Probability Set Function

Transcription

STAT 516: Basic Probability and its Applications
Lecture 2: Probability Set Function
Prof. Michael Levine
January 20, 2015
Levine
Axioms of Probability
I
For all C ∈ B, P(C ) ≥ 0
I
P(C) = 1
I
If {Cn } is a sequence of events in B and Cm ∩ Cn = ∅ for all
m 6= n,
∞
X
C
)
=
P(Cn )
P (∪∞
n
n=1
n=1
Levine
Countable Additivity Axiom
I
The last axiom above is called the countable additivity
axiom
I
The following is implied by the countable additivity axiom...
I
Let A1 ⊃ A2 ⊃ A3 ⊃ · · · be an infinite family of subsets of a
sample space Ω s.t. An ↓ A
I
Then, P(An ) → P(A) as n → ∞
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Countable Additivity Axiom
I
To prove, note if Bi = Aci , we have B1 ⊂ B2 ⊂ B3 · · · .. and
Bn ↑ B
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It is enough to show that P(Bn ) → P(B) as n → ∞
I
For any fixed n, Bn = ∪ni=1 (Bi − Bi−1 )
I
c
In the above, B0 = ∅ and Bi − Bi−1 = Bi ∩ Bi−1
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The needed result immediately follows...
I
Note that if finite additivity and the above result are assumed
as axioms, then countable additivity follows as a consequence
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Simple and Compound Events
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An event is simple if it consists of exactly one outcome and
compound otherwise.
I
Example: consider recording the freeway exit (L or R) for
each of the 3 vehicles at the end of the exit ramp. The eight
possible outcomes form the sample space
B = {LLL, RLL, LRL, LLR, LRR, RLR, RRL, RRR}.
Examples of simple events would be:
I
I
I
E1 = LLL
E2 = LRL
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Examples of Compound Events
I
Examples of compound events would be:
I
I
Exactly one vehicle turns left A = {LRR, RLR, RRL}
All three vehicles turn in the same direction B = {LLL, RRR}
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Disjoint Events I
I
When A and B have no outcomes in common, they are said
to be mutually exclusive or disjoint events.
I
Example: when rolling a die, the event A = {2, 4, 6} (evens )
and B = {1, 3, 5} (odds) are mutually exclusive
I
Example: when pulling a single card from a standard deck of
cards, events A =heart, diamond(red) and B =spade,
club(black) are mutually exclusive.
Levine
Disjoint Events II
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A collection of events {Cn } is exhaustive if the union of these
events is a sample space so that
∞
X
P(Cn ) = 1
n=1
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A mutually exclusive and exhaustive collection of events forms
a partition of C
I
Example: events A and B from the card example form a
partition of the sample space
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Simplest Properties of Probability
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For each C ∈ B
0
P(C ) = 1 − P(C )
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The probability of the null set is zero:
P(∅) = 0
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If C1 ⊂ C2 we have
P(C1 ) ≤ P(C2 )
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For each event C ∈ B
0 ≤ P(C ) ≤ 1
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Equilikely Case I
I
Let C1 , . . . , Ck be mutually exclusive and exhaustive
subsets.
I
Let each of Ci ,i = 1, . . . , k have the same probability
P(Ci ) = k1
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For any E = C1 ∪ C2 ∪ . . . ∪ Cr with r ≤ k
P(E ) =
r
X
i=1
Levine
P(Ci ) =
r
k
Example I
I
During off-peak hours a commuter train has five cars.
Suppose a commuter is twice as likely to select the middle car
#3 as to select either adjacent car (#2 or #4) , and is twice
as likely to select either adjacent car as to select either end
car (#1 or #5).
I
If the probability of selecting car i is pi , we have
p3 = 2p2 = 2p4 , and p2 = 2p1 = 2p5 = p4
P
Therefore, 1 =
pi = 10p1 and p1 = p5 = 0.1,
p2 = p4 = 0.2 and p3 = 0.4.
I
I
The probability that one of the three middle cars is selected is,
then, p2 + p3 + p4 = 0.8.
Levine
Counting rules
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A multiplication rule or the mn-rule
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The number of permutations of k elements out of n is
Pkn = n(n − 1) · · · (n − (k − 1)) =
n!
(n − k)!
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The number of combinations of k elements out of n is
n
n!
=
k
k!(n − k)!
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The binomial expansion
n X
n k n−k
(a + b) =
a b
k
n
k=0
Levine
Probability set function for a finite set and loaded dice
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Let C consist of k distinct points (outcomes)
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Assign the probability of
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For any event C ∈ C define
P(C ) =
1
k
to each of these outcomes
number of points in C X
=
f (x)
k
x∈C
with f (x) =
1
k
for x ∈ C
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With loaded dice, with C = {1, 2, 3, 4, 5, 6} we may have
x
f (x) = 21
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Then, for C = {1, 2, 3} we have
P(C ) =
1
2
3
2
+
+
=
21 21 21
7
Levine
Counting rules
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The number of ways of linearly arranging n distinct objects
with order of arrangement matters is n!
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The number of ways of choosing r objects from n distinct
objects if the same object could be chosen repeatedly is nr
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The number of ways of distributing n distinct objects into k
distinct categories when order in which the distributions are
made is not important, ni being the number of objects
allocated to the ith category
is
n n−n1
n−n1 −n2 −···−nk−1
n!
···
n1 !n2 !···nk ! = n1
n2
nk
Levine
Example
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A carton of eggs has 12 eggs of which three are bad. We
don’t know that fact in advance
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We want to make a three-egg omelet; we can select three
eggs in 12
3 = 220 ways
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Assume that three eggs are selected completely at random;
the
number of ways to select them out of 9 good eggs is 93 = 84
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The probability of the favorable outcome is
