Download LECTURE 1: Probability models and axioms Readings: Sections 1.1

LECTURE 1: Probability models and axioms
• Sections
Sample 1.1,
space
Readings:•
1.2
Sample
space
• Probability
Probability laws
laws
•
– Axioms
Axioms
–
Lecture outline
– Properties
Some
properties
–
that follow from the axioms
Some properties
• Examples
Examples
•
– Discrete
Discrete
–
– Continuous
Continuous
–
• Discussion
Discussion
•
– Countable
Countable additivity
additivity
–
– Mathematical
Mathematical subtleties
subtleties
–
Interpretations of
of probabilities
probabilities
••• Interpretations
Interpretations
Sample space
• List (set) of possible outcomes, Ω
• List must be:
– Mutually exclusive
– Collectively exhaustive
Sample
space
– At the
“right” granularity
ssible outcomes
• Two steps:
– Describe possible outcomes
sive
– Describe beliefs about likelihood of outcomes
haustive
granularity
Sample space
Sample space
Sample space
ssible outcomes
• List (set) of possible outcomes, Ω
• List (set) of possible outcomes, Ω
• List must be:
sive
• – List
must be:
Mutually
exclusive
haustive
Mutually exclusive
– Collectively
exhaustive
granularity
– At
Collectively
exhaustive
the “right”
granularity
– At the “right” granularity
Sample space: discrete/finite example
Sample space: discrete/finite example
Sample space: discrete/finite example
space: die
discrete/finite example
• Two rolls of aSample
tetrahedral
• Two rolls of a tetrahedral die
rolls of a tetrahedral die
• Two rolls
of a vs.
tetrahedral
–dieSample
space vs. sequential description
– space
Sample
space
sequential
description
mple
vs. sequential
description
Die roll
example
– Sample space vs. sequential
description
1,1
1,2
1,3
1,4
1
4
4
2
44
Y = Second 3
3
Y = roll
Second
2
roll
1
1
Y = Second
2
1,1
1,2
4 1,3
1,4
3
1
Y = Second 33
2
3
12
2
X = First roll
4
3
3
1
3
2
1
4
1
2
1
3
44,4
3
4
4
X = First roll
4,4
4
X = First roll
4,4
3 •1 A
2
3
4 continuous
2
ntinuous sample space: 1
4 sample space:
4
(x,
y)
such
that
0
≤
x,
y
≤
1
•
A
continuous
sample
space:
) such that 0 ≤ x, y ≤ 1
X = First roll X = First roll
(x, y) such
y
y that 0 ≤ x, y ≤ 1
y Y)=2
• A continuous
space:
• Let B be1thesample
event: min(X,
(x, •y)Let
such
0 Y≤) x, y1≤ 1
M =that
max(X,
4,4
1
y
• P(M = 1 | B) =
1
x1
1,1
1,2
1,3
1,4
2
2
3
Y = Second
roll
roll
1
2
roll
2
1
1
1,1
1,2
1,3
1,4
1
x
mple space: continuous example
mple space: Sample space: continuous example
Sample
space: sample
continuous
• A continuous
space:example
ntinuous sample space:
Sample space: continuous example
• (x, y) such that 0 ≤ x, y ≤ 1
inuous sample space:
such
that 0space:
≤ x, y ≤continuous
1
Sample
example
0 ≤ x, y ≤ 1
y
us sample space:
uch that 0 ≤ x, y ≤ 1
1y
that 0 ≤ x, y ≤ 1
y
1
1
Sample space:
continuous
example example
Sample
space: continuous
x sample space:
• A continuous
sample
• A continuous
1 space:
1
x
• (x,
y) 0such
• (x, y) such
that
≤ x, that
y ≤y10 ≤ x, y ≤ 1
y
y11
1
1
xy
1
1
x
Probability axioms
Probability axioms
et of the•sample
space
Event:
a subset of the sample space
assigned –
to Probability
events
Probability
axioms
is assignedProbability
to
events axioms
Probability
Probability axioms
axioms
• •Event:
a subset
of of
thethe
sample
space
Event:
a subset
subset
sample
space
•
Event:
a
of
the
sample
Axioms:
• Event: a subset of the sample space
space
– –Probability
is is
assigned
to to
events
Probability
assigned
events
Probability
is
1. P–
≥0
–(A)
Probability
is assigned
assigned to
to events
events
• •Axioms:
Axioms:
• Axioms:
1
Axioms: = 1
2. •P(universe)
1. P
(A)
≥0
–
Nonnegativity:
P(A)
(A) ≥
≥ 00
–
Nonnegativity:
P
–
Nonnegativity:
P
(A)
≥
0
3. If A ∩ B = Ø,
P
(universe)
=1
–
Normalization:
P(Ω)
= 11
= P(A)2.
