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# note 2 - MA2216/ST2131 Probability Notes 2 Â 1 Axioms of...

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Unformatted text preview: MA2216/ST2131 Probability Notes 2 Â§ 1. Axioms of Probability Consider an experiment whose sample space is S . The objective of probability is to assign to each event A a number IP( A ), in [0 , 1], called the probability of the event A , which will give a precise measure of the chance that A will occur. 1. Consider the collection of all events and denote it by A . For each event A of the sample space S , we assume that a number IP( A ), which is called the probability of the event A , is defined and satisfies the following three axioms: Axiom 1: â‰¤ IP( A ) â‰¤ 1 Axiom 2: IP( S ) = 1 Axiom 3: If A 1 ,A 2 ,... are mutually exclusive (disjoint) (i.e., A i A j = âˆ… when i 6 = j ), then IP Ë† âˆž [ i =1 A i ! = âˆž X i =1 IP( A i ) . 1 Notes: Axiom 1 says that the probability of an event is some number between 0 and 1. Axiom 2 states that, with probability 1, the sample space itself will surely occur, (and hence S is called a sure event ). Axiom 3 says that IP is Ïƒ- additive * . In particular, when A 1 , A 2 , ..., A n are mutually exclusive (disjoint) (that is, A i A j = âˆ… when i 6 = j ), then, ( Cf. 3(ii) below), IP Ë† n [ i =1 A i ! = n X i =1 IP( A i ) . Take note that IP can be regarded as a function defined on A , the collec- tion of all events of the sample space S , taking values in the unit interval [0 , 1], IP : A 7â†’ [0 , 1] such that IP satisfies the above 3 axioms. 2 2. Examples: (a) Consider an experiment of tossing a coin. Find IP( H ) if (i) the coin is fair; (ii) if the coin is biased and a head is twice as likely to appear as a tail. Ans. IP( H ) = 1 2 for a fair coin. If a head is twice as likely to appear as a tail, then IP( H ) = 2IP( T ) and IP( H ) + IP( T ) = 1 . Solving for IP( H ), one arrives at 2 3 . (b) A fair die is tossed. Let A be the event that an even number turns up and let B be the event that a number divisible by 3 occurs. Find IP( A ), IP( B ), IP( A âˆª B ) and IP( AB ). Ans. Trivially, IP( A ) = 3 6 = 1 2 and IP( B ) = 2 6 = 1 3 . Observe that A âˆ© B = { 6 } with IP( A âˆ© B ) = 1 6 . Thus, IP( A âˆª B ) = IP( A ) + IP( B )- IP( A âˆ© B ) = 2 3 . (See 3(v) below.) 3 3. Some Important Properties: (i) IP( âˆ… ) = 0. To see it, take A i = âˆ… for all i â‰¥ 1 and observe that, by Axiom 3, IP( âˆ… ) = IP Ë† âˆž [ i =1 A i ! = âˆž X i =1 IP( A i ) = âˆž X i =1 IP( âˆ… ) . Since 0 â‰¤ IP( âˆ… ) â‰¤ 1, it implies that the infinite series on the right is convergent, it forces IP( âˆ… ) = 0. (ii) If A 1 ,A 2 ,...,A n are mutually exclusive, then IP Ë† n [ i =1 A i ! = n X i =1 IP( A i ) . (This says that IP is finitely additive .) To establish the finite additiveness, refer to Axiom 3 and let A i = âˆ… for all i â‰¥ n + 1. Obviously, âˆž [ i =1 A i = n [ i =1 A i and note also that IP( A i ) = IP( âˆ… ) = 0 for i â‰¥ n + 1....
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note 2 - MA2216/ST2131 Probability Notes 2 Â 1 Axioms of...

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