Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

(Jacob Rumans) #1
1.3. The Probability Set Function 19

Theorem 1.3.6.Let{Cn}be a nondecreasing sequence of events. Then

lim
n→∞
P(Cn)=P( lim
n→∞
Cn)=P

(∞

n=1

Cn

)

. (1.3.8)


Let{Cn}be a decreasing sequence of events. Then

lim
n→∞
P(Cn)=P( lim
n→∞
Cn)=P

(∞

n=1

Cn

)

. (1.3.9)


Proof. We prove the result (1.3.8) and leave the second result as Exercise 1.3.20.
Define the sets, called rings, asR 1 =C 1 and, forn>1,Rn=Cn∩Cnc− 1 .It
follows that

⋃∞
n=1Cn=

⋃∞
n=1Rnand thatRm∩Rn =φ,form^ = n.Also,
P(Rn)=P(Cn)−P(Cn− 1 ). Applying the third axiom of probability yields the
following string of equalities:


P

[
lim
n→∞
Cn

]
= P

(∞

n=1

Cn

)
=P

(∞

n=1

Rn

)
=

∑∞

n=1

P(Rn) = lim
n→∞

∑n

j=1

P(Rj)

= lim
n→∞




P(C 1 )+

∑n

j=2

[P(Cj)−P(Cj− 1 )]




= lim
n→∞
P(Cn).(1.3.10)

This is the desired result.


Another useful result for arbitrary unions is given by

Theorem 1.3.7(Boole’s Inequality).Let{Cn}be an arbitrary sequence of events.
Then


P

(∞

n=1

Cn

)

∑∞

n=1

P(Cn). (1.3.11)

Proof: LetDn=


⋃n
i=1Ci.Then{Dn}is an increasing sequence of events that go
up to


⋃∞
n=1Cn. Also, for allj,Dj=Dj−^1 ∪Cj. Hence, by Theorem 1.3.5,
P(Dj)≤P(Dj− 1 )+P(Cj),

that is,
P(Dj)−P(Dj− 1 )≤P(Cj).


In this case, theCis are replaced by theDis in expression (1.3.10). Hence, using
the above inequality in this expression and the fact thatP(C 1 )=P(D 1 ), we have


P

(∞

n=1

Cn

)
= P

(∞

n=1

Dn

)
= lim
n→∞




P(D 1 )+

∑n

j=2

[P(Dj)−P(Dj− 1 )]




≤ lim
n→∞

∑n

j=1

P(Cj)=

∑∞

n=1

P(Cn).
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