+ Pthen
(B)
Normalization:
P
=
P
(A
∪
B)
= P(A)
+P
– Normalization:
P(Ω)
(Ω)
=(B)
1
3. If– A(Finite)
∩ B = Ø,
additivity:
additivity:
–
(Finite)
additivity:
(to
be strengthened later)
thenIf P
(A
∪
B)
=
P
(A)
+
P
(B)
Ø, then
then P
P(A
(A ∪
∪ B)
B) =
=P
P(A)
(A)+
+P
P(B)
(B)
If A
A∩
∩B
B=
= Ø,
Ø, then
P
(A ∪
B) =
P
(A) +
P
(B)
k }) =
P({s
+ 1· ,· s· 2+
}) = P({s1}) + · · · + P({s })
• 1})
P({s
, .P
. .({s
, skk})
k
= P(s1) + · · · + P(sk )
= P(s1) + · · · + P(sk )
• P({s
==
P({s
++
· · ·· +
P({s
1 , s2
1 })})
k })})
({s
, ,s.22., .. ,. s. k, })
skk})
P
({s
····+
P
({s
•• P
})
=
P
({s
})
+
·
+
P
({s
})
1
1
1
1
P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({skkk})
==
P(s
)+
· · ·· +
P(s
) )
1
k
P
(s
)
+
·
·
+
P
(s
1
=
P
(s
)
+
·
·
·
+
P
(s
= P(s11) + · · · + P(skkk))
ds strengthening
– Axiom 3 needs strengthening
• Axioms
• Axioms
• Consequences
• Axioms
• Consequences
• Consequences
• Axioms
For disjoint sets:
Probability axioms
axioms • Axioms
For disjoint sets:
Probability
• Consequences• Axioms
For
disjoint sets:
• Consequences
Probability
axioms
•
Consequences
a subset
subset of
of
the sample
sample
space
P(A
∪ B) = P(A) + P(B)
a
the
Some
simple space
consequences
of the
axioms
P(A
∪ B) = P(A) + P(B)
• Axioms
For
disjoint
sets:
ility
assigned
to
events
∪ B) = P
(A) + P(B)
Pility
({s1is
,issof
, .the
. . , ssample
=P
({s
}) disjoint
2assigned
1}) + · · · + P({sP
For
sets:
k }) to
k(A
bset
space
events
For disjoint
sets:
• Consequences
• Axioms
Some simple consequences of the ax
Some simple consequences of the a
is assigned to events
P
(A
∪ ·B)
=PP(s
(A)
+
P
(B)
=
P
(s
)
+
·
·
+
)
1
k
• Consequences
Some
consequences of the axioms
P(A ∪•B)P=
P(A)
(B)
(A)
≤
1 + Psimple
gativity:
P
(A)
≥
0
•
P
(A)
≤
1
ativity: •P(A)
≥
0
P(A ∪For
B) disjoint
= P(A) sets:
+ P(B)
Axioms
• P(A)
1 consequences
Some
simple
of the axioms
• ≤
P
(Ø)
=0
ization:
Pdisjoint
(Ω)
=1
vity:
PFor
(A)
≥ 0=
sets:
•
P
(Ø) =
0
ization:
P
(Ω)
1
• Consequences
Some
simple consequences of the axioms
consequences
of+the
axioms
• P(Ø)
=
P
(A
∪0
B)CSome
=disjoint:
P(A)simple
