Probability basics
Probability theory offers a mathematical language for describing
uncertainties:
p(disease d) = 0.35
Probability theory and
assessment
0
0.5
1
In a medical problem, a probability p(disease d) = 0.35 for a
patient is interpreted as:
• in a population of 1000 similar patients, 350 have the
disease d;
• the attending physician assesses the likelihood that the
patient has the disease d at 0.35.
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A frequentist’s probabilities
• in theory, a probability Pr(d) is the relative frequency with
which d occurs in an infinitely repeated experiment;
• in practice, the probability is estimated from (sufficient)
experimental data.
Sources of information
Sources of probabilistic information are:
Example
In a clinical study, 10000 men over 40 years of age have been
examined for hypertension:
• (experimental) data;
• literature;
• human experts.
hypertension no hypertension
1983
8017
The first two sources are the most reliable.
The probability of hypertension in a man aged 45 is estimated to
be
1983
p(hypertension) =
= 0.20
10000
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Probability scales — an example
For their new soda, an expert from Colaco has to assess the
probability that the soda will turn out a national success:
Subjective probabilities
For assessing probabilities with experts, various tools are
available:
certain
(almost)
• probability scales;
• the expert is asked, using
mathematical notation, to assess:
• probability wheels;
• betting models;
p(national success) = ?
• lottery models.
A subjective probability is based upon personal knowledge and
experience.
• the expert is asked to indicate the
probability on a scale from 0 to
100% certainty:
probable
100
85
75
expected
fifty-fifty
50
uncertain
25
improbable
(almost)
impossible
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15
0
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Betting models — an example
For their new soda, an expert from Colaco is asked to assess
the probability of a national success:
• the expert is offered two bets:
Probability wheels
A probability wheel is composed of two coloured faces and a
hand:
national success
x euro
d
national failure
national success
d¯
national failure
The expert is asked to adjust the area of the red face so that the
probability of the hand stopping there, equals the probability of
interest.
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−y euro
−x euro
y euro
¯ then
• if the expert is indifferent between d and d,
x · Pr(n) − y · (1 − Pr(n)) = y · (1 − Pr(n)) − x · Pr(n)
y
from which we find Pr(n) =
.
x+y
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Lottery models — an example
For their new soda, an expert from Colaco is asked to assess
the probability of a national success:
• the expert is offered two lotteries:
national success
Hawaiian trip
• subjective probabilities are coherent if they adhere to the
d
national failure
p(outcome)
d¯
p(not outcome)
Coherence and calibration
postulates of probability theory;
• subjective probabilities are well calibrated if they reflect true
frequencies.
chocolate bar
Hawaiian trip
chocolate bar
¯ then
• if the expert is indifferent between d and d,
Pr(n) = p(outcome).
• the second lottery is termed the reference lottery;
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Overconfidence and underconfidence
Heuristics
Upon assessing probabilities, people tend to use simple
cognitive heuristics:
• representativeness: the probability of an outcome is based
• a human expert is an overconfident assessor if, compared
with the true frequencies, his subjective probabilities show a
tendency towards the extremes;
• a human expert is an underconfident assessor if, compared
with the true frequencies, his subjective probabilities show a
tendency away from the extremes.
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upon the similarity with a stereotype outcome;
• availability: the probability of an outcome is based upon the
ease with which similar outcomes are recalled;
• anchoring-and-adjusting: the probability of an outcome is
assessed by adjusting an initially chosen anchor probability:
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Pitfalls — cntd.
Pitfalls
Using the representativeness heuristic upon assessing
probabilities, can introduce biases:
• the prior probabilities, or base rates, are insufficiently taken
Using the availability heuristic upon assessing probabilities, can
introduce biases:
• the ease of recall from memory is influenced by recency,
rareness, and the past consequences for the assessor;
• the ease of recall is further influenced by external stimuli:
Example
into account;
• the assessments are based upon insufficient samples;
• the weights of the characteristics of the stereotype outcome
are insufficiently taken into consideration;
• ...
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• ...
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Pitfalls — cntd.
Continuous chance variables
Using the anchoring-and-adjusting heuristic upon assessing
probabilities, can introduce biases:
• the assessor does not choose an appropriate anchor;
• the assessor does not adjust the anchor to a sufficient
extent:
Example
In decision trees, chance variables are discrete:
• a discrete variable C takes a single value from a non-empty
finite set of values {c1 , . . . , cn }, n ≥ 2;
• the distribution associated with C is a probability mass
function, which assigns a probability to each value ci of C.
In reality, chance variables can also be continuous:
• a continuous variable C takes a single value within a
non-empty range of values [a, b], a < b;
• the distribution associated with C is a probability distribution
function, which defines a probability for any interval
[x, y] ⊆ [a, b].
• ...
