Download IV SAVAGE`S THEORY OF RATIONAL DECISION (Experts in

Savage’s
3. THEORY OF RATIONAL DECISION
(Experts in Uncertainty ch 6)
Roger M Cooke
ShortCourse on Expert Judgment
National Aerospace Institute
April 15,16, 2008
1. Basic Concepts
S: the set of possible worlds or set of states of the world
C: the set of consequences or states of an acting subject
F: set of available acts
≤: preference relation on available acts
Where;
S <> 
#C < ∞
F = Cs = {f | f: S → C}
  F × F.
Write f ~ g if f ≤ g and f ≥ g; and f > g if f ≥ g and not f ≤ g.
Example
available acts:
f = "buy a car"
g = "buy a motorcycle"
factors influencing the outcomes of acts:
N: get new job further from home
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A: have a driving accident
R: price of gasoline rises
N'; not get new job further from home
ect.
(reduced) states of the world:
NAR
NA'R
NAR'
NA'R'
N'AR
N'A'R
N'AR'
N'A'R'
Idea: State of world is not known with certainty; hence outcomes of acts are uncertain. The value
of the outcome of an act can vary in different circumstances. The value of an act is to be
represented as expect utility:
EU(f) = ∑cC U(c) ×P{sS | f(s) = c}
Further, we want:
*
f ≥ g if and only if U(f) ≥ U(g).
Problem: Is P independent of f and/or g?? Is P(A) the same regardless whether we buy a car or a
motorcycle? If this does not hold, then the problem has not been modeled properly for
representation in Savage's formalism.
Technical fix: Can always be applied to yield a model in which probability is independent of act
which is performed. E.G. if probability of A is influenced by choice between f and g, then write
A = Af  Ag;
Af = would have an accident if driving a car
Ag = would have an accident if driving a motorcycle.
Now repeat the analysis with events N, Af, Ag, R.
For representation * to go through, we must assign probabilities to events, and numerical utilities to
outcomes.
We seek conditions on ≥ which
- are true for all "rational" preference
behavior
- admit * for all available acts
- the P in * is unique and the U is unique
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up to U' = aU+b, a > 0.
2. Axioms
1.Weak Order:
 f,g  F, f ≥g or g ≥ f or both (≥ is connected)
 f,g,h  F, if f ≥ g, and g ≥ h, then f ≥ h (transitivity)
n.b. if c  C, then we may let c  F denote the 'constant act" c(s) = c. If c,d  C, then we may write
'c > d' if this holds for the corresponding constant acts.
2. Principle of definition
g,b  C such that g > b (g = 'good', b = 'bad');
A  S, define: rA(s) = g if s  A; and = b otherwise. rA  F;
If g*, b*  C with g* > b*, let r*A (s) =g* if s  A; and = b* otherwise.
then
 A,B,  S; rA > rB if and only if r*A > r*B.
Definition 1: For A, B  S, A ≥. B if rA ≥ rB. We write >. and <. with the obvious meaning.
3. Sure thing principle
 A  S, let f,h,f*, h*  F such that
f(A) = f*(A)
h(A) = h*(A)
f(A') = h(A')
f*(A') = h*(A')
then
f  h if and only if f*  h*.
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Allais paradox
Choice 1
Act 1
get $500,000
Act 2 get $2,500,000 with probability 0.1
get $500,000 with probability 0.89
get 0 with probability 0.01
Choice 2
Act 3
get $500,000 with probability 0.11
get 0 with probability 0.89
Act 4
get $2,500,000 with probability 0.1
Get 0 with probability 0.9.
Many people say Act 1 > Act 2, but Act 3 < Act 4.
Lemma 1; '≥.' is additive that is; A,B,C  S, if AC = BC = , then
A ≥. B if and only if AC ≥. BC
Proof: Exercise (hint, use the sure thing principle).
Definition 2: A  S is null if rA ~ r
4. Dominance:
If  sS, f(s) ≥ g(s), then f ≥ g. If also f(s) > g(s) for all s  B, B non-null, then f > g.
Lemma 2: A  S.  ≤. A ≤. S.