+P
(B)
P(A) ≤ 1
)ion:
additivity:
(to
be• strengthened
later)
•
A,
B,
P
(A
∪
B
∪
C)
=
P
(A)
P
(B)
+ P(C
P
(Ω)
=
1
additivity: (to be strengthened later)
• ≤
A,1B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P(
• P(A)
c ) = 1 for k disjoint events
B = Ø,Pthen
P
(A=
∪P
B)
=+
P(A)
+ P(B)• P(A)
(A
∪P
B)
(A)
P(B)
and
similarly
(A) +
≤
1 (A
For
disjoint
events:
P
=
Ø,
then
(A
∪
B)
=
P
(A)
+
P
(B)
•
P
(Ø)
=
0
and
similarly
for k disjoint events
ditivity: (to be strengthened later)
Some simple consequences of the ax
• P(Ø) = 0
Ø, then P(A ∪ B) =• PA,
(A)
+CPdisjoint:
(B)
• ∪CP
({s
, .P
.P
.(A)
,(A
sk })
=
({s
+
· ·+
· +PP
({s+
• A,
P(A
(Ø)
=
0 C)
B,
disjoint:
∪
B
∪P
C)
=
P
(A)
(B)
P(C)
1,,ss2
1})
k })
B,
P
B
∪
=
+
P
(B)
+
P
(C)
•
P
({s
,
.
.
.
,
s
})
=
P
({s
})
+
·
·
·
+
P
({s
})
1
2
1
k
k
Some
simple
consequences
of
the
axioms
P(A ∪ B) = P(A) + P(B) • and
P(A)
≤1
A, •
B,similarly
C
disjoint:
Pdisjoint
(A ∪ B ∪events
C) = P(A) + P(B) + P(C)
for kP
and similarly for
disjoint
events
• kA,
B,
C
disjoint:
(A
∪
B
∪
C)
=
P·(A)
+PP
(B)
+ P(C)
=
P
(s
)
+
·
·
+
(s
)
and
similarly
for
k
disjoint
events
1
k
• P(A) ≤ 1
= P(s1) + · · · + P(sk )
• similarly
P
(Ø)
=
0})k=disjoint
and
for
P
({s
,
s
,
.
.
.
,
s
P
({s
+ · · · + P({sk })
1
2
1 })events
• PSome
({s1, s2simple
, . . . , sk•})
=
P
({s
})
+
·
·
·
+
P
({s
k
1
consequences
of
the
axioms
k })
• P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk })
• P(Ø) = 0
• A,
B, C disjoint:
P(A})
∪ B ·∪· ·C)
P(A)
+ P(B) + P(C
• P({s
= P({s
+=
P({s
1 , s2 , . . . , sk }) =
1+ +
k })
P
(s
)
·
·
·
+
P
(s
)
• P(A) ≤ 1
= P(s
· · · + P(s
k
1) +
k )k 1disjoint events
and
for
• A, B, C disjoint: P(A ∪ B ∪ C) = P(A)
+similarly
P(B) +=
P
(C)
P(s1) + · · · + P(sk )
= P(s1) + · · · + P(sk )
•and
P(Ø)
= 0 for k disjoint events
similarly
• P({s1, s2, . . . , sk }) =
P({s1}) + · · · + P({sk })
– Axiom
3 needs strengthening
P(A)
+ P(Ac) = 1
Axiom
• •P({s
P({s
1 , s2 , . . . , sk }) = P({s1 }) + · · · +–
k }) 3 needs strengthening
= P(s
– Axiom
3 needs
strengthening
1 ) + · · · + P(sk )
–
Do
weird
sets
have
probabilities?