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Pivoting on values
For a real-estate agency,
the demand for luxury
apartments is a
continuous chance
variable A:
The following procedure provides for modelling a continuous
variable C in a decision tree:
1
construct a cumulative distribution function for C:
• pivoting on the values of C, or
• pivoting on the cumulative probabilities for C;
2
approximate the probability distribution function for C by a
probability mass function for a discrete chance variable C ′ :
• using the Pearson-Tukey method, or
• using bracket medians.
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
50
60
70
80
90
100
no. of appartments
110
120
130
A cumulative distribution function is constructed by
• assessing the cumulative probabilities for a number of
values of A:
Pr(A ≤ 65) = 0.08 Pr(A ≤ 90) = 0.90
Pr(A ≤ 80) = 0.65 Pr(A ≤ 110) = 0.97
• and drawing a curve
through them:
1
cum. distribution
Using continuous chance variables
0.8
0.6
0.4
0.2
0
50
60
70
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80
90
100
no. of appartments
110
120
130
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Pivoting on cumulative probabilities
Pivoting on values — cntd.
0.35
For assessing cumulative probabilities, by pivoting on the values
of a variable under study, lottery models can be used.
Example
Reconsider the demand
for luxury apartments,
modelled by a continuous
chance variable A:
0.3
0.25
0.2
0.15
0.1
0.05
0
50
d
A > 90 apartments
p(outcome)
d¯
p(not outcome)
70
80
90
100
no. of appartments
110
120
130
A cumulative distribution function is constructed by
• finding the values for A that give a number of cumulative
probabilities:
Hawaiian trip
chocolate bar
Pr(A ≤ a1 ) = 0.05 Pr(A ≤ a3 ) = 0.50
Pr(A ≤ a2 ) = 0.35 Pr(A ≤ a4 ) = 0.95
Hawaiian trip
• and drawing a curve
chocolate bar
through them:
Other elicitation tools can be used as well.
1
cum. distribution
A ≤ 90 apartments
60
0.8
0.6
0.4
0.2
0
50
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60
70
80
90
100
no. of appartments
110
120
130
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The Pearson-Tukey method
Pivoting on cumulative probabilities — cntd.
The Pearson-Tukey method approximates the distribution
function for a continuous variable C by a probability mass
function over a discrete variable C ′ :
For assessing cumulative probabilities, by pivoting on these
cumulative probabilities, lottery models can be used.
Example
• find the values c1 , c2 , c3 for which
A ≤ x apartments
Pr(C ≤ c1 ) = 0.05 Pr(C ≤ c3 ) = 0.95
Pr(C ≤ c2 ) = 0.50
Hawaiian trip
d
A > x apartments
p(outcome) = 0.35
d¯
chocolate bar
• construct the discrete chance variable C ′ with values
Hawaiian trip
c1 , c2 , c3 and probabilities
p(not outcome) = 0.65
chocolate bar
Pr(C ′ = c1 ) = 0.185 Pr(C ′ = c3 ) = 0.185
Pr(C ′ = c2 ) = 0.63
Other elicitation tools can be used as well.
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Bracket medians (Clemen)
The Pearson-Tukey method — an example
For the real-estate agency, the demand for luxury apartments is
a continuous chance variable A, for which probabilities need to
be assessed:
• for the cumulative probabilities
Pr(A ≤ a1 ) = 0.05 Pr(A ≤ a3 ) = 0.95
Pr(A ≤ a2 ) = 0.50
we have found
a1 = 62
a2 = 77
The method of bracket medians approximates the distribution
function for a continuous variable C by a probability mass
function over a discrete variable C ′ :
1 for a number of equally likely intervals, for example five, find
the values of C for which
Pr(C ≤ c1 ) = 0
Pr(C ≤ c4 ) = 0.60
Pr(C ≤ c2 ) = 0.20 Pr(C ≤ c5 ) = 0.80
Pr(C ≤ c3 ) = 0.40 Pr(C ≤ c6 ) = 1.00
2
a3 = 99
for each interval [ci , ci+1 ], i = 1, . . . , 5, establish the bracket
median mi such that
Pr(ci ≤ C ≤ mi ) = Pr(mi ≤ C ≤ ci+1 )
• the variable A′ is constructed with
3
Pr(A′ = 62) = 0.185 Pr(A′ = 99) = 0.185
Pr(A′ = 77) = 0.63
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construct the discrete chance variable C ′ with
Pr(C ′ = m1 ) = Pr(C ′ = m2 ) = Pr(C ′ = m3 ) =
Pr(C ′ = m4 ) = Pr(C ′ = m5 ) = 0.20
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Brackets medians — an example
Bracket medians – more general and practical
With the bracket medians method, using n equally likely
intervals, steps 1 and 2 basically provide for finding values mi ,
i = 1, . . . , n, of the chance variable C, such that
Pr(C ≤ mi ) =
1
+ k · n,
2n
k = 0, . . . , n − 1
1
,
n
Pr(A ≤ m1 ) = 0.10 Pr(A ≤ m4 ) = 0.70
Pr(A ≤ m2 ) = 0.30 Pr(A ≤ m5 ) = 0.90
Pr(A ≤ m3 ) = 0.50
we have found
m1 = 65 m3 = 77 m5 = 89
m2 = 73 m4 = 81
to construct, in step 3, the discrete chance variable C ′ with
Pr(C ′ = mi ) =
For the real-estate agency, the demand for luxury apartments is
a continuous chance variable A, for which probabilities need to
be assessed:
• for the five cumulative probabilities
• the variable A′ is constructed with
i = 1, . . . , n
Pr(A′ = 65) = 0.20 Pr(A′ = 81) = 0.20
Pr(A′ = 73) = 0.20 Pr(A′ = 89) = 0.20
Pr(A′ = 77) = 0.20
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M.C. Airport: probability assessments Ia
The real-estate agency revisited
For each of the identified chance variables, the outcome will
depend on the chosen strategy, that is, on activity and site in
1975, 1985 and 1995.
