Presentation 1-2-3: Axioms and Consequences Primary reference: Casella-Berger 2 nd Edition
Presentation 1-2-3: Axioms and Consequences 1. P( π΅ ) β₯ 0 for all π΅ β β¬ 2. P( β¦ ) = 1 β π΅ π ) = 3. If sets π΅ 1 , π΅ 2 , π΅ 3 , β¦ β β¬ are pairwise disjoint, then P( Ϊ π=1 β π(π΅ π ) Ο π=1 Why am I showing you this again? Because, dear reader, there is yet more these beauties can do! Letβs use them to prove some other basic properties. 2
Presentation 1-2-3: Axioms and Consequences Of course! Throughout this lesson, weβll look at properties dealing with the complements A and π΅ π· , and with the more general events A and B. So for A and π΅ π· , letβs use the coins: π΅ π· = { } A = { } H T For more general events A and B, letβs use faces of a die as shown below. Note, however, that these events are not disjoint and their union doesnβt include every element in the sample space. Neither of these properties is necessary. Weβre just trying to generalize: B = { } A ={ } B C = { A C ={ } } 3
Presentation 1-2-3: Axioms and Consequences π π΅ π· = 1 β π(π΅) : the probability of the complement of A is just I. 1 minus the probability of A Remember Axiom 2: P( β¦ ) = 1, and the axiom of finite additivity: If sets π΅ ββ¬ and πΆ ββ¬ are disjoint, then π ( π΅ βͺ πΆ )= π ( π΅ )+ π ( πΆ ) Well, π΅ ββ¬ and π΅ π· ββ¬ are disjoint, so π ( π΅ βͺ π΅ π· )= π ( π΅ )+ π ( π΅ π· ), and also remember that π΅ βͺ π΅ π· = β¦ . Put it all together and: 1 = P( β¦ ) = π ( π΅ βͺ π΅ π· )= π ( π΅ )+ π ( π΅ π· ) β π ( π΅ )+ π ( π΅ π· ) =1 β π ( π΅ π· ) =1 - π ( π΅ ) 4
Presentation 1-2-3: Axioms and Consequences ) : the probability of the complement of A is just I. π( ) = 1 β π( T H H 1 minus the probability of A H Remember Axiom 2: P( β¦ ) = 1, and the axiom of finite additivity: If sets π΅ ββ¬ and πΆ ββ¬ are disjoint, then π ( π΅ βͺ πΆ )= π ( π΅ )+ π ( πΆ ) Well ββ¬ and ββ¬ are disjoint, so π ( βͺ )= π ( )+ π ( ), and also T H H T H T remember that βͺ = β¦ . Put it all together and: H T 1 = P( β¦ ) = π ( βͺ )= π ( )+ π ( ) β π ( )+ π ( ) =1 β π ( ) =1 - π ( ) H T H H T T T H 5
Presentation 1-2-3: Axioms and Consequences II. π(π΅) β€ 1 Well, we just proved that π π΅ π· = 1 β π(π΅) , and since π΅ π· ββ¬ then by axiom 1, π π΅ π· β₯ 0 , and so P(A) cannot be greater than 1, since if P(A) was greater than 1, P( π΅ π· ) would have to be negative. 6
Presentation 1-2-3: Axioms and Consequences II. π( ) β€ 1 H ) , and since ββ¬ then by Well, we just proved that π( ) = 1 β π( T T H axiom 1, π( ) β₯ 0 , and so P( ) cannot be greater than 1, since if T H P(A ) was greater than 1, P( ) would have to be negative. T H 7
Presentation 1-2-3: Axioms and Consequences III. π β = 0 This one is my favorite because even though itβs super easy, I still choked when asked the question during my graduate exam. β β© Ξ© = β , so β and Ξ© are disjoint (and again both are elements of β¬ ). By the axiom of finite additivity, then, P β βͺ Ξ© = π β + π Ξ© . But at the same time, β βͺ Ξ© = Ξ© , and by axiom 2, π Ξ© = 1. Put it all together: π Ξ© = P β βͺ Ξ© = π β + π Ξ© = π β + 1 β π Ξ© = 1 = π β + 1 β 1 = π β + 1 β π β = 0 8
Presentation 1-2-3: Axioms and Consequences Youβre right: theyβre pretty simple , but they donβt need to be axioms, since weβre able to prove them using axioms we already have. And trust me on this: things are gonna get a little more complicated with these next three properties. Iβm gonna call these properties a, b, and, c, because thereβs a lot of callbacks in these three. 9