Do+weird
• A, B, C disjoint: P(A ∪ B ∪ C) = –
P(A)
P(B)sets
+ Phave
(C) probabilities?
= P(s1) + ·–· · +
P(s
)
Do
weird
sets
have
probabilities?
k
– Axiom
3 needsevents
strengthening
and similarly
for k disjoint
– Axiom 3 needs strengthening
sets})
have
–
Axiom
needs
strengthening
• P({s1, s2, .–. . ,Do
sk })weird
= P({s
· · probabilities?
·+
P3({s
weird
sets
1 – +Do
k }) have probabilities?
Axiom 3 needs strengthening
– Do weird sets have probabilities?
Probability axioms
axioms
Probability
Probability
axioms
a subset
subset of
of
the
sample
space
a
the
sample
space
Some simple consequences of the axioms
ility
assigned
to
events
Pility
({s1is
,issof
, .the
. . , ssample
=P
({s
2assigned
1}) + · · · + P({sk })
k }) to
bset
space
events
• Axioms
is assigned to events
= P(s1) + · · · + P(sk )
• Consequences
gativity:
P
(A)
≥
0
ativity: •P(A)
≥
0
Axioms
ization:
Pdisjoint
(Ω)
=1
vity:
PFor
(A)
≥ 0=
ization:
1 sets:
•P(Ω)
Consequences
)ion:
additivity:
(to
be strengthened later)
P(Ω) = (to
1 be
additivity:
strengthened later)
B = Ø,Pthen
P
(A=
∪P
B)
= P(A)
+ P(B)
(A
∪P
B)
(A)
P(B)
For
disjoint
events:
= Ø, then
(A
B)
=+
P(A)
+later)
P(B)
ditivity:
(to
be∪strengthened
Ø, then P(A ∪ B) = P(A) + P(B)
Some
P(A ∪ B) = P
(A) +simple
P(B) consequences of the axioms
• P(A) ≤ 1
Some simple consequences of the axioms
• P(Ø) = 0
• P(A) ≤ 1
• A, B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P(C)
•and
P(Ø)
= 0 for k disjoint events
similarly
P(A)
+ P(Ac) = 1
• •P({s
1 , s2 , . . . , sk }) = P({s1 }) + · · · + P({sk })
• A, B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P(C)
= P(s1) + · · · + P(s )
and similarly for k disjoint events k
• P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk })
Axiom 3 needs strengthening
P(A ∪ B) = P(A) + P(B)
• Axioms
Some simple consequences of the axioms
• Consequences
• P(A) ≤ 1
•ForP(Ø)
= 0sets:
disjoint
Some
simple consequences of the axioms
c =1
P({s1, s2•, . .P
. ,(A)
sk })+=PP(A
({s)1})
+ · · · + P({sk })
P(A ∪ B) = P(A) + P(B)
• A, B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P(C)
= P(s1) + · · · + P(sk )
and similarly for k disjoint events
Some simple consequences of the axioms
• P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk })
• P(A) ≤ 1
• P(Ø) = 0
= P(s1) + · · · + P(sk )
• P(A) + P(Ac) = 1
• A, B, C disjoint: P(A ∪ B ∪ C) = P(A) + P(B) + P(C)
– Axiom 3 needs strengthening
and similarly for k disjoint events
– Do weird sets have probabilities?
• P({s1, s2, . . . , sk }) = P({s1}) + · · · + P({sk })
= P(s1) + · · · + P(sk )
– Axiom 3 needs strengthening
– Do weird sets have probabilities?
