For each event E E E of the sample space S S S a number P ( E ) P(E) P(E) is de fined and satisfies the following three axioms:
Axiom 1 0 ⩽ P ( E ) ⩽ 1 0 \leqslant P(E) \leqslant 1 0⩽P(E)⩽1Axiom 2 P ( S ) = 1 P(S) = 1 P(S)=1Axiom 3 For any sequence of events E 1 E_1 E1 , E 2 E_2 E2 , … that are mutually exclusive, that is, events for which E i E j = Φ E_iE_j=\varPhi EiEj=Φ when i ≠ j i \ne j i=j (where the Φ \varPhi Φ is the null set), P ( ⋃ i = 1 ∞ E i ) = ∑ i = 1 ∞ P ( E i ) P\left( \bigcup_{i=1}^{\infty}{E_i} \right) =\sum_{i=1}^{\infty}{P\left( E_i \right)} P(i=1⋃∞Ei)=i=1∑∞P(Ei)We refer to P ( E ) P(E) P(E) as the probability of the event of E, and we can deduce some simple consequences of three axioms.
If E ⊂ F E\subset F E⊂F, then P ( E ) ⩽ P ( F ) P(E) \leqslant P(F) P(E)⩽P(F). P ( E c ) = 1 − P ( E ) P(E^c) = 1- P(E) P(Ec)=1−P(E), where E c E^c Ec is the complement of E. P ( ⋃ i = 1 n E i ) = ∑ i = 1 n P ( E i ) P\left( \bigcup_{i=1}^{n}{E_i} \right) =\sum_{i=1}^{n}{P\left( E_i \right)} P(⋃i=1nEi)=∑i=1nP(Ei), when the E i E_i Ei are mutually exclusive. P ( ⋃ i = 1 ∞ E i ) ⩽ ∑ i = 1 ∞ P ( E i ) P\left( \bigcup_{i=1}^{\infty}{E_i} \right) \leqslant \sum_{i=1}^{\infty}{P\left( E_i \right)} P(⋃i=1∞Ei)⩽∑i=1∞P(Ei)If { E n , n ⩾ 1 } \left\{ E_n,n\geqslant 1 \right\} {En,n⩾1} is either an increasing or decreasing sequence of events, then lim n → ∞ P ( E n ) = P ( lim n → ∞ E n ) \underset{n\rightarrow \infty}{\lim}P\left( E_n \right) =P\left( \underset{n\rightarrow \infty}{\lim}E_n \right) n→∞limP(En)=P(n→∞limEn)
Proof Suppose, first, that { E n , n ⩾ 1 } \left\{ E_n,n\geqslant 1 \right\} {En,n⩾1} is an increasing sequence, and define events F n , n ⩾ 1 F_n, n\geqslant 1 Fn,n⩾1 by F 1 = E 1 F n = E n ( ⋃ 1 n − 1 E i ) c = E n E n − 1 c , n > 1 \begin{aligned} F_1 &= E_1 \\ F_n &= E_n\left( \bigcup_1^{n-1}{E_i} \right) ^c = E_nE_{n-1}^{c},\,\ n>1 \end{aligned} F1Fn=E1=En(1⋃n−1Ei)c=EnEn−1c, n>1 That is, F n F_n Fn consists of those points in E n E_n En that are not in any of the earlier E i , i < n E_i, i<n Ei,i<n It is easy to verify that the F n F_n Fn are mutually exclusive events such that ⋃ i = 1 ∞ F i = lim n → ∞ E n a n d ⋃ i = 1 n F i = E n f o r a l l n ⩾ 1. \bigcup_{i=1}^{\infty}{F_i=\underset{n\rightarrow \infty}{\lim}E_n\,\,\ and}\,\,\ \bigcup_{i=1}^n{F_i=E_n\,\,}\,for\,\,all\,\,n\geqslant 1. i=1⋃∞Fi=n→∞limEn and i=1⋃nFi=Enforalln⩾1. Thus, lim n → ∞ P ( E n ) = lim n → ∞ P ( ⋃ 1 n F i ) ( f o r t h e d e f i n i t i o n o f e v e n t s F ) = lim n → ∞ ∑ 1 n P ( F i ) ( f o r A x i o m 3 ) = ∑ 1 ∞ P ( F i ) = P ( ⋃ 1 ∞ F i ) = P ( lim n → ∞ E n ) ( f o r t h e d e f i n a t i o n o f e v e n t s F ) \begin{aligned} \underset{n\rightarrow \infty}{\lim}P\left( E_n \right) &=\underset{n\rightarrow \infty}{\lim}P\left( \bigcup_1^n{F_i} \right)\,\ \left( for\,\,the\,\,definition\,\,of\,\,events\,\,F \right) \\ &=\underset{n\rightarrow \infty}{\lim}\sum_1^n{P\left( F_i \right)}\,\ \left( for\,\,Axiom\,\,3 \right) \\ &=\sum_1^{\infty}{P\left( F_i \right)} \\ &=P\left( \bigcup_1^{\infty}{F_i} \right) \\ &=P\left( \underset{n\rightarrow \infty}{\lim}E_n \right)\,\ \left( for\,\,the\,\,defination\,\,of\,\,events\,\,F \right) \end{aligned} n→∞limP(En)=n→∞limP(1⋃nFi) (forthedefinitionofeventsF)=n→∞lim1∑nP(Fi) (forAxiom3)=1∑∞P(Fi)=P(1⋃∞Fi)=P(n→∞limEn) (forthedefinationofeventsF) which proves the result when { E n , n ⩾ 1 } \left\{ E_n,n\geqslant 1 \right\} {En,n⩾1} is increasing. Second, if { E n , n ⩾ 1 } \left\{ E_n,n\geqslant 1 \right\} {En,n⩾1} is a decreasing sequence, then { E n c , n ⩾ 1 } \left\{ E^c_n, \,\ n\geqslant 1 \right\} {Enc, n⩾1} is an increasing sequence, hence, lim n → ∞ P ( E n c ) = P ( lim n → ∞ E n c ) lim n → ∞ [ 1 − P ( E n ) ] = P ( ⋃ i = 1 ∞ E i c ) ( f o r t h e d e f i n i t i o n o f i n c r e a sin g s e q u e n c e ) 1 − lim n → ∞ P ( E n ) = P [ ( ⋂ i = 1 ∞ E i ) c ] = 1 − P ( ⋂ i = 1 ∞ E i ) = 1 − P ( E n ) ( f o r t h e d e f i n i t i o n o f d e c r e a sin g s e q u e n c e ) lim n → ∞ P ( E n ) = P ( E n ) \begin{aligned} \underset{n\rightarrow \infty}{\lim}P\left( E_{n}^{c} \right) &=P\left( \underset{n\rightarrow \infty}{\lim}E_{n}^{c} \right) \,\, \\ \underset{n\rightarrow \infty}{\lim}\left[ 1-P\left( E_n \right) \right] &=P\left( \bigcup_{i=1}^{\infty}{E_{i}^{c}} \right) \,\,\left( for\,\,the\,\,definition\,\,of\,\,incre\text{a}\sin g\,\,sequence \right) \\ 1-\underset{n\rightarrow \infty}{\lim}P\left( E_n \right) &=P\left[ \left( \bigcap_{i=1}^{\infty}{E_i} \right) ^c \right] \\ &=1-P\left( \bigcap_{i=1}^{\infty}{E_i} \right) \\ &=1-P\left( E_n \right) \left (for\,\,the\,\,definition\,\,of\,\,decre\text{a}\sin g\,\,sequence \right) \\ \underset{n\rightarrow \infty}{\lim}P\left( E_n \right) &=P\left( E_n \right) \end{aligned} n→∞limP(Enc)n→∞lim[1−P(En)]1−n→∞limP(En)n→∞limP(En)=P(n→∞limEnc)=P(i=1⋃∞Eic)(forthedefinitionofincreasingsequence)=P[(i=1⋂∞Ei)c]=1−P(i=1⋂∞Ei)=1−P(En)(forthedefinitionofdecreasingsequence)=P(En) which proves the result.
Let E 1 E_1 E1, E 2 E_2 E2, … denote a sequence of events. If ∑ i = 1 ∞ P ( E i ) < ∞ \sum_{i=1}^{\infty}{P\left( E_i \right)}<\infty i=1∑∞P(Ei)<∞ then, P { a n i n f i n i t e n u m b e r o f t h e E i o c c u r } = 0 P\left\{ an\,\,infinite\,\,number\,\,of\,\,the\,\,E_i\,\,occur \right\} =0 P{aninfinitenumberoftheEioccur}=0
Proof The event that an infinite number of the E i E_i Ei occur, called the lim s u p i → ∞ E i \underset{i\rightarrow \infty}{\lim sup}E_i i→∞limsupEi, can be expressed as lim s u p i → ∞ E i = ⋂ n = 1 ∞ ⋃ i = n ∞ E i \underset{i\rightarrow \infty}{\lim sup}E_i = \bigcap_{n=1}^{\infty}{\bigcup_{i=n}^{\infty}{E_i}} i→∞limsupEi=n=1⋂∞i=n⋃∞Ei As { ⋃ i = n ∞ E i , n ⩾ 1 } \left\{ \bigcup_{i=n}^{\infty}{E_i}, n \geqslant 1 \right\} {⋃i=n∞Ei,n⩾1} is a decreasing sequence of events, it follows from Proposition 1 that P ( ⋂ n = 1 ∞ ⋃ i = n ∞ E i ) = P ( lim n → ∞ ⋃ i = n ∞ E i ) ( f o r t h e d e f i n i t i o n o f d e c r e a sin g e v e n t s ) = lim n → ∞ P ( ⋃ i = n ∞ E i ) ( f o r t h e p r o p o s i t i o n 1 ) ⩽ lim n → ∞ ∑ i = n ∞ P ( E i ) ( f o r t h e c o n s e q u e n c e 4 ) = 0 \begin{aligned} P\left( \bigcap_{n=1}^{\infty}{\bigcup_{i=n}^{\infty}{E_i}} \right) & =P\left( \underset{n\rightarrow \infty}{\lim}\bigcup_{i=n}^{\infty}{E_i} \right) \,\, \left( for\,\,the\,\,definition\,\,of\,\,decre\text{a}\sin g\,\,events \right) \\ & =\underset{n\rightarrow \infty}{\lim}P\left( \bigcup_{i=n}^{\infty}{E_i} \right) \,\, \left( for\,\,the\,\,proposition\,\,1 \right) \\ & \leqslant \underset{n\rightarrow \infty}{\lim}\sum_{i=n}^{\infty}{P\left( E_i \right)}\,\, \left( for\,\,the\,\,consequence\,\,4 \right) \\ & =0 \end{aligned} P(n=1⋂∞i=n⋃∞Ei)=P(n→∞limi=n⋃∞Ei)(forthedefinitionofdecreasingevents)=n→∞limP(i=n⋃∞Ei)(fortheproposition1)⩽n→∞limi=n∑∞P(Ei)(fortheconsequence4)=0 and then result is proven.