Presentation 1-2-3: Axioms and Consequences a. π πΆ β© π΅ π = π πΆ β π(πΆ β© π΅) Oh man, what are we gonna do?! Our axioms donβt even HAVE intersections in them! Okay, donβt panic, lets just arrange this in a different way and hope something comes to usβ¦ π πΆ β© π΅ + π πΆ β© π΅ π = π πΆ Hey, wait a minute, isnβt there something funny about πΆ β© π΅ and πΆ β© π΅ π ? Letβs take the intersection of these two and investigate. By that ol β associative property, we have: πΆ β© π΅ β© πΆ β© π΅ π = (πΆ β© π΅ β© πΆ β© π΅ π 10
Presentation 1-2-3: Axioms and Consequences But then, by that ol β commutative property, we got: π΅ β© πΆ β© π΅ π = πΆ β© πΆ β© π΅ β© π΅ π = πΆ β© πΆ β© πΆ β© β = πΆ β© β = β So πΆ β© π΅ β© πΆ β© π΅ π = β and thus πΆ β© π΅ and πΆ β© π΅ π are disjoint! That means we get to use that beautiful axiom of finite additivity! (though we are kind of going in reverse this time) π πΆ β© π΅ + π πΆ β© π΅ π = π( πΆ β© π΅ βͺ πΆ β© π΅ π ) = π(πΆ β© π΅ βͺ π΅ π ) 11
Presentation 1-2-3: Axioms and Consequences But we remember (hopefully) that an event and its complement partition the sample space: π πΆ β© π΅ βͺ π΅ π = π πΆ β© Ξ© = π(πΆ) Ok, that was tough and took a LOT of steps, so weβre gonna combine them all into one succinct proof, and you can try to follow along: 12
Presentation 1-2-3: Axioms and Consequences What we want to prove: π πΆ β© π΅ π = π πΆ β π(πΆ β© π΅) β π πΆ β© π΅ π + π πΆ β© π΅ = π πΆ How we prove it: π πΆ β© π΅ π + π πΆ β© π΅ = π( πΆ β© π΅ π βͺ πΆ β© π΅) = π(πΆ β© π΅ π βͺ π΅ ) = π πΆ β© Ξ© = π(πΆ) β π πΆ β© π΅ π + π πΆ β© π΅ = π(πΆ) β π πΆ β© π΅ π = π πΆ β π(πΆ β© π΅) 13
Presentation 1-2-3: Axioms and Consequences a. π πΆ β© π΅ π = π πΆ β π(πΆ β© π΅) Weβre not going to do the proof with dice because I value my sanity, but we will illustrate the property: 14
Presentation 1-2-3: Axioms and Consequences b. π π΅ βͺ πΆ = π π΅ + π(πΆ) β π(π΅ β© πΆ) π π΅ βͺ πΆ = π π΅ βͺ (πΆ β© Ξ©) = π π΅ βͺ (πΆ β© π΅ βͺ π΅ π ) = π π΅ βͺ (πΆ β© π΅) βͺ πΆ β© π΅ π = π π΅ βͺ πΆ β© π΅ π All right, so we established that π π΅ βͺ πΆ = π π΅ βͺ πΆ β© π΅ π , and if youβre getting used to this pattern, you might be wanting to see if maybe A and πΆ β© π΅ π are disjoint so letβs take the intersection: π΅ β© πΆ β© π΅ π = (π΅ β© π΅ π ) β© πΆ = β β© πΆ = β Aha! Theyβre disjoint, so we can use axiom 3!!! 15
Presentation 1-2-3: Axioms and Consequences π π΅ βͺ πΆ = π π΅ βͺ πΆ β© π΅ π = π π΅ + π πΆ β© π΅ π Ok, so it looks like weβre already pretty close , but weβre stuck with this weird π πΆ β© π΅ π term. Or, rather, we would be if we hadnβt proven earlier that π ( πΆ β© π΅ π )= π ( πΆ )β π ( πΆ β© π΅ )! Letβs take it on home! π π΅ βͺ πΆ = π π΅ + π πΆ β© π΅ π = π π΅ + π πΆ β π(πΆ β© π΅) Oh yeah, we did it! 16
Presentation 1-2-3: Axioms and Consequences b. π π΅ βͺ πΆ = π π΅ + π(πΆ) β π(π΅ β© πΆ) Notice how this property accounts for the double-counting caused by the overlapping
Presentation 1-2-3: Axioms and Consequences c. π½π π΅ β πΆ, π’βππ π(π΅) β€ π(πΆ) Whoa, set notation? Since when are we combining THAT with probability functions? Well, it doesnβt really matter, this is pretty simple: If A is a subset of B, then every element of A is also an element of B so π΅ β© πΆ = π΅ ! 0 β€ π πΆ β© π΅ π = π πΆ β π πΆ β© π΅ = π πΆ β π π΅ β 0 β€ π πΆ β π π΅ β π π΅ β€ π πΆ 18
Presentation 1-2-3: Axioms and Consequences c. π½π π΅ β πΆ, π’βππ π(π΅) β€ π(πΆ) For this one we need to quickly define a set of which A is a subset, so lets define C: C = { } A ={ } πππππ π΅ β π·, π’βππ π(π΅) β€ π(π·) 19
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