Reconsider the decision problem for the real-estate agency:
• with the Pearson-Tukey
method, the agency’s
decision tree includes:
the expected demand is 78
apartments;
Pr(A′ = 62) = 0.185
It is assumed that the chance variables are all probabilistically
independent of each other, therefore
Pr(A′ = 77) = 0.63
Pr(A′ = 99) = 0.185
Pr(C) = Pr(C1 ∧ C2 ∧ C3 ∧ C4 ∧ C5 ∧ C6 ) =
Pr(A′ = 65) = 0.20
• using bracket medians, the
agency’s tree includes:
the expected demand is 77
apartments;
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Pr(A′ = 73) = 0.20
6
Y
Pr(Ci )
i=1
Probabilities are to be assessed for
• the outcome of each Ci , i = 1, . . . , 6 for each of the ∼ 100
decision alternatives of D
(where D captures the sequence D75 , D85 , D95 )
Pr(A′ = 77) = 0.20
Pr(A′ = 81) = 0.20
Pr(A′ = 89) = 0.20
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M.C. Airport: probability assessments Ib
M.C. Airport: probability assessments IIa
For each of the identified chance variables, the outcome will
depend on the chosen strategy, that is, on activity and site in
1975, 1985 and 1995.
Alternatively, probabilities can be assessed for
• the outcome of each Ci , i = 1, . . . , 6, for the 16 decision
alternatives of each D j , j = 75, 85, 95.
This second option
• requires less assessments;
• requires easier assessments (Cij vs Ci );
• assumes that Cik is independent of Cij given Dk , j = 75, 85,
k = 85, 95;
• requires that for some functions fi , i = 1, . . . , 6,
Pr(Ci ) = fi (Pr(Ci75 ), Pr(Ci85 ), Pr(Ci95 )).
The required probabilities were established from
• information from previous studies;
• government administrators (group consensus).
For each of the Cij (i = 1, . . . , 6, j = 75, 85, 95) cumulative
distributions were assessed using
• the fractile method, and
• consistency checks
Distributions for Ci were derived from Cij by defining
Ci ≡
Ci75 + Ci85 + Ci95
3
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M.C. Airport: probability assessments IIb
Example
Consider the 1975 noise impact of the ’all activity at Texcoco’
alternative. To establish Pr(C675 | D 75 = T-IDMG), the following
numbers are assessed:
min #people = ?
max #people = ?
Pr(#people ≤ a1 ) = 0.5 ⇒ a1 = ?
Pr(#people ≤ a2 ) = 0.25 ⇒ a2 = ?
Pr(#people ≤ a3 ) = 0.75 ⇒ a3 = ?
etc.
400.000
800.000
640.000
540.000
700.000
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Introduction to utility theory
and assessment
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The appraisal of consequences
For various decision problems, the fundamental objectives have
a natural numerical scale:
• money;
• percentages;
• length of life;
• ...
For other decision problems, not all objectives have such a
scale:
• reputation;
• attractiveness;
• quality of life;
• ...
For such objectives, a proxy scale may be used, or a new
numerical scale need be designed.
The gangrene problem
A 68-year old man is suffering from diabetic gangrene at an
injured foot. The attending physician has to choose between two
decision alternatives:
• to amputate the leg below the knee,
which involves a small risk of death;
• to wait:
• if untreated, the gangrene may cure;
• if the gangrene expands, an amputation above the
knee becomes necessary, which involves a larger risk
of death.
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The gangrene problem — continued
The gangrene problem — continued
The elements of the gangrene problem are organised in the
following decision tree:
survive
amputation
below knee
amputated
below knee
p = 0.99
0.89
amputation
die
death
p = 0.01
below knee
survive
0.9
p = 0.99
die
0
p = 0.01
recover
wait
Reconsider the gangrene problem and compare the following
appraisals of the consequences
cured
p = 0.70
survive
worsen
amputation
p = 0.30
above knee
p = 0.90
die
p = 0.10
0.92
amputated
above knee
wait
death
recover
1.0
p = 0.70
survive
worsen
amputation
p = 0.30
above knee
0.72
Before the decision tree can be evaluated, the various
consequences need be assigned numerical appraisals.
