Information Technology Reference
In-Depth Information
We construct a chain of
n
-element sets
X
S
,
X
1
,X
2
, ...X
k
,X
R
so that each set
X
and its successor
Y
satisfy the conditions of Lemma 2. We retain elements of
X
Y
and
replace others in turn, substituting the largest in
Y
for the largest in
X
and continuing
in descending order. For example, if
X
=
∩
{
9,8,6,5
}
and
Y
=
{
8,4,3,1
}
, the intermediate
sets are
.
Since
X
S
contains the largest weights,
w
x
>w
y
is satisfied for each pair in the
chain. By Lemma 2, for each pair
{
4,8,6,5
}
,
{
4,8,3,5
}
X,Y
,
P
(
X
)
≥
P
(
Y
)
. Hence
P
(
X
S
)
≥
P
(
X
R
)
,a
contradiction. So no such better strategy exists, and we conclude that
.
Suppose we relax the assumption that the persuadee knows nothing about the options
in the initial situation, and instead starts with a set of initial (perhaps unverified) beliefs
that lead to a preference for one of the options, but where there still remain attributes
whose value for the options is unknown. This was the case for Bert in the Burger World
example above. Corollary 1 demonstrates that
S
is optimal.
S
is an optimal strategy in this case also.
Corollary 1:
If the utilities of
O
1
and
O
2
are initially unequal,
S
remains an optimal
strategy.
Proof of Corollary 1:
Let the current utilities be
U
1
,U
2
with
U
1
>U
2
and
W
=
{
w
1
.., w
m
}
be the set of weights.
Suppose, for contradiction, that a better strategy
R
than
S
exists, expected to termi-
nate in
n
(
R
)
steps with
n
(
R
)
<n
(
S
)
.Let
L
be a number greater than any weight in
W
.
Suppose there were additional attributes
b
1
,
b
2
with weights
L
+
U
1
,L
+
U
2
.Now
initiate a dialogue with weights
W
∪{
L
+
U
1
,L
+
U
2
}
and the initial utilities zero for
both options.
Apply the following strategy: Step 1: Inquire about
b
1
. Step 2: Inquire about
b
2
.After
Step 2: if Step 1 assigns TF and Step 2 assigns FT, (difference between utilities is
U
1
− U
2
), continue as for
, otherwise continue in descending order. For this strategy
the expected number of steps is 2 + 1/4
n
(
R
)
+3/4
n
(
S
)
. The expected number of steps
for the descending order strategy is 2 +
n
(
S
)
which is greater, contradicting Theorem
1. So no such strategy
R
Next consider partial persuasion, where nothing improving the current estimated utility
of the preferred option can be learned, since the persuader will choose to remain silent.
Now all inquiry will take place concerning the option advocated by the persuader, values
for the persuadee's preferred option being supplied by current beliefs and expectations.
That
R
exists.
S
remains an optimal strategy in the case is established by Theorem 2.
Theorem 2:
If
τ
1
j
is known in advance for all attributes
a
j
,
S
is an optimal strategy for
inquiring about the
τ
2
j
.
Sketch of Proof of Theorem 2:
In this case
U
1
is constant throughout. After examining
attributes with weights in
X
W
,
U
2
=
w
j
∈X
⊆
τ
2
j
w
j
.
The algorithm terminates when
either
U
2
>U
1
or
U
1
− U
2
>
w
j
∈W\X
w
j
=
w
j
∈W
w
j
−
w
j
.
w
j
∈X
