2.3. Value of Information: Decision Trees and Backward Induction

2.3. Value of Information: Decision Trees andBackward Induction

2.1 Basic concepts: Preferences, and utility2.2 Choice under uncertainty: Lotteries and risk

aversion2.3 Value of information: Decision trees and

backward induction

Outline

Complex decision scenarios usually involve both the choiceof an alternative (action) and the influence of uncertainevents (lotteries) on the final outcome of the decisions

In many cases, reducing the “degree” of the uncertainty willhelp the decision process.

In this sense, we will see that Information is a valuable element

Decision Trees

Decision Trees (as we have already seen) are useful diagrammatic representations of decision problems thathelp in the search for the best decision

Usually, a Decision Tree consists of a sequence of actionbranches (decisions) and chance branches (lotteries) thatrepresent the problem at hand

Solving a Decision Tree means computing a value for each action branch (Expected Value, Expected Utility) and finally decide what is the best action based on these values.

Example: The garden

apples

oranges

apricots

0.25

0.75

0.25

0.75

0.25

0.75

100

40

-20

140

80

60

E(apples) = 0.25 x 100 + 0.75 x 40 = 55

E(apricots) = 0.25 x 80 + 0.75 x 60 = 65

E(oranges) = 0.25 x (-20) + 0.75 x 140 = 100

E(u(apples)) = 0.25 x 1002 + 0.75 x 402 = 3700

E(u(oranges)) = 0.25 x -(20)2 + 0.75 x 1402 = 14,600

E(u(apricots)) = 0.25 x 802 + 0.75 x 602 = 4300

Risk Neutral:Risk Lover:

u(x)=xu(x)=x2

The method of “solving the tree” moving from rightto left is called Backward Induction (or tree rollback)

In this example, it is clear that knowing before handwhat the weather is going to be like would be avaluable information for taking the best decision

apples

oranges

apricots0.25

0.75

apples

oranges

apricots

100

40

-20

140

80

60

Snows

Does notSnow

apples

oranges

apricots0.25

0.75

apples

oranges

apricots

100

40

-20

140

80

60

Snows

Does notSnow

What is the value of information ?

If we do not have any information regarding the weatherin the future, our best choice is to plant Oranges as it isthe alternative that has the highest Expected Value

E(Oranges) = 100

But if we had full information about the weather, that is,if we knew before hand if it is going to snow or not, ourbest choice would be:

Apples if it is going to snow Value = 100 Oranges if it is not going to snow Value = 140

What is the value of information ?

Since the ”real” probability of snowing is of 0.25, the probabilitythat you are informed that it is going to snow is also of 0.25.

Hence, your decision

Apples will be taken with probability 0.25Oranges will be taken with probability 0.75

Therefore, the Expected Value using this information is

E(Information) = 0.25 x 100 + 0.75 x 140 = 130

What is the value of information ?

Thus,

Expected revenue without infomation E(Oranges) = 100

Expected revenue with information

E(Information) = 130

The information has a value of 30

Nevertheless, having full information is rarely the case

In most situations, some partial information (forinstance, a weather forecast) is all we can get

This partial information, though, can be useful to obtaina more accurate measure (updated probabilities) of the uncertain events

Asian Import Company (i) Asian Import Company is a small Chicago based firm specialized on the import and distribution of Asian collectibles.AIC is about to close a deal with a large Chinese company consisting of the acquisition a large collection of Ming porcelain for reselling in the U.S. market. Such operation would have a cost of $500,000 for AIC and would produce a revenueof $800,000 (net profit of $300,000)However, the U.S. Government is currently negotiating with China over a special trade agreement for “arts and crafts” items.If the negotiations are successful, AIC will be allowed to freely import the Ming collection at zero cost, but if an agreement isnot reached, AIC will be forced to pay a 50% tariff on the sale of anyimported item. Before hand, without any specific information, thechances that the negotiation is successful are 50%.

(For simplicity, we will assume that AIC is risk neutral, thatis, AIC uses Expected Value (Expected Profit in this case)as the decision criterion)

Should AIC buy the Ming collection ?

Buy

Don't buy

Agreement

Disagreement

(0.5)

(0.5)

$0

$8 - $5 = $3

$4 - $5 = -$1

E(Profit) = 0.5 x 3 + 0.5 x (-1) = $1

Should AIC buy the Ming collection ?

