Primary reference: Casella-Berger 2 nd Edition Presentation 1-2-3: - - PowerPoint PPT Presentation

Presentation 1-2-3: Axioms and Consequences

Primary reference: Casella-Berger 2nd Edition

Presentation 1-2-3: Axioms and Consequences

• 1. P(𝐵) ≥ 0 for all 𝐵 ∈ ℬ
• 2. P(Ω) = 1
• 3. If sets 𝐵1, 𝐵2, 𝐵3, … ∈ ℬ are pairwise disjoint, then P(ڂ𝑗=1

∞ 𝐵𝑗) =

σ𝑗=1

∞ 𝑄(𝐵𝑗)

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Why am I showing you this again? Because, dear reader, there is yet more these beauties can do! Let’s use them to prove some other basic properties.

Presentation 1-2-3: Axioms and Consequences

Of course! Throughout this lesson, we’ll look at properties dealing with the complements A and 𝐵𝐷, and with the more general events A and B. So for A and 𝐵𝐷, let’s use the coins:

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For more general events A and B, let’s use faces of a die as shown below. Note, however, that these events are not disjoint and their union doesn’t include every element in the sample space. Neither of these properties is necessary. We’re just trying to generalize:

A = { }

H

𝐵𝐷 = { }

T

A ={ } B ={

}

AC ={ } BC ={

}

Presentation 1-2-3: Axioms and Consequences

I. 𝑄 𝐵𝐷 = 1 − 𝑄(𝐵): the probability of the complement of A is just 1 minus the probability of A Remember Axiom 2: P(Ω) = 1, and the axiom of finite additivity: If sets

𝐵∈ℬ and 𝐶∈ℬ are disjoint, then 𝑄(𝐵∪𝐶)=𝑄(𝐵)+𝑄(𝐶) Well, 𝐵∈ℬ and 𝐵𝐷∈ℬ are disjoint, so 𝑄(𝐵 ∪ 𝐵𝐷)=𝑄(𝐵)+𝑄(𝐵𝐷), and also remember that 𝐵 ∪ 𝐵𝐷= Ω. Put it all together and: 1 = P(Ω) = 𝑄(𝐵 ∪ 𝐵𝐷)=𝑄(𝐵)+𝑄(𝐵𝐷) ⇒ 𝑄(𝐵)+𝑄(𝐵𝐷) =1 ⇒ 𝑄(𝐵𝐷) =1 - 𝑄(𝐵)

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Presentation 1-2-3: Axioms and Consequences

I. 𝑄( ) = 1 − 𝑄( ): the probability of the complement of A is just 1 minus the probability of A Remember Axiom 2: P(Ω) = 1, and the axiom of finite additivity: If sets

𝐵∈ℬ and 𝐶∈ℬ are disjoint, then 𝑄(𝐵∪𝐶)=𝑄(𝐵)+𝑄(𝐶) Well ∈ℬ and ∈ℬ are disjoint, so 𝑄( ∪ )=𝑄( )+𝑄( ), and also remember that ∪ = Ω. Put it all together and: 1 = P(Ω) = 𝑄( ∪ )=𝑄( )+𝑄( ) ⇒ 𝑄( )+𝑄( ) =1 ⇒ 𝑄( ) =1 - 𝑄( )

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T H T H H T H T T H T H T H H T T H H H

Presentation 1-2-3: Axioms and Consequences

• II. 𝑄(𝐵) ≤ 1

Well, we just proved that 𝑄 𝐵𝐷 = 1 − 𝑄(𝐵), and since 𝐵𝐷∈ℬ then by

axiom 1, 𝑄 𝐵𝐷 ≥ 0, and so P(A) cannot be greater than 1, since if

P(A) was greater than 1, P(𝐵𝐷) would have to be negative.

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Presentation 1-2-3: Axioms and Consequences

• II. 𝑄(

) ≤ 1 Well, we just proved that 𝑄( ) = 1 − 𝑄( ), and since ∈ℬ then by

axiom 1, 𝑄(

) ≥ 0, and so P( ) cannot be greater than 1, since if P(A ) was greater than 1, P( ) would have to be negative.

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H H H H T T T T

Presentation 1-2-3: Axioms and Consequences

• III. 𝑄 ∅ = 0

This one is my favorite because even though it’s super easy, I still choked when asked the question during my graduate exam. ∅ ∩ Ω = ∅, so ∅ and Ω are disjoint (and again both are elements of ℬ). By the axiom of finite additivity, then, P ∅ ∪ Ω = 𝑄 ∅ + 𝑄 Ω . But at the same time,∅ ∪ Ω = Ω, and by axiom 2, 𝑄 Ω = 1. Put it all together:

𝑄 Ω = P ∅ ∪ Ω = 𝑄 ∅ + 𝑄 Ω = 𝑄 ∅ + 1

⇒ 𝑄 Ω = 1 = 𝑄 ∅ + 1

⇒ 1 = 𝑄 ∅ + 1 ⇒ 𝑄 ∅ = 0

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Presentation 1-2-3: Axioms and Consequences

You’re right: they’re pretty simple, but they don’t need to be axioms, since we’re 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. I’m gonna call these properties a, b, and, c, because there’s a lot of callbacks in these three.

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Presentation 1-2-3: Axioms and Consequences

• a. 𝑄 𝐶 ∩ 𝐵𝑑 = 𝑄 𝐶 − 𝑄(𝐶 ∩ 𝐵)

Oh man, what are we gonna do?! Our axioms don’t even HAVE intersections in them! Okay, don’t panic, lets just arrange this in a different way and hope something comes to us…

𝑄 𝐶 ∩ 𝐵 + 𝑄 𝐶 ∩ 𝐵𝑑 = 𝑄 𝐶

Hey, wait a minute, isn’t there something funny about 𝐶∩𝐵 and 𝐶∩𝐵𝑑 ? Let’s take the intersection of these two and investigate. By that ol’ associative property, we have:

𝐶 ∩ 𝐵 ∩ 𝐶 ∩ 𝐵𝑑 = (𝐶 ∩ 𝐵 ∩ 𝐶 ∩ 𝐵𝑑

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Presentation 1-2-3: Axioms and Consequences

But then, by that ol’ commutative property, we got:

𝐶 ∩ 𝐵 ∩ 𝐶 ∩ 𝐵𝑑 = 𝐶 ∩ 𝐶 ∩ 𝐵 ∩ 𝐵𝑑 = 𝐶 ∩ 𝐶 ∩ ∅ = 𝐶 ∩ ∅ = ∅

So 𝐶 ∩ 𝐵 ∩ 𝐶 ∩ 𝐵𝑑 = ∅ and thus 𝐶∩𝐵 and 𝐶∩𝐵𝑑 are disjoint! That means we get to use that beautiful axiom of finite additivity! (though we are kind of going in reverse this time)

𝑄 𝐶 ∩ 𝐵 + 𝑄 𝐶 ∩ 𝐵𝑑 = 𝑄( 𝐶 ∩ 𝐵 ∪ 𝐶 ∩ 𝐵𝑑) = 𝑄(𝐶 ∩ 𝐵 ∪ 𝐵𝑑 )

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Presentation 1-2-3: Axioms and Consequences

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𝑄 𝐶 ∩ 𝐵 ∪ 𝐵𝑑 = 𝑄 𝐶 ∩ Ω = 𝑄(𝐶)

But we remember (hopefully) that an event and its complement partition the sample space:

Ok, that was tough and took a LOT of steps, so we’re gonna combine them all into

• ne succinct proof, and you can try to follow along:

Presentation 1-2-3: Axioms and Consequences

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What we want to prove: 𝑄 𝐶 ∩ 𝐵𝑑 = 𝑄 𝐶 − 𝑄(𝐶 ∩ 𝐵) ⇒ 𝑄 𝐶 ∩ 𝐵𝑑 +𝑄 𝐶 ∩ 𝐵 = 𝑄 𝐶 How we prove it:

𝑄 𝐶 ∩ 𝐵𝑑 +𝑄 𝐶 ∩ 𝐵 = 𝑄( 𝐶 ∩ 𝐵𝑑 ∪ 𝐶 ∩ 𝐵) = 𝑄(𝐶 ∩ 𝐵𝑑 ∪ 𝐵 ) = 𝑄 𝐶 ∩ Ω = 𝑄(𝐶)

⇒ 𝑄 𝐶 ∩ 𝐵𝑑 +𝑄 𝐶 ∩ 𝐵 = 𝑄(𝐶) ⇒ 𝑄 𝐶 ∩ 𝐵𝑑 = 𝑄 𝐶 − 𝑄(𝐶 ∩ 𝐵)

Presentation 1-2-3: Axioms and Consequences

• a. 𝑄 𝐶 ∩ 𝐵𝑑 = 𝑄 𝐶 − 𝑄(𝐶 ∩ 𝐵)

We’re not going to do the proof with dice because I value my sanity, but we will illustrate the property:

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Presentation 1-2-3: Axioms and Consequences

• b. 𝑄 𝐵 ∪ 𝐶 = 𝑄 𝐵 + 𝑄(𝐶) − 𝑄(𝐵 ∩ 𝐶)

𝑄 𝐵 ∪ 𝐶 = 𝑄 𝐵 ∪ (𝐶 ∩ Ω) = 𝑄 𝐵 ∪ (𝐶 ∩ 𝐵 ∪ 𝐵𝑑 ) = 𝑄 𝐵 ∪ (𝐶 ∩ 𝐵) ∪ 𝐶 ∩ 𝐵𝑑 = 𝑄 𝐵 ∪ 𝐶 ∩ 𝐵𝑑 All right, so we established that 𝑄 𝐵 ∪ 𝐶 = 𝑄 𝐵 ∪ 𝐶 ∩ 𝐵𝑑 , and if you’re getting used to this pattern, you might be wanting to see if maybe A and 𝐶 ∩ 𝐵𝑑 are disjoint so let’s take the intersection: 𝐵 ∩ 𝐶 ∩ 𝐵𝑑 = (𝐵 ∩ 𝐵𝑑) ∩ 𝐶 = ∅ ∩ 𝐶=∅

Aha! They’re disjoint, so we can use axiom 3!!!

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Presentation 1-2-3: Axioms and Consequences

𝑄 𝐵 ∪ 𝐶 = 𝑄 𝐵 ∪ 𝐶 ∩ 𝐵𝑑 = 𝑄 𝐵 + 𝑄 𝐶 ∩ 𝐵𝑑

Ok, so it looks like we’re already pretty close, but we’re stuck with this weird 𝑄 𝐶 ∩ 𝐵𝑑 term. Or, rather, we would be if we hadn’t

proven earlier that 𝑄(𝐶∩𝐵𝑑 )=𝑄(𝐶)−𝑄(𝐶∩𝐵)! Let’s take it on home! 𝑄 𝐵 ∪ 𝐶 = 𝑄 𝐵 + 𝑄 𝐶 ∩ 𝐵𝑑 = 𝑄 𝐵 + 𝑄 𝐶 − 𝑄(𝐶 ∩ 𝐵) Oh yeah, we did it!

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Presentation 1-2-3: Axioms and Consequences

• b. 𝑄 𝐵 ∪ 𝐶 = 𝑄 𝐵 + 𝑄(𝐶) − 𝑄(𝐵 ∩ 𝐶)

Notice how this property accounts for the double-counting caused by the overlapping

Presentation 1-2-3: Axioms and Consequences

• c. 𝐽𝑔 𝐵 ⊂ 𝐶, 𝑢ℎ𝑓𝑜 𝑄(𝐵) ≤ 𝑄(𝐶)

Whoa, set notation? Since when are we combining THAT with probability functions? Well, it doesn’t 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 𝐵 ∩ 𝐶 = 𝐵!

0 ≤ 𝑄 𝐶 ∩ 𝐵𝑑 = 𝑄 𝐶 − 𝑄 𝐶 ∩ 𝐵 = 𝑄 𝐶 − 𝑄 𝐵 ⇒ 0 ≤ 𝑄 𝐶 − 𝑄 𝐵 ⇒ 𝑄 𝐵 ≤ 𝑄 𝐶

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Presentation 1-2-3: Axioms and Consequences

• c. 𝐽𝑔 𝐵 ⊂ 𝐶, 𝑢ℎ𝑓𝑜 𝑄(𝐵) ≤ 𝑄(𝐶)

For this one we need to quickly define a set of which A is a subset, so lets define C:

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C ={

}

A ={ }

𝑇𝑗𝑜𝑑𝑓 𝐵 ⊂ 𝐷, 𝑢ℎ𝑓𝑜 𝑄(𝐵) ≤ 𝑄(𝐷)