Proof: exercise.
Lemma 3:  A,B  S, A ≥. B if and only if B' ≥. A'.
Proof: exercise.
Verify:
A,B,C  S with AC = , if A ≥. B then AC ≥. BC.
find a counter example for the converse.
5. Refinement B,C  S, if B <. C then there exists a partition {Ai}i=1..n; such that
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B  Ai <. C; for i = 1,...n.
Definition 3: a relation <* on PS × PS is a qualitative probability if
1) <* is a weak order
2) <* is additive
3) A  S,  * A * S.
Lemma 4: The relation '>.' from definition 1 is a qualitative probability.
Proof: exercise.
3. Probability
Theorem 1. If > on F  F satisfies axioms 1 through 5, then there exists a unique probability P on
(S, P(S)), such that for all A, B  S,
P(A)  P(B) if and only if A . B.
Proof sketch
We use
Lemma 5. AS there exist A1, A2  A such that A1A2 = , A1A2 = A and A1=. A2 .
Proof: see supplement Chpt. 6.
With this we can show that for all n there exists a uniform partition {Ai}i=1...2n, i.e. a partition
such that Ai ~. Aj, i,j=1...2n.
Pick one such uniform partition {Ai}i=1...2n; and let k(A,n) be the largest integer k such that
k
 Ai ≤. A.
i=1
Claim, k(A,n) is the same for any 2n-fold uniform partition.
Proof: The proof is by induction on n; we illustrate for n = 2. Suppose {A1,A2} and {B1,B2} are
uniform partitions; then we show Bi =. Ai. Suppose to the contrary A1 >. B1, then A2 =. A1 >. B1 =.
B2. But by Lemma 3, B2 = B1' >. A1' = A2. 
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k(A, 2n)/2n is an increasing function of n, and is < 1; therefore the limit exists and we define:
P(A) := limn->∞ k(A,2n)/2n
Verify; P is a probability.
We show that for all A, B, P(A) ≥ P(B) if and only if A ≥. B. Suppose A ≥. B, then also k(A,n) ≥
k(B,n), for all n, so that P(A) ≥ P(B). Suppose that A >. B, then there exists a partition {Ci}i=1...n*
such that BCi <. A, for each i = 1,...n*. Pick j such that  <. B’Cj; and pick partition {Dj}j=1...n'
such that Dj <. B’Cj. We can find a uniform partition {Ai}i=1...2n such that Ai <. Max {Dj} i. For
this partition we can show
k(B,2n) < k(B,2n)+1  k(BCj, 2n) ≤ k(A,2n)
If we consider a 2n+m uniform partition then
k(B,2n+m) < k(B,2n+m)+2m  k(A,2n+m)
P(B)= limm [k(B,2n+m)+2m] / 2(n+m)
P(B)+ 2-n < P(A).
Verify: The P defined above is unique, that is, if P* such that
A,B  S, P*(A) ≥ P*(B) if and only if A ≥. B
then P* = P.

Lemma 6: If '' satisfies axioms 1 through 5, then #S = ∞. Moreover, for any r  (0,1) and A  S
there is Ar  A such that P(Ar) = rP(A). (See Experts in Uncertanty ch. 6).
4. UTILITY
Note that the foregoing derivation of P used the 'utility' of consequences only in the Principle of
Definition. We need a utility function for the representation of preference. For this purpose we need
one more axiom on the preference relation, and this axiom explicitly uses the P which are derived
from the previous five axioms:
6. Strengthend refinement principle:
f,g  F, if f > g, then there exists an α, α > 0, such that for all AS with P(A) < α, and for all f', g'
with
f(A') = f'(A'); g(A') = g'(A')
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we have f' > g'.
Theorem 2: If '' satisfies axioms 1 through 6, then
1) There exists a U*: C → R such that f,g  F
f > g  U*(f) > U*(g)
where U*(f) = ΣcC U*(c)P(sS|f(s) = c}, etc.
2) If U: C → R is another function for which f > g  U(f) > U(g), then there exist a,b  R, a > 0,
such that
U = aU* + b.