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p = 0.90
die
p = 0.10
0.8
0
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Lotteries
D EFINITION
A lottery is simply a probability distribution over a known, finite
set of outcomes, that is, it is a pair L = (R, Pr) with
• R = {r1 , . . . , rn }, n ≥ 2, is a set of rewards;
• Pr is a probability distribution over R, with
P
i=1,...,n Pr(ri ) = 1.
A lottery is graphically
represented as:
A lottery L is commonly denoted
L = [ p 1 , r1 ; . . . ; p i , ri ; . . . ; p n , rn ]
p1
Types of lottery
There exist different types of lottery:
• a lottery is termed a certain lottery if it has a single reward r
with Pr(r) = 1;
• a lottery is called a simple lottery if all its rewards are
certain lotteries;
• a lottery is coined a compound lottery if at least one of its
rewards is not a certain lottery.
r1
where pi = Pr(ri ), i = 1, . . . , n;
pi
···
occasionally, we write [R] for short.
pn
···
ri
rn
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The gangrene problem revisited
A preference ordering
Reconsider the gangrene problem. In the problem, several types
of lottery occur:
• the certain lottery [1.0, amputated above knee];
• the simple lottery [0.99, amputated below knee; 0.01, death]:
p = 0.99
amputated below knee
p = 0.01
• for all Li , Lj , Lk ∈ L, if Li Lj and Lj Lk , then Li Lk ;
0.7, cured; 0.3, [0.9, amputated above knee; 0.1, death] :
p = 0.70
If Li Lj , we say that Li is preferred over Lj .
cured
p = 0.90
If Li Lj and Lj Li , we say that Li and Lj are equivalent; we
then write Li ∼ Lj .
amputated above knee
p = 0.30
p = 0.10
Let L be a set of lotteries. A binary relation is termed a
preference ordering on L if adheres to the following
properties:
• for all Li , Lj ∈ L, we have that Li Lj or Lj Li ;
death
• the
compound lottery
D EFINITION
death
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The gangrene problem revisited
The gangrene problem — continued
Reconsider the gangrene problem:
• a preference ordering on the set of all lotteries of the
problem specifies a total ordering on the four certain
lotteries:
L1 = amputated below knee
L2 = death
L3 = cured
L4 = amputated above knee
it seems evident that L3 L1 L4 L2 ;
• the ordering also specifies a total ordering on the more
complex lotteries:
p = 0.99
p = 0.90
amputated
below knee
L5
Reconsider the gangrene problem, with L3 L1 L4 L2 :
L1
L2
L3
L4
death
p = 0.99
amputated
above knee
p = 0.90
amputated
above knee
p = 0.30
L7
p = 0.10
death
death
Does L7 L8 hold, or is L8 L7 ?
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The continuity axiom (aka Archimedean axiom)
for all Li , Lj , Lk ∈ L with Li Lj Lk , there is
probability p such that [p, Li ; (1 − p), Lk ] ∼ Lj
Consider three (certain) lotteries Li Lj Lk with rewards
ri , rj , rk . The axiom states that there exists a probability p such
that
p
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The independence axiom (or: substitutability)
Let L be a set of lotteries and let be a preference ordering on
L. Then, the continuity axiom asserts:
rj
L8
amputated
below knee
cured
death
it seems evident that L5 L6 .
1.0
p = 0.70
p = 0.01
p = 0.10
amputatied below knee
death
cured
amputated above knee
For
L6
p = 0.01
=
=
=
=
ri
Let L be a set of lotteries and let be a preference ordering on
L. Then, the independence axiom asserts:
for all Li , Lj , Lk ∈ L with Li ∼ Lj and for each
probability p, we have that
[p, Li ; (1 − p), Lk ] ∼ [p, Lj ; (1 − p), Lk ]
Consider three lotteries Li , Lj , Lk . The independence axiom
states that if Li and Lj are equivalent, then so are
p
∼
1−p
Li
p
Lj
rk
1−p
• p is termed the calibration probability for Lj and
[p, Li ; (1 − p), Lk ];
• Lj is termed the certainty equivalent for [p, Li ; (1 − p), Lk ].
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Lk
1−p
Lk
for all probabilities p.