Axiom 3 needs strengthening
More consequences of the axioms
More consequences of the axioms
, then P(A) ≤ P(B)
• If A ⊂ B, then P(A) ≤ P(B)
) = P(A) + P(B) − P(A ∩ B)
• P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
) ≤ P(A) + P(B)
• P(A ∪ B) ≤ P(A) + P(B)
∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C)
More consequences of the axioms
• P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C)
• If A ⊂ B, then P(A) ≤ P(B)
• P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
• P(A ∪ B) ≤ P(A) + P(B)
More consequences of the axioms
• P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C)
• If A ⊂ B, then P(A) ≤ P(B)
• P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
• P(A ∪ B) ≤ P(A) + P(B)
• P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C)
More consequences of the axioms
• If A ⊂ B, then P(A) ≤ P(B)
• P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
More consequences of the axioms
• P(A ∪ B) ≤ P(A) + P(B)
, then P(A) ≤ P(B)
• P(A ∪ B ∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C)
) = P(A) + P(B) − P(A ∩ B)
) ≤ P(A) + P(B)
∪ C) = P(A) + P(Ac ∩ B) + P(Ac ∩ B c ∩ C)
Example
4
4
3
Y = Second
roll
3
Y = Second
2
roll
1
2
11
Probability calculation: discrete/finite example
Sample space: discrete/finite example
2
3
2
X =1First roll
4
3
4
X = First roll
Exampledie
• Two rolls of a tetrahedral
• Let
possible
Letevery
Z = min(X,
Y )outcome have probability 1/16
every
– Sample space vs. 4sequential
description
• P•(XLet
= 1)
= possible outcome have probability 1/16
Die roll
example
• P(Z = 1) =
• P(X = 1) 1,1
=
3
Y = Second
1
2
1,4
4
1
Y = Second 3
1,2
• P(Z = 2) =
1,3
roll
Let Z =2 min(X, Y )
• P(Z = 3) =
Z ==min(X,
•LetP(Z
1) = Y )
3 4) =
• P(Z =
•• P
P(Z
(Z =
= 1)
2) =
=
4
Y = Second 31
2
3
4
roll
roll
2
2
X = First roll
1
t every possible outcome have
3
2
1
1
X = First roll
bability 1/16
3
2
1
4
X = 1)•=A continuous sample space:
(x, y) such that X
0 =≤First
x, yroll
≤1
y
4
• Let B be the event: min(X, Y ) = 2
• Let M = max(X, Y )
• P(M = 1 | B) =
1
4
•• P
P(Z
(Z =
= 2)
3) =
=
4,4
•• P
P(Z
(Z =
= 3)
4) =
=
• P(Z = 4) =
Discrete uniform law
Discrete uniform law
Discrete uniform law
− Assume
Ω consists
of n equally
Discrete
uniform
law likely element
Discrete uniform law
− Assume A consists of m elements
number of elements of A
Then−: Assume
P(A) =
e points be equally likely
Ωnumber
consistsofofelements
n equallyoflikel
Ω
− Assume Ω consists
of
Discrete uniform law
− Assume Ω consists of n equally likely elements
− Assume A consists of m elements
− Assume A consists of
Assume
Aofconsists
of of
klikely
elements
• Let all −
sample
points
be
equally
number
elements
A • Just count. . .
number of eleme
P(A) =
num
number
of
elements
of
A
k
total number of sample points
Then =
: P(A) =
Then
:
P
(A)
=
Then
: P(A)
=
• Then,
number
of eleme
number of elements of Ω
n
num
number of elements of A
P(A) =
• Just count. . . total number of sample points
1
• Just count. . .
= count. . .
•prob
Just
n
• Just count. . .
Continuous uniform law
prob =
m” numbers in [0, 1].
y
Continuous uniform law
1
• Two “random” numbers in [0, 1].
: Probability = Area
y
1
1
x
1
n
Probabilitynumbers
calculation:
example
• Two “random”
in [0,continuous
1].
Probability calculation: continuous exam
y
• Two “random” numbers in [0, 1].
1 1].
y in [0,
• Two “random” numbers
mple space: continuous example
mple space:
Sample space: continuous
1
1
example
Probability calculation: continuous example
• A continuous sample space:
Two “random” numbers in [0, 1].