Proof sketch:
1) Order C such that cn > cn-1 >...c1. Put U*(cn) = 1, U*(c1) = 0. For each event A we can consider
the act cn|A  c1|A'.
Lemma 1: For ci, we can find Ai such that
ci ~ cn|Ai + c1|Ai’ ;
moreover, any B with P(B) = P(Ai) will satisfy the above equation in preference (which compares
the constant act ci with another act).
Lemma 2: For each f, we have f ~ f where
f=
c1|f-1(c1)B2..Bn-1) + cn|f-1(cn)G2...Gn-1)
where
BiGi= f-1(ci);BiGi = ; Gi = f-1(ci)P(Ai); i = 2,...n-1;
for the Ai from Lemma 1.
Proof (see Experts in Uncertainty chapt. 6)
2). Verify: For i = 2,...n-1, putting U*(ci) = P(Ai); U* satisfies (1) of the theorem.
Proof sketch: Use the notation Gi(f), Gi(g), as in Lemma 2, and put
G(f) =  Gi(f); G(g) =  Gi(g).
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Verify that: f ≥ g  f ≥ g  G(f) ≥. G(g)  rG(f) ≥ rG(g)  P(G(f)) ≥ P(G(g))  U*(f) ≥ U*(g).
3) Verify: that if U satisfies (1) of the theorem then so does aU+b, with a,b  R, a > 0.
4) Suppose U satisfies (1) of the theorem. Choose a, b such that
aU(cn) + b = 1; aU(c1) + b = 0.
We show that aU + b = U*. Indeed, since aU+b satisfies (1):
(aU + b)(ci) = (aU+b)(cn1Ai + c11Ai’) =
(aU(cn)+b)P(Ai) = P(Ai)
= U*(ci))
5. Observation
In light of the previous theorem we may simply assmue that our acts are utility valued, and we
write Ef instead of U(f) for the expected utility of f.
Uncertainty concerns the outcomes of potential observations, uncertainty is reduced or removed by
observation. A representation of uncertainty must also represent the role of observations, and
explain the value of observation for decision making.
For a subjectivist the question "why observe" is not trivial. Isn't one subjective probability just as
rational as every other? The value of observation does not devolve from the "rationality" of the
preference relation as defined by the axioms 1 through 6. Rather it depends on the acts at our
disposal and on the utilities of consequences. The set F contains all functions from S to C. One of
these acts takes the best consequence in all possible worlds. If this act is also available to us, there
is no reason to observe anything.
Example
Suppose we are considering whether to invest in gold. There are two acts at our disposal, invest or
not invest. Suppose we restrict attention to one uncertain event, namely the event that the price of
gold rises next year. Let B1 denote the event "the price of gold rises next year" and let B2 = B1'.
Suppose before choosing whether to invest, we can perform a cost-free observation. For example,
we may ask an expert what (s)he thinks the gold price next year will be. The result of this
observation is of course uncertain before it is performed. An observation must therefore be
represented as a random variable X: S R. Let x1,...xn denote the possible values of X.
Before performing the observation, our degree of belief in B1 is P(B1). After the observing X = x i,
our probability is P(B1|xi). If the observation of xi is definitive, then P(B1|xi) = 0 or 1. We assume
that none of the possible values xi are definitive. By Bayes' theorem:
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P(B1|xi) = P(xi|B1)P(B1)/P(xi).
Dividing by a similar equation for B2:
P(B1|xi)
────── =
P(B2|xi)
P(xi|B1)P(B1)/P(xi)
─────────── .
P(xi|B2)P(B2)/P(xi)
The idea is this; if B1 is the case, then we should expect that after the observation of X, the LHS >
P(B1)/P(B2). Define
R(X) =
P(X|B1)
─────.
P(X|B2)
In other words, E(R(X)|B1) > 1. To prove this we recall Jensen's inequality.
Definition: A function q: D → R, D an interval, is convex (concave) if α(0,1), and x,y  D:
q(αx + (1-α)y) ≤(≥) αq(x) + (1-α)q(y).
q is strictly convex (concave) if the inequality above is strict.