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The compound lottery axiom
The unequal probability axiom (or: monotonicity)
Let L be a set of lotteries and let be a preference ordering on
L. Then, the unequal probability axiom asserts:
for all Li , Lj ∈ L with Li Lj and for all probabilities
p, p′ with p ≥ p′ , we have
[p, Li ; (1 − p), Lj ] [p′ , Li ; (1 − p), Lj ]
Let L be a set of lotteries and let be a preference ordering on
L. Then, the compound lottery axiom asserts:
for all Li , Lj ∈ L, Lj = [q, Lm ; (1 − q),Ln ], Lm , Ln ∈ L,
0 ≤ q ≤ 1, and for each probability p, we have
[p, Li ; (1−p), Lj ] ∼ [p, Li ; (1−p)·q, Lm ; (1−p)·(1−q), Ln ].
Consider two lotteries Li , Lk with Lk = [q, Li ; (1 − q), Lj ], Lj ∈ L,
0 ≤ q ≤ 1. The compound lottery axiom states that
Consider two lotteries Li , Lj with Li Lj . The unequal
probability axiom states that
p
p
p′
Li
1−p
Lj
Li
q
1−p
1 − p′
for all probabilities p ≥ p′ .
1−q
Lj
p + (1 − p) · q
Li
∼
Li
(1 − p) · (1 − q)
Li
Lj
Lj
for all probabilities p. The axiom is also termed “no fun in
gambling”.
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The assumption of finiteness
A rational preference ordering
Consider the following decision problem:
D EFINITION
lose all your money
Let L be a set of lotteries. A preference ordering on L is a
rational preference ordering if it adheres to the Von Neumann –
Morgenstern utility axioms:
• the continuity axiom;
• the independence axiom;
• the unequal probability axiom;
• the compound lottery axiom.
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win billions
gamble
p = 0.99999
everything bad
p = 0.00001
bankruptcy
riches
infinite misery
Bayes criterion for choosing between decision alternatives does
not help much if the problem involves consequences of infinite
appraisal.
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The main theorem — a sketchy proof
Consider two lotteries L and L′ with the rewards r1 · · · rn ,
n ≥ 1:
The main theorem of utility theory
L′ = [p′1 , r1 ; · · · ; p′n , rn ]
L = [p1 , r1 ; · · · ; pn , rn ]
T HEOREM
Let L be a set of lotteries and let be a rational preference
ordering on L. Then, there exists a real function u on L such that
• for all Li , Lj ∈ L, we have that Li Lj iff u(Li ) ≥ u(Lj );
X
i=1,...,n
[ui , r1 ; (1 − ui ), rn ] ∼ ri
For the two lotteries, we then have that
#
"
X
X
pi · ui ), rn
pi · ui , r1 ; (1 −
L ∼
• for each [p1 , L1 ; . . . ; pn , Ln ] ∈ L, we have that
u([p1 , L1 ; . . . ; pn , Ln ]) =
For each reward ri we can find a value 0 ≤ ui ≤ 1, such that
pi · u(Li ).
i=1,...,n
i=1,...,n
L
The function u is termed a utility function on L.
′
∼
"
X
i=1,...,n
p′i
· ui , r1 ; (1 −
So, L L′ if and only if
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X
i=1,...,n
X
· ui ), rn
X
p′i · ui .
i=1,...,n
p i · ui ≥
#
p′i
i=1,...,n
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Utility versus expected reward
Example
Consider the set of euro rewards R = {0, 10, 20, 50}, and the
following two lotteries over R:
Some notes
The main theorem of utility theory implies that:
L1 = [0.0, 0; 1.0, 20],
• a lottery with highest utility is a most preferred lottery;
and
L2 = [0.7, 10; 0.3, 50]
Although IEr(L2 ) > IEr(L1 ), the decisionmaker may express the
preference L1 ≻ L2 without being irrational! Consider two
possible utility functions over the rewards, expressing that more
money is preferred to less:
• any rational preference ordering on a set of lotteries is
encoded uniquely by the utilities of its certain lotteries.
u1 (0) = 0, u1 (10) = 0.2, u1 (20) = 0.4, u1 (50) = 1
The main theorem does not imply that:
• a lottery with highest expected reward is a most preferred
lottery
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u2 (0) = 0, u2 (10) = 0.1, u2 (20) = 0.6, u1 (50) = 1
Note that u1 (L1 ) < u1 (L2 ), but u2 (L1 ) > u2 (L2 ).
Note: even if objectives have a natural numerical scale,
preferences (over lotteries) may be such that a utility function is
required!
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Strategic equivalence
D EFINITION
Let L be a set of lotteries. Two utility functions ui , uj on L are
strategically equivalent, written ui ∼ uj , if they imply the same
preference ordering on L.
Example: Consider the following two utility functions for the
gangrene problem:
u1 (amputated below knee) = 0.9
u1 (death) = 0
u1 (cured) = 1.0
u1 (amputated above knee) = 0.8
and
u2 (amputated below knee) = 2.98
u2 (death) = 1
u2 (cured) = 3.2
u2 (amputated above knee) = 2.76
A linear transformation of utilities
T HEOREM
Let L be a set of lotteries and let ui , uj be two utility functions on
L, then
ui ∼ uj
⇐⇒
uj = a · ui + b,
for some constants a, b with a > 0
A utility function is unique up to a positive linear transformation.