At this point the decision is clear. Since the ExpectedProfit is positive ($100,000), AIC should purchasethe Ming collection

Asian Import Company (ii) Reducing the risk

After “cheap talking “ to an analyst, AIC realize that they could wait until the uncertainty is resolved.

The problem is that then it might be too late and that AIC looses the Chinese deal to a competitor firm ! Based on previous experiences, AIC knows that the chances that thishappens are of 70 %

Should AIC wait ?

Buy

Don't buy

Agreement

Disagreement

(0.5)

(0.5)

$0

$3

-$1

E(Profit) = 0.5 x 3 + 0.5 x (-1) = $1

WaitAgreement

Disagreement

(0.5)

(0.5)$0

Still Available

Not available

(0.3)

(0.7)

$3

$0

Backward Induction

Solving the tree by Backward Induction consists of:

Start at the top-rightmost end of the tree

For each set of chance branches, find the corresponding Expected Value (or Expected Utility)

Chop off those branches and replace them by the computed Expected Value (or Expected Utility)

Backward Induction (continued)

For each set of choice branches, find the best choice

Chop off those branches and replace them by the value that corresponds to that best choice

Proceed left wise in the same way until all chance branches Are removed

Buy

Don't buy

Agreement

Disagreement

(0.5)

(0.5)

$0

$3

-$1

E(Profit) = 0.5 x 3 + 0.5 x (-1) = $1

WaitAgreement

Disagreement

(0.5)

(0.5)$0

Still Available

Not available

(0.3)

(0.7)

$3

$0

Remove thesechance branches

Replace them bythe Expected Value

Buy

Don't buy$0

$1

WaitAgreement

Disagreement

(0.5)

(0.5)$0

Still Available

Not available

(0.3)

(0.7)

$3

$0

Buy

Don't buy$0

$1

WaitAgreement

Disagreement

(0.5)

(0.5)$0

Still Available

Not available

(0.3)

(0,.7)

$3

$0

E(Profit) = = 0.3 x 3 + 0.7 x 0 = $0.9

Buy

Don't buy$0

$1

WaitAgreement

Disagreement

(0.5)

(0.5)$0

$0.9

Buy

Don't buy$0

$1

WaitAgreement

Disagreement

(0.5)

(0.5)$0

$0.9

E(Profit) = 0.5 x 0.9 + 0.5 x 0 = $0.45

Buy

Don't buy$0

$1

Wait

$0.45

Should AIC wait ?

Clearly not. The expected profit of waiting is of only$45.000 (because of the high chances of loosing thedeal with the Chinese firm). Buying the collection without knowing the outcome of the negotiations is a better choice ($100.000 expected profit)

Asian Import Company (iii) Full information

Suppose now that AIC has the opportunity to access a source of full information. That is, AIC could know right away whether the trade agreement is going to be signed or not.

How much would AIC pay for such information ?

Buy

Don't buy $0

$1

Wait $0.45

Agreement

Disagreement

(0.5)

(0.5)

Buy

Don't buy

$3

$0

Buy

Don't buy

-$1

$0

FullInformation

usingBackward Induction

Buy

Don't buy $0

$1

Wait $0.45

Agreement

Disagreement

(0.5)

(0.5)

Buy

Don't buy

$3

$0

Buy

Don't buy

-$1

$0

FullInformation

Buy

Don't buy $0

$1

Wait $0.45

Agreement

Disagreement

(0.5)

(0.5)

$3

$0

FullInformation

Buy

Don't buy $0

$1

Wait $0.45

Agreement

Disagreement

(0.5)

(0.5)

$3

$0

E(Profit) = 0.5 x 3 + 0.5 x 0 = $1 .5

FullInformation

Buy

Don't buy $0

$1

Wait $0.45

$1 .5

FullInformation

How much would AIC pay for such information ?

Expected gain without infomation (best option)

E(Buy) = $100,000

Expected gain with information

E(Full Information) = $150,000

Thus, the information has a value of

$50,000

Asian Import Company (iv) Partial information

Suppose, finally, that AIC can hire an analyst that has good connections with the Washington bureaucracy. For a fee he can tