Movatterモバイル変換


[0]ホーム

URL:


Jump to content
WikipediaThe Free Encyclopedia
Search

Probability axioms

From Wikipedia, the free encyclopedia
Foundations of probability theory
Part of a series onstatistics
Probability theory

The standardprobability axioms are the foundations ofprobability theory introduced by Russian mathematicianAndrey Kolmogorov in 1933.[1] Theseaxioms remain central and have direct contributions to mathematics, the physical sciences, and real-world probability cases.[2]

There are several other (equivalent) approaches to formalising probability.Bayesians will often motivate the Kolmogorov axioms by invokingCox's theorem or theDutch book arguments instead.[3][4]

Kolmogorov axioms

[edit]

The assumptions as to setting up the axioms can be summarised as follows: Let(Ω,F,P){\displaystyle (\Omega ,F,P)} be ameasure space such thatP(E){\displaystyle P(E)} is theprobability of someeventE{\displaystyle E}, andP(Ω)=1{\displaystyle P(\Omega )=1}. Then(Ω,F,P){\displaystyle (\Omega ,F,P)} is aprobability space, with sample spaceΩ{\displaystyle \Omega }, event spaceF{\displaystyle F} andprobability measureP{\displaystyle P}.[1]

First axiom

[edit]

The probability of an event is a non-negative real number:

P(E)R,P(E)0EF{\displaystyle P(E)\in \mathbb {R} ,P(E)\geq 0\qquad \forall E\in F}

whereF{\displaystyle F} is the event space. It follows (when combined with the second axiom) thatP(E){\displaystyle P(E)} is always finite, in contrast with more generalmeasure theory. Theories which assignnegative probability relax the first axiom.

Second axiom

[edit]

This is the assumption ofunit measure: that the probability that at least one of theelementary events in the entire sample space will occur is 1.

P(Ω)=1{\displaystyle P(\Omega )=1}

Third axiom

[edit]

This is the assumption ofσ-additivity:

Anycountable sequence ofdisjoint sets (synonymous withmutually exclusive events)E1,E2,{\displaystyle E_{1},E_{2},\ldots } satisfies
P(i=1Ei)=i=1P(Ei).{\displaystyle P\left(\bigcup _{i=1}^{\infty }E_{i}\right)=\sum _{i=1}^{\infty }P(E_{i}).}

Some authors consider merelyfinitely additive probability spaces, in which case one just needs analgebra of sets, rather than aσ-algebra.[5]Quasiprobability distributions in general relax the third axiom.

Consequences

[edit]

From the Kolmogorov axioms, one can deduce other useful rules for studying probabilities. The proofs[6][7][8] of these rules are a very insightful procedure that illustrates the power of the third axiom, and its interaction with the prior two axioms. Four of the immediate corollaries and their proofs are shown below:

Monotonicity

[edit]
ifABthenP(A)P(B).{\displaystyle \quad {\text{if}}\quad A\subseteq B\quad {\text{then}}\quad P(A)\leq P(B).}

If A is a subset of, or equal to B, then the probability of A is less than, or equal to the probability of B.

Proof of monotonicity

[edit]

Source:[6]

In order to verify the monotonicity property, we setE1=A{\displaystyle E_{1}=A} andE2=BA{\displaystyle E_{2}=B\setminus A}, whereAB{\displaystyle A\subseteq B} andEi={\displaystyle E_{i}=\varnothing } fori3{\displaystyle i\geq 3}. From the properties of theempty set ({\displaystyle \varnothing }), it is easy to see that the setsEi{\displaystyle E_{i}} are pairwise disjoint andE1E2=B{\displaystyle E_{1}\cup E_{2}\cup \cdots =B}. Hence, we obtain from the third axiom that

P(A)+P(BA)+i=3P(Ei)=P(B).{\displaystyle P(A)+P(B\setminus A)+\sum _{i=3}^{\infty }P(E_{i})=P(B).}

Since, by the first axiom, the left-hand side of this equation is a series of non-negative numbers, and since it converges toP(B){\displaystyle P(B)} which is finite, we obtain bothP(A)P(B){\displaystyle P(A)\leq P(B)} andP()=0{\displaystyle P(\varnothing )=0}.

The probability of the empty set

[edit]
P()=0.{\displaystyle P(\varnothing )=0.}

In many cases,{\displaystyle \varnothing } is not the only event with probability 0.

Proof of the probability of the empty set

[edit]

P()=P(){\displaystyle P(\varnothing \cup \varnothing )=P(\varnothing )} since={\displaystyle \varnothing \cup \varnothing =\varnothing },

P()+P()=P(){\displaystyle P(\varnothing )+P(\varnothing )=P(\varnothing )} by applying the third axiom to the left-hand side (note{\displaystyle \varnothing } is disjoint with itself), and so

P()=0{\displaystyle P(\varnothing )=0} by subtractingP(){\displaystyle P(\varnothing )} from each side of the equation.