If E 1 E_1 E1, E 2 E_2 E2, … are indenpent events such that ∑ i = 1 ∞ P ( E i ) = ∞ \sum_{i=1}^{\infty}{P\left( E_i \right)}=\infty i=1∑∞P(Ei)=∞ then P { a n i n f i n i t e n u m b e r o f t h e E i o c c u r } = 1 P\left\{ an\,\,infinite\,\,number\,\,of\,\,the\,\,E_i\,\,occur \right\} = 1 P{aninfinitenumberoftheEioccur}=1
Proof P ( ⋂ n = 1 ∞ ⋃ i = n ∞ E i ) = P ( lim n → ∞ ⋃ i = n ∞ E i ) ( f o r t h e d e f i n i t i o n o f d e c r e a sin g e v e n t s ) = lim n → ∞ P ( ⋃ i = n ∞ E i ) ( f o r t h e p r o p o s i t i o n 1 ) = lim n → ∞ [ 1 − P ( ⋂ i = n ∞ E i c ) ] ( T i c k S t e p ) = 1 − lim n → ∞ P ( ⋂ i = n ∞ E i c ) \begin{aligned}P\left( \bigcap_{n=1}^{\infty}{\bigcup_{i=n}^{\infty}{E_i}} \right) & =P\left( \underset{n\rightarrow \infty}{\lim}\bigcup_{i=n}^{\infty}{E_i} \right) \,\, \left( for\,\,the\,\,definition\,\,of\,\,decre\text{a}\sin g\,\,events \right) \\ & =\underset{n\rightarrow \infty}{\lim}P\left( \bigcup_{i=n}^{\infty}{E_i} \right) \,\, \left( for\,\,the\,\,proposition\,\,1 \right) \\ & =\underset{n\rightarrow \infty}{\lim}\left[ 1-P\left( \bigcap_{i=n}^{\infty}{E_{i}^{c}} \right) \right] \,\,\left( Tick\,\,Step \right) \\ & =1-\underset{n\rightarrow \infty}{\lim}P\left( \bigcap_{i=n}^{\infty}{E_{i}^{c}} \right)\end{aligned} P(n=1⋂∞i=n⋃∞Ei)=P(n→∞limi=n⋃∞Ei)(forthedefinitionofdecreasingevents)=n→∞limP(i=n⋃∞Ei)(fortheproposition1)=n→∞lim[1−P(i=n⋂∞Eic)](TickStep)=1−n→∞limP(i=n⋂∞Eic)now, lim n → ∞ P ( ⋂ i = n ∞ E i c ) = ∏ i = n ∞ P ( E i c ) ( b y i n d e p e d e n c e ) = ∏ i = n ∞ [ 1 − P ( E i ) ] ⩽ ∏ i = n ∞ e − P ( E i ) ( b y t h e i n e q u a l i t y 1 − x ⩽ e − x ) = e − ∑ n ∞ P ( E i ) = 0 ( f o r t h e p r i m a r y c o n d i t i o n ) \begin{aligned}\underset{n\rightarrow \infty}{\lim}P\left( \bigcap_{i=n}^{\infty}{E_{i}^{c}} \right) & =\prod_{i=n}^{\infty}{P\left( E_{i}^{c} \right)}\,\,\left( by\,\,indepedence \right) \\ & =\prod_{i=n}^{\infty}{\left[ 1-P\left( E_i \right) \right]} \\ & \leqslant \prod_{i=n}^{\infty}{e^{^{-P\left( E_i \right)}}}\,\,\left( by\,\,the\,\,inequality\,\,1-x\leqslant e^{-x} \right) \\ & =\text{e}^{-\sum_n^{\infty}{P\left( E_i \right)}} \\ & =0 \left( for\,\,the\,\,primary\,\,condition \right) \end{aligned} n→∞limP(i=n⋂∞Eic)=i=n∏∞P(Eic)(byindepedence)=i=n∏∞[1−P(Ei)]⩽i=n∏∞e−P(Ei)(bytheinequality1−x⩽e−x)=e−∑n∞P(Ei)=0(fortheprimarycondition) Hence the result follows.