Sample space: continuous example
y 0 ≤ x, y ≤
• space:
(x, y) such that
uous sample
Sample space: continuous example
≤ x, y ≤ 1
y
1y
hat 0 ≤ x, y ≤ 1
1
x
1 x
• Uniform probability law: Probability =x Area
1
1
• Uniform probabilty law: Probability = Area
1
s sample space:
h that 0 ≤ x, y ≤ 1
y
�
�
P {(x, y) |probability
x + y ≤ 1/2}
� law:=Probability = Area
• �Uniform
P {(x, y) | x + y ≤ 1/2} =
y
1
� �
P {0.5,
P� {(x,
y) | 0.3)}
x �+ y
x
1
�
P {0.5, 0.3)} =
1
�
�
Uniform law: Probability
= Area
Sample
space: continuous
example
Sample space:
continuous
example
P {(0.5, 0.3)} =
x
1
• A continuous
sample space:
• A continuous
sample space:
1
x
• (x,
y)0such
that
• (x, y) such
that
≤ x, y
≤ 10 ≤ x, y ≤ 1
y1
1
xy
y1
1
1
x
x
1
1
x
�
≤=1/2} =
Probability calculation steps
Probability calculation steps
he sample space
• Specify the sample space
probability law
• Specify a probability law
n event of interest
• Identify an event of interest
...
• Calculate...
ability calculation: discrete but infinite sample space
Probability calculation: discrete but infinite sample space
pace: {1, 2, . . .}
• Sample
space: {1, 2, . . .}
1
given P(n) = n , n = 1, 2, . . .
1
2
– We are given P(n) = n , n = 1, 2, . . .
2
me is even) =
• P(outcome is even) =
p
p
1/2
1/2
1/4
1/8
1/16
1/4
Probability calculation: discrete
Probability
discrete but
but infinite
infinite sample
sample space
space
Sample space:
space: {1, 2, . . .}
•• Sample
We are
are given P(n) = 2−n, n
–– We
n=
= 1,
1,2,
2,......
Find P
P(outcome
(outcome
Find
is even)
Probability calculation: ••discrete
but infinite
sample space
Probability calculation: discrete but infinite sample space
discrete but pinfinite sample space
• Sample Probability
space: {1, 2, calculation:
. . .}
• Sample space: {1, 2,−n
. . .}
• are
Sample
– We
givenspace:
P(n) ={1,
2 2, ,. .n.}= 1, 2, . . .
1/2
1/2
– We are given P(n) = 2−n
, n = 1, 2, . . .
1
• Find –P(outcome
is
even)
We are given P(n) = n , n = 1, 2, . . .
• Find
P(outcome
is even)2 discrete but infinite sample
Probability
calculation:
space
1/4
1/4
p
• Find P(outcome
is even)
• Sample space: {1,p 2, . . .}
1
p1/2
– We are given P(n)
= 1/2
, n = 1, 2, . . .
2n
• P(outcome is even) =
1
22
33
1/16
1/16
…..
…..
44
1/4 1/2
1/4
Solution:
1/8
•• Solution:
1/16
1/4 1/8
p
1
1/8
1/8
2
1
1/2
3
2
41/8
3
…..
P({2,
4, 6, . . .})
…..
…..