The complement rule

[edit]

P(A)=P(ΩA)=1P(A){\displaystyle P\left(A^{\complement }\right)=P(\Omega -A)=1-P(A)}

Proof of the complement rule

[edit]

GivenA{\displaystyle A} andA{\displaystyle A^{\complement }} are mutually exclusive and thatAA=Ω{\displaystyle A\cup A^{\complement }=\Omega }:

P(AA)=P(A)+P(A){\displaystyle P(A\cup A^{\complement })=P(A)+P(A^{\complement })}... (by axiom 3)

and,P(AA)=P(Ω)=1{\displaystyle P(A\cup A^{\complement })=P(\Omega )=1} ...(by axiom 2)

P(A)+P(A)=1{\displaystyle \Rightarrow P(A)+P(A^{\complement })=1}

P(A)=1P(A){\displaystyle \therefore P(A^{\complement })=1-P(A)}

The numeric bound

[edit]

It immediately follows from the monotonicity property that

0P(E)1EF.{\displaystyle 0\leq P(E)\leq 1\qquad \forall E\in F.}

Proof of the numeric bound

[edit]

Given the complement ruleP(Ec)=1P(E){\displaystyle P(E^{c})=1-P(E)} andaxiom 1P(Ec)0{\displaystyle P(E^{c})\geq 0}:

1P(E)0{\displaystyle 1-P(E)\geq 0}

1P(E){\displaystyle \Rightarrow 1\geq P(E)}

0P(E)1{\displaystyle \therefore 0\leq P(E)\leq 1}

Further consequences

[edit]

Another important property is:

P(AB)=P(A)+P(B)P(AB).{\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B).}

This is called the addition law of probability, or the sum rule.That is, the probability that an event inAorB will happen is the sum of the probability of an event inA and the probability of an event inB, minus the probability of an event that is in bothAandB. The proof of this is as follows:

Firstly,

P(AB)=P(A)+P(BA){\displaystyle P(A\cup B)=P(A)+P(B\setminus A)}.(by Axiom 3)

So,

P(AB)=P(A)+P(B(AB)){\displaystyle P(A\cup B)=P(A)+P(B\setminus (A\cap B))} (byBA=B(AB){\displaystyle B\setminus A=B\setminus (A\cap B)}).

Also,

P(B)=P(B(AB))+P(AB){\displaystyle P(B)=P(B\setminus (A\cap B))+P(A\cap B)}

and eliminatingP(B(AB)){\displaystyle P(B\setminus (A\cap B))} from both equations gives us the desired result.

An extension of the addition law to any number of sets is theinclusion–exclusion principle.

SettingB to the complementAc ofA in the addition law gives

P(Ac)=P(ΩA)=1P(A){\displaystyle P\left(A^{c}\right)=P(\Omega \setminus A)=1-P(A)}

That is, the probability that any event willnot happen (or the event'scomplement) is 1 minus the probability that it will.

Simple example: coin toss

[edit]

Consider a single coin-toss, and assume that the coin will either land heads (H) or tails (T) (but not both). No assumption is made as to whether the coin is fair or as to whether or not any bias depends on how the coin is tossed.[9]

We may define:

Ω={H,T}{\displaystyle \Omega =\{H,T\}}
F={,{H},{T},{H,T}}{\displaystyle F=\{\varnothing ,\{H\},\{T\},\{H,T\}\}}

Kolmogorov's axioms imply that:

P()=0{\displaystyle P(\varnothing )=0}

The probability ofneither headsnor tails, is 0.

P({H,T}c)=0{\displaystyle P(\{H,T\}^{c})=0}

The probability ofeither headsor tails, is 1.

P({H})+P({T})=1{\displaystyle P(\{H\})+P(\{T\})=1}

The sum of the probability of heads and the probability of tails, is 1.

See also

[edit]

References

[edit]
  1. ^abKolmogorov, Andrey (1950) [1933].Foundations of the theory of probability. New York, US: Chelsea Publishing Company.
  2. ^Aldous, David."What is the significance of the Kolmogorov axioms?".David Aldous. RetrievedNovember 19, 2019.
  3. ^Cox, R. T. (1946). "Probability, Frequency and Reasonable Expectation".American Journal of Physics.14 (1):1–10.Bibcode:1946AmJPh..14....1C.doi:10.1119/1.1990764.
  4. ^Cox, R. T. (1961).The Algebra of Probable Inference. Baltimore, MD: Johns Hopkins University Press.
  5. ^Hájek, Alan (August 28, 2019)."Interpretations of Probability".Stanford Encyclopedia of Philosophy. RetrievedNovember 17, 2019.
  6. ^abRoss, Sheldon M. (2014).A first course in probability (Ninth ed.). Upper Saddle River, New Jersey. pp. 27, 28.ISBN 978-0-321-79477-2.OCLC 827003384.{{cite book}}: CS1 maint: location missing publisher (link)
  7. ^Gerard, David (December 9, 2017)."Proofs from axioms"(PDF). RetrievedNovember 20, 2019.
  8. ^Jackson, Bill (2010)."Probability (Lecture Notes - Week 3)"(PDF).School of Mathematics, Queen Mary University of London. RetrievedNovember 20, 2019.
  9. ^Diaconis, Persi; Holmes, Susan; Montgomery, Richard (2007)."Dynamical Bias in the Coin Toss"(PDF).SIAM Review.49 (211–235):211–235.Bibcode:2007SIAMR..49..211D.doi:10.1137/S0036144504446436. Retrieved5 January 2024.

Further reading

[edit]
Retrieved from "https://en.wikipedia.org/w/index.php?title=Probability_axioms&oldid=1286322081"
Categories:
Hidden categories:

[8]ページ先頭

©2009-2025 Movatter.jp