Jensen's inequality: If X is an random variable, and q: D → R, D an interval, D  Rng(X), is
convex (concave), then
E(q(X)) ≥ (≤) q(E(X)),
with strict inequality if q is strictly convex (concave).
Theorem 3: E(R(X)|B1) > 1, and E(R(X)|B1) = 1 if and only if P(xi|B1)/P(xi|B2) = 1 for i = 1,...n.
Proof:
log(P) is concave, and -log(P) is convex. Hence
log(E(R(X)|B1)) ≥ E(logR(X)|B1)
= E(-logR(X)-1| B1) ≥ -log(E(R(X)-1|B1)).
Further
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P(xi|B2)
E(R(X) | B1) = Σxi
-1
P(xi | B1) = 1
P(xi|B1)
so that
-log(E(R(X)-1| B1)) = -log(1) = 0.
log(E(R(X)|B1)) ≥ 0; E(R(X)|B1) ≥ 1.
Since log is strictly concave, log(E(R(X)|B1)) = E(logR(X)|B1) only if R(X) is constant, given B1.
That is, if s  B1, then P(Xi(s)|B1) = KP(Xi(s)|B2). Summing this equation over i = 1,...n, and
recalling that no observation is definitive, we find that 1 = K. 
This theorem says that we expect to become more certain of B1, if B1 is the case, by observing X.
It shows how we expect our beliefs to change as a result of observation, but it does not answer the
question why observe? To answer this question we must consider the value of an observation is
determined by the acts at our disposal.
Suppose F*  F is a set of available acts, and we have to choose a single element of F*.
Definition: For F*  F the value of v(F*) of F* is
v(F*) = maxfF* Ef.
Before choosing an element of F*, we may consider the option of observing X. After observing X =
xi our probability changes from P(•) to P(•|X=xi), and the value of F* would be computed with
respect P(•|X=xi). Before observing we don't know which value of X will be observed, but we can
consider the value of F* as conditional on the value of X. Recall that the consequences are assumed
to be utility values.
Definition: For F*  F the value of v(F*|X=xi) of F* given X = xi is
v(F*|X=xi) = maxfF* ΣcC c P(f-1(c)|X=xi)) = maxfF* E(f|xi);
The value of v(F*|X) of F* given X is
v(F*|X) = Σ v(F*|X=xi)P(X=xi).
Theorem 4: For any X: S → R, and any F*  F, v(F*|X) ≥ v(F*).
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Proof:
v(F*|X) = Σ maxfF* [E(f|xi)] P(X=xi) ≥ Σ E(f|xi)P(X=xi) = Ef.
Since this holds fF*, we have v(F*|X) ≥ maxfF* Ef = v(F*).
Exercise: Let X take two possible values, 1, and 2; let pi = P(B1|i), i = 1,2. Let F* be the available
acts, and suppose that if f e F*, then f is constant on B1 and on B2 = B1'. Let p = P(B1), and let
v(F*)(p) denote the value of F as a function of p.
• Show that the expectation of f  F* is a linear function of p.
• Show that v(F*)(p) is convex.
• Show that if p1 = p2, then v(F*|X) = v(F*)
• Show that if v(F*)(p) is linear in the interval (p1,p2), then v(F*|X) = v(F*).
• Let X' be another two-valued observation with p1' and p2'. Show that if |p1' - p2'| > |p1 - p2|,
then v(F*|X') > v(F*|X).
Example
Value of problem F
15
Exp(act(p))
10
f
5
g
0
-5 0
0.2
0.4
0.6
0.8
1
h
j
-10
V(F)
-15
-20
probability
f(p) = 20p-12
g(p) = -30p+13
h(p) = 2p
j(p) = -3p-2
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Value of Problem
Value of problem F
Value of Perfect Information
Value of
Observation X
15
Exp(act(p))
10
f
5
g
0
-5 0
0.2
0.4
0.6
0.8
1
h
j
-10
V(F)
-15
-20
P(A|x1)=0.25
probability
12
P(A|x2)=0.7