1/16
4
1/16
=
= P
P(2)
(2) +
+P
P(4)
(4)+
+······
11
11
11
11
=
= 22 +
+ 44 +
+ 66 +
+······=
=
22
22
22
33
1
2
3
4
• Solution:
• Solution:
1/4 needed:
•• Axiom
Axiom
needed:
P({2, 4, 6, . . .}) =
P(2)
···
If
,, .P
. .(4)
are+disjoint
events, then:
then:
If A
A
A22+
events,
1/8
• Solution:
1,, A
P({2, 4, 6, . . .}) 11=
P1(2) +1/16
P
(4)
+ ··1
·
1
+ 4 + 16 + ·1· · =
2=1
P({2, 4, 6,=
. . .})
(2)2+
P(4)
P(A
· )) =
=P
P(A
(A11))+
+P
P(A
(A22))+
+· ·· ·· ·
A2·3·∪· · · ·1
1 ∪+
2=
2P+
1
2
321
4 1+ 61+ · · · =
2
2
2
31
=
+
+
+
·
·
·
=
• Axiom needed:
22
24
26
3
•
Axiom
needed:
If •A1Solution:
, A2, . . . are disjoint events, then:
• If Axiom
A1, A2,needed:
. . . are disjoint events, then:
({2,
4,∪6,· ·. ·. ).})
=(AP
P(4)
+· · · ·
∪ Adisjoint
=P
)(2)
++
P(A
If A1, A2P
, .(A
. .P1are
events,
2
1then:
2) +
…..
Countable additivity axiom
Countable additivity axiom
Countable additivity axiom
ens the finite additivity axiom
• Strengthens the finite additivity axiom
• Strengthens the finite additivity axiom
A3,. . . is an infinite sequence of events,
A3),.+
. . Pis(Aan) infinite
sequence of events,
2,(A
∪ A2 ∪ A3 ∪If· ·A
· )1,=AP
1
2 + P(A3 ) + · · ·
Countable
Axiom:
then
P(A1 ∪ AAdditivity
2 ∪ A3 ∪ · · · ) = P(A1) + P(A2) + P(A3) + · · ·
If A1, A2, A3,. . . is an infinite sequence of disjoint events,
then P(A1 ∪ A2 ∪ A3 ∪ · · · ) = P(A1) + P(A2) + P(A3) + · · ·
Countable additivity axiom
•Mathematical
Strengthens subtleties
the finite additivity axiom
Countable Additivity Axiom:
If A1, A2, A3,. . . is an infinite sequence of disjoint events,
then P(A1 ∪ A2 ∪ A3 ∪ · · · ) = P(A1) + P(A2) + P(A3) + · · ·
Mathematical
••
•
••
•
•
•
•
Mathematical
subtleties
Mathematical
subtleties
Mathematical subtleties
• Additivity holds only for “countable”
Additivity
sequences
of events
Additivity holds
holdsonly
onlyfor
for“countable”
“countable”
sequences
of events
Additivity holds only for “countable” sequences of events
• The unit square (simlarly, the real line
(itscountable
elements cannot be arranged in a
The
thethe
real
line,
etc.)
is not
countable
The unit
unit square
square(simlarly,
(similarly,
real
line,
etc.)
is not
The unit square (simlarly, the real line, etc.) are not countable
(its
in in
a sequence)
(its elements
elementscannot
cannotbebearranged
arranged
a sequence)
• “Area” is a legitimate probability law
“Area”
as long as we do not try to assign pro
“Area” is
is a
a legitimate
legitimate probability
probability law
law on
on the
the unit
unit square,
square,
“Area”
is
a
legitimate
probability
law
on
the
unit
square,
as
toto
assign
probabilities/areas
to to
“very
strange”
“very
strange” sets
as long
long as
aswe
wedo
donot
nottry
try
assign
probabilities/areas
as
long
as
we
do
not
try
to
assign
probabilities/areas
sets
to “very strange” sets
to “very strange” sets
Interpretations of probability theor
Interpretations ofInterpretations
probability theory
of
theory
• probability
A narrow view:
a branch of math
Interpretations
theory
Interpretationsofofprobability
probability
theory
–
Axioms
⇒
theorems
ow view: •a branch
of
math
A narrow view: a branch of math
• –A
narrow
view:
aabranch
ms ⇒ theorems
“Thm:” “Frequency” of event A “is” P(A)
AAxioms
narrow⇒
view:
branchofofmath
math
theorems
–
⇒
– Axioms
Axioms
⇒Atheorems
theorems
“Frequency”
of event
“is” P(A)
“Thm:”
“Frequency”
of event A “is”
P(A)
• Are
probabilities frequencies?
