3 Suppose 1 . Then 1 & 3 ⇒∈ eCC eCC CC Suppose 2 . Then ∩∩ Hence CC C CC CC 3 .
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probability-part-1 - ARE 210 BASIC CONCEPTS FOR THE ALGEBRA...

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