Levine
84
220
= 0.38
Example
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Six different cookies are distributed completely at random to
six children. It is possible for a child to get more than one
cookie
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This implies that there are 66 = 46, 656 possible outcomes
(sample points).
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Let A = {Each child receives exactly one cookie}. There are
720
= 0.015
6! = 720 ways of doing so and thus P(A) = 46656
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Then, P(A ) = 1 − 0.015 = 0.985
0
Levine
Example
I
I
I
Out of 5 pairs of shoes in the closet, we select 4 shoes at
random. What is the probability that we will have at least one
complete pair among them?
The total number of sample points is 10
4 = 210
We can have either two complete pairs, selected in 52 = 10
ways or...
I
Or one
pair and two nonconforming shoes selected
complete
in 51 42 × 2 × 2 = 120 ways
I
The probability of the event of interest is, then,
10 + 120
= .62
210
Levine
Example-a five card poker
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A player is given 5 cards from a full deck of 52 cards at
random
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Events of interest are A = { two pairs } and B = a flush
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Two pairs is a hand with two cards each of two different
denominations and the fifth card of some other denomination
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Thus,
h 4i2
P(A) =
13
2
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2
52
5
44
1
= 0.04754
Example - a 5 card poker
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A flush is a hand with 5 cards of the same suit that are not in
a sequence
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There are 10 ways to select five cards from a suit such that
the cards are in a sequence
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Thus, The probability of flushes is
13
4
−
10
1
5
= .00197
P(B) =
52
5
Levine
Inclusion-Exclusion Formula
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P(C1 ∪ C2 ) = P(C1 ) + P(C2 ) − P(C1 ∩ C2 )
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The general inclusion-exclusion formula
P(C1 ∪ C2 ∪ . . . ∪ Ck ) =
n
X
i=1
+
X
P(Ci ) −
X
P(Ci ∩ Cj )
1≤i<j≤n
P(Ci ∩ Cj ∩ Ck ) − · · · + (−1)k+1 P(C1 ∩ C2 ∩ · · · ∩ Ck )
1≤i<j<k≤n
Levine
Example II
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In a residential suburb, 60% of all households get Internet
service from a local cable company, 80% get television service
from that company, and 50% get both services from that
company
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The probability that a randomly selected household subscribes
to at least one of these two services from the local company is
P(C1 ∪ C2 ) = P(C1 ) + P(C2 ) − P(C1 ∩ C2 ) = 0.9
Levine
Example II
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The probability that a household subscribes only to TV service
is
0
P(C1 ∩ C2 ) = P(C1 ∪ C2 ) − P(C1 ) = 0.9 − 0.6 = 0.3,
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The probability that a household subscribes only to Internet is
0
P(C1 ∩ C2 ) = P(C1 ∪ C2 ) − P(C2 ) = 0.1,
I
The probability that a household subscribes to exactly one of
these services from the local company is
0
0
P(C1 ∩ C2 ) + P(C1 ∩ C2 ) = 0.1 + 0.3 = 0.4
Levine
Missing faces in dice rolls
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A fair die is rolled n times. What is the probability that at
least one of six sides never shows up in these n rolls?
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Let Ai be the side that never shows up, 1 ≤ i ≤ 6
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If all of 6n outcomes are equally likely, P(Ai ) =
n
n
P(Ai ∩ Aj ) = 46n , P(Ai ∩ Aj ∩ Ak ) = 36n etc
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Thus, the needed probability is
n n
6
5
6
3
pn =
−
+ ···
1
6n
2
6n
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Manual calculation gives p10 = 0.73, p12 = 0.56, p13 = 0.49,
p15 = 0.36
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Note that it takes 13 rolls of a fair die to have better than
50% chance (1 − pn ) that each of the six sides shows up
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5n
6n ,
Bonferroni bounds and inequality
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Pn
P
Denote
PS1 = i=1 P(Ci ), S2 = 1≤i<j≤n P(Ci ∩ Cj ),
S3 = 1≤i<j<k≤n P(Ci ∩ Cj ∩ Ck ) etc
I
Then, the inclusion-exclusion formula becomes
P(∪ni=1 Ci ) = S1 − S2 + S3 − · · ·
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Let pn = P(∪ni=1 Ci ). The following series of bounds is true:
pn ≤ S1 ; pn ≥ S1 − S2 ; pn ≤ S1 − S2 + S3 ; · · ·
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Also, (Bonferroni inequality)
P(∩ni=1 Ci )
≥1−
n
X
0
P(Ci )
i=1
Levine
Example
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Let P(Ci ) ≥ 0.95 for i = 1, . . . , 10. This means that
0
P(Ai ) ≤ 0.05 for each i
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By Bonferroni inequality, P(∩ni=1 Ai ) ≥ 1 − 10 ∗ 0.05 = 0.5
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Therefore, despite individual probabilities of at least 95%, we
can only be 50% sure that all 10 of events will occur
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This is a rather crude lower bound - better options are
available
Levine
Continuity theorem for probabilities
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A sequence of events Cn is non-decreasing if Cn ⊂ Cn+1 for
all n
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Call limn→∞ Cn = ∪∞
n=1 Cn
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Theorem (this is called the continuity from below):
lim P(Cn ) = P( lim Cn ) = P (∪∞
n=1 Cn )
n→∞
n→∞
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For a sequence An s.t. An ⊃ An+1 (a non-increasing set
sequence) the limit is limn→∞ An = ∩∞
i=1 Ai
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The continuity from above:
lim P(An ) = P( lim An ) = P (∩∞
n=1 An )
n→∞
n→∞
Levine
Bool’s inequality
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Let Cn be an arbitrary sequence of events.
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P (∪∞
n=1 Cn ) ≤
∞
X
P(Cn )
n=1
Levine

Probability Set Function

Transcription

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