“Thm:” “Frequency” of event A “is”
P(A)
“Thm:” “Frequency” of event A “is”
P
(A) toss yields heads) = 1/2
–
P
(coin
obabilities•frequencies?
Are probabilities frequencies?
• Are probabilities frequencies?
n toss yields
heads)
=
1/2yields
•–
probabilities
frequencies?
– Are
P(coin
(coin
toss
heads) =
= 1/2
1/2 – P(sole shooter of JFK) = 0.92
P
toss
yields
heads)
e shooter of
=shooter
0.92
– P(a piece of equipment aboard the space shuttle fa
– JFK)
P(the
(coin
toss yields
= 1/2
(sole
of
= be
0.92
–
P
president
of JFK)
.heads)
. . will
reelected) = 0.7
iece of equipment
aboard
the of
space
shuttle
fails)
= 10−8
−8
(the
president
. . . will
be reelected)
=
0.7
– P(a
piece
of equipment
aboard
the
space
shuttle
fails)
=a10
•
Probability
models
are
framework
for
• Probabilities are often intepreted as:
describing uncertainty
ility models
are
a framework
for
Betting
preferences
Probabilities
are often
•– Probability
models
are aintepreted
frameworkas:
for
ing uncertainty
– Use for consistent reasoning
–
Description
of
uncertainty
–describing
Description
ofbeliefs
beliefs
or consistent reasoning
– Use for predictions, decisions
for consistent
reasoning
•–
framework
for dealing
with experiments that have uncertain outcomes
–AUse
Betting
preferences
or predictions, decisions
Use for
– Rules
forpredictions,
consistent decisions
reasoning
•– AUsed
framework
for analyzing
phenomena with uncertain outcomes
for predictions
and decisions
Rules
consistent
reasoning
– P
(sole for
shooter
of JFK)
= 0.92
– P
Used
for predictions
and
decisions
–
(a piece
of equipment
aboard
the space shuttle fails) = 10−8
– P(sole shooter of JFK) = 0.92
• Probability models are a framework for
–describing
P(a pieceuncertainty
of equipment aboard the space shuttle fails) = 10−8
ory
The role of probability th
• Are probabilities frequencies?
– P(coin toss yields heads) = 1/2
The
role of probability theory
world
– P(the president of . . . will be reelected) Real
= 0.7
The role of proba
The role of probability theory
• Probabilities are often intepreted
Real worldas:
Data
Real world
– Betting preferences
The role of probabili
Real
world
The
role of probability
theory
– Description
of beliefs Data
Inference/Statistics
Data
Real
world
• A frameworkData
for analyzing phenomena with uncertain
outcomes
Inference/Statistics
Analysis
– Rules for consistent reasoning
Inference/Statistics
Data
– Used for predictions
and
decisions
Inference/Statistics
(Analysis)
Predictions
– P(sole shooter of JFK)
= 0.92
Analysis
The role of probability
theory −8
Inference/Statistics
– P(a piece of
equipment
aboard
the
space
shuttle
fails)
= 10
(Analysis)
Predictions
Probability theory
Predictions
The role
theory
• Probability
models are a framework
for of probability
Real world
The
role of
probability theory
(Analysis)
Predictions
Decisions
describing uncertainty
Probability theoryModel building
– Use
for
consistent
Probability theory
Real
world reasoning
Data
Real world
Predictions
Probability
theory
– Use for predictions,
decisions
Models building
Models building
Data
Inference
Data
Probability theory
Models building
Inference/Statistics
Analysis
Inference/Statistics
Analysis
Predictions
Analysis
Predictions
Probability
theoryPredictions
Probability
Probability theory
Model
buildingtheory
Models building