The functions u1 and u2 are strategically equivalent.
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The gangrene problem revisited
Normalisation
Let X be an attribute with values x1 . . . xn , n > 1.
Often, utility functions are normalised such that, for example,
Reconsider the gangrene problem with
u(amputated below knee) = 0.9
u(death) = 0
u(cured) = 1.0
u(amputated above knee) = 0.8
For the two lotteries
u(x1 ) = 0 and u(xn ) = 1
p = 0.99
L5
These values only set the origin of u(X) and the unit of
measurement.
amputated
above knee
L6
p = 0.01
If we decide to consider, for example, xi ≺ x1 or xj ≻ xn , then
u(xi ) < 0, and u(xj ) > 1, respectively.
p = 0.90
amputated
below knee
death
p = 0.10
death
we have that
u(L5 ) = 0.99 · u(amputated below knee)+
+0.01 · u(death) = 0.99 · 0.9 = 0.89
u(L6 ) = 0.90 · u(amputated above knee)+
+0.10 · u(death) = 0.90 · 0.8 = 0.72
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So, L5 L6 .
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The gangrene problem — continued
Reconsider the gangrene problem with
u(amputated below knee) = 0.9
u(death) = 0
u(cured) = 1.0
u(amputated above knie) = 0.8
For the two lotteries
p = 0.99
• subjective assessment
• direct methods
• magnitude estimation/production
p = 0.70
amputated
below knee
L7
p = 0.01
Utility assessment
L8
cured
p = 0.90
• ratio estimation/production
amputated
above knee
p = 0.30
death
p = 0.10
• indirect (behavioural) methods
• based on reference gambles
• ”objective” assessment
death
we have that
u(L7 ) = 0.99 · 0.9 = 0.89
• choose a mathematical function
u(L8 ) = 0.70 · 1.0 + 0.3 · 0.9 · 0.8 = 0.92
So, L8 L7 .
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Direct methods
Direct methods
Ratio estimation or production:
Magnitude estimation or production can be done using
a utility scale:
1.0
0.5
Example
Reconsider the diabetic gangrene treatment
example. To assess the utilities for the different
treatment consequences, a patient is asked one
of the following types of question:
• ”How do you value life after an above-knee
amputation?” (estimation);
• ”Which consequence do you associate with a
utility of 0.2?” (production)
0
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Example
My sister is interested in buying a new car. To assess the
utilities for the different car options, she is asked one of the
following types of question:
• ”How much more do you value a Volvo than a Fiat?”
(estimation);
• ”Which car seems to you twice as valuable as a Fiat?”
(production)
This method was used to assess the empirical utility of money
(Galanter, 1962).
√
empirical utility function for monetary gain: ∼ x
empirical utility function for monetary loss: ∼ −x2
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Assessment using certainty equivalents
Reference gamble
Consider a set of consequences for which utilities are to be
assessed. Let ci , cj , ck be consequences from that set such that
ci cj ck .
A reference gamble is a choice between two lotteries:
1 the certain lottery L = [1.0, cj ]
′
2 the simple lottery L = [p, ci ; (1 − p), ck ]
The utilities of a decision maker for the possible consequences
in a decision problem can be assessed using several certainty
equivalents:
1
2
elicit a preference order on consequences from most
preferred (c+ ) to least preferred (c− );
assign the first two points of the utility function:
u(c+ ) = 1 and u(c− ) = 0
• L′ is the reference lottery of the ”gamble” — if u(ci ) = 1.0
3
• for fixed ci , cj , and ck , the probability p for which L ∼ L′ is
5
and u(ck ) = 0, then L is called a standard reference lottery;
′
the indifference probability for L and L′ ;
• for fixed p and ci , ck , the consequence cj for which L ∼ L′ is
the certainty equivalent, or indifference point, for L′ .
4
6
7
create a standard reference gamble with p = 0.5;
elicit the certainty equivalent cCE for the reference lottery;
compute the utility of this indifference point;
create two reference gambles with p = 0.5 and
consequences c+ and cCE , and cCE and c− , respectively;
repeat steps 4 – 6 until enough points are found to draw the
utility curve.
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Computing a utility
An example
Consider the lotteries L and L′ of a reference gamble:
′
L = [1.0, cj ] and L = [p, ci ; (1 − p), ck ]
where ci cj ck . Assume that the utilities u(ci ) and u(ck ) for
consequences ci and ck are known.
Question:
How do we compute u(cj ) for consequence cj ?
Suppose you have an old computer for which components are
bound to need replacement in the near future. You have a
number of replacement alternatives that will cost you between
e 50 and e 500. What is your utility function?
Two points can be fixed:
u(e 500) = 0 and u(e 50) = 1
Answer:
If either
You are then presented with the
following gamble:
p is the indifference probability for L and L′ ,
or
cCE
0.5
0.5
cj is the certainty equivalent of L′
500
Suppose that cCE = 200 is the indifference point for this gamble.
We then have that
then L ∼ L′ and consequently
1.0 · u(cj ) = p · u(ci ) + (1 − p) · u(ck ).
From this equation we can solve u(cj ).
50
1.0 · u(200) = 0.5 · u(50) + 0.5 · u(500) = 0.5
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An example
Assessment using probability equivalents
The utilities of a decision maker for the possible consequences
in a decision problem can be assessed using several probability
equivalents:
1
2
elicit a preference order on consequences from most
preferred (c+ ) to least preferred (c− );
assign the first two points of the utility function:
4
5
amputated
below knee
The patient is given the
following gamble:
u(c+ ) = 1 and u(c− ) = 0
3
Reconsider the decision problem for treatment of diabetic
gangrene. The possible consequences of treatment are:
cured
1.0
amputated below knee
?
amputated above knee
?
death
0.0
create a standard reference gamble for enough
intermediate outcomes;
elicit the indifference probability for the lotteries in each of
the gambles;
compute the utility of an intermediate outcome using this
indifference probability.
p
cured
1−p
death
Suppose that p = 0.9 is the indifference probability for the
lotteries.
We then have that
u(amputated below knee) = 0.9·u(cured)+0.1·u(death) = 0.9
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Example: extreme utilities
Direct vs indirect methods
direct:
• roots in psychophysics
• inferior in both validity & reliability
• easily applied to risky tasks with complex consequences
indirect:
• roots in utility axioms
• time consuming
• irrelevant ”gaming” effect
• distasteful / unethical
• unsuitable for measuring very small or very large utilities
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For the diabetic gangrene decision problem, we could use the
following gambles:
p
1−p
amputated
above knee
amputated
below knee
death
amputated
below knee
q
cured
1−q
amputated
above knee
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Subjective utility assessments
• can change overtime
Risk attitudes
• are required for unknown / unexperienced outcomes
• are influenced by framing and certainty effects
• are influenced by third parties
• are not comparable from person to person
• ...
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Risk-neutral preferences
Let X be an attribute with values x1 . . . xn , n ≥ 2, measured
in some unit. Let L = [p, xi ; (1 − p), xj ] be a lottery over X.
An Example
Suppose you are given the choice between the following two
“games”:
0.5
0.5
0.5
0.5
A decision maker is risk-neutral if each additional unit of X is
valued with the same increase in utility:
Example
e5
D EFINITION
A decision maker is risk-neutral,
if
e −1
u(p · xi + (1 − p) · xj ) =
e6
p · u(xi ) + (1 − p) · u(xj )
e −2
that is, the utility function is
linear.
Do you have a clear preference for playing one or the other?
Note that u′′ (X) = 0.
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Risk-averse preferences
An Example
Let X be an attribute with values x1 . . . xn , n ≥ 2, measured
in some unit. Let L = [p, xi ; (1 − p), xj ] be a lottery over X.
Suppose you are given the choice between the following two
“games”:
0.5
0.5
0.5
0.5
A decision maker is risk-averse if each additional unit of X is
valued with a smaller increase in utility:
Example
e 52
D EFINITION
A decision maker is risk-averse,
if
u(p · xi + (1 − p) · xj ) >
e −2
e 5000
p · u(xi ) + (1 − p) · u(xj )
e −4950
that is, the utility function is
concave.
Which game would you prefer to play?
Note that u′′ (X) < 0.
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Risk-prone preferences
Let X be an attribute with values x1 . . . xn , n ≥ 2, measured
in some unit. Let L = [p, xi ; (1 − p), xj ] be a lottery over X.
An Example
Suppose you are given the choice between the following two
“games”:
A decision maker is risk-prone, or risk-seeking, if each
additional unit of X is valued with a larger increase in utility:
Example
0.2
0.8
0.8
0.2
D EFINITION
A decision maker is risk-prone,
if
e 10
e 0, 10
e 2, 50
u(p · xi + (1 − p) · xj ) <
e1
p · u(xi ) + (1 − p) · u(xj )
that is, the utility function is
convex.
Which game would you prefer to play?
Note that u′′ (X) > 0.
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Discounting
Risk attitudes
• the process of translating future rewards to their present
value;
• necessary to perform ”cost-benefit” analyses now.
The zero-illusion curve:
u(X)
u(X)
RA
RA
Examples
RS
• the time value of money
value of an euro depends on when it is available
euros can be invested to yield more euros
RA
RS
0
X
0
• the time value of life
X
life years in the future less valuable than today (?)
life years are valued relative to money
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An example
Consider the following ‘standard reference gamble’:
A discountingfactor
0.50
A discountingfactor indicates how much less a patient values
each successive year of life, compared to the previous year:
The utility function for length of life
with a constant discountingfactor δ
is approximated by
Z x
u(x) =
e−δ·t dt
t=0
for life-expectancy x.
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1.0
x years
∼
0.50
25 years
0 years
For a utility function for length of life with a constant
discountingfactor δ = 0.02, we find:
u(25 years) =
u(0 years) =
and
Z 25
Z 0
−0.02·t
e−0.02·t dt = 19.67
e
dt = 0
t=0
t=0
The patient should be indifferent about the choice between the
two lotteries for a life-expectancy x for which
u(x) = 0.5 · 19.67 + 0.5 · 0 = 9.835
Z x
We find from
e−0.02·t dt = 9.835 that x ≈ 11 years.
t=0
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An example
The risk-premium
Let X be an attribute with values x1 . . . xn , n ≥ 2, measured
in some unit. Let u(X) be a utility function over X.
Consider a lottery L = [p, xi ; (1 − p), xj ] over X. Let xC be the
certainty equivalent of lottery L and xE the expected value of L.
Let u(x) = −e
for all x ∈ [0 . . . 50] Consider the lottery
L = [0.5, 0; 0.5, 10]. Compute the risk premium for this lottery.
−0.2x
The expected value xE of the lottery is
0.5 · 0 + 0.5 · 10 = 5
The certainty equivalent xC of the lottery is determined from
The risk premium RP of L is
defined as:
xE − xC if u(X) is increasing
RP =
xC − xE if u(X) is decreasing
u(xC ) = 0.5 · u(0) + 0.5 · u(10) = −0.5 − 0.5 · e−2 ≈ −0.57
and equals 2.83.
As u(x) is an increasing function, the risk premium for L is
RP = 5 − 2.83 = 2.17
Exercise
Let u(x) = − log(x + 30) for all x > −30. Consider the lottery
L = [0.5, −20; 0.5, −10]. What is the lottery’s risk premium? Is
the decision maker risk averse or risk prone?
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Risk averseness & risk premium
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Risk proneness & risk premium
A decision maker is
• risk averse iff his/her risk premium is positive for all
nondegenerate lotteries;
• decreasingly risk averse iff he/she is risk averse and
• his/her risk premium for any lottery
[0.5, x − h; 0.5, x + h] decreases (↓ 0) as x increases;
• increasingly risk averse iff he/she is risk averse and
• his/her risk premium for any lottery
[0.5, x − h; 0.5, x + h] increases (↑ ∞) as x increases;
• constantly risk averse iff he/she is risk averse and
• his/her risk premium for any lottery
[0.5, x − h; 0.5, x + h] remains constant for all x.
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A decision maker is
• risk prone iff his/her risk premium is negative for all
nondegenerate lotteries;
• decreasingly risk prone iff he/she is risk prone and
• his/her risk premium for any lottery
[0.5, x − h; 0.5, x + h] increases (↑ 0) as x increases;
• increasingly risk prone iff he/she is risk prone and
• his/her risk premium for any lottery
[0.5, x − h; 0.5, x + h] decreases (↓ −∞) as x increases;
• constantly risk prone iff he/she is risk prone and
• his/her risk premium for any lottery
[0.5, x − h; 0.5, x + h] remains constant for all x.
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The risk-aversion function
The degree of risk aversion / proneness – Introduction
Information regarding a decision maker’s risk attitude is given by
RP: sign indicates aversion vs proneness;
magnitude captures the degree of this behaviour,
for one specific lottery!
u′′ (X): sign indicates aversion vs proneness;
magnitude conveys no relevant information,
since strategically equivalent functions capture
same risk attitude:
u(X)
u(x) = −3e−x
u(X)
u(x) = 1 − e−x
D EFINITION
Consider a utility function u(X), and let σ ∈ {+, −} denote the
sign of u′ (x). The risk-aversion function R(X) for u(X) is then
defined by
R(X) = −σ
u′′ (X)
u′ (X)
• if R(X) > 0 then the decision maker is risk averse;
• if R(X) < 0 then the decision maker is risk prone;
• if R(X) = 0 then the decision maker is risk neutral;
T HEOREM
RP
x
(xE )
x−h
R(X) is increasing (decreasing, constant) iff the decision
maker’s risk premium for any lottery [0.5, x − h; 0.5, x + h] is
increasing (decreasing, constant) for increasing x.
RP
x+h
x−h
X
certainty
equivalent
x
(xE )
x+h
X
certainty
equivalent
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An example
Let u(x) = 1 − e
, x > 0, be a utility function. Find the
risk-aversion function for X.
−x/900
We have that
u′ (x) =
1
· e−x/900 ,
900
x>0
and
−1
· e−x/900 , x > 0
9002
Since u(x) is an increasing function in x (u′ (x) > 0 for all x > 0),
the risk-aversion function is defined as
u′′ (X)
R(X) = − ′
u (X)
u′′ (x) =
1
and that u(X) models constant
We conclude that R(X) =
900
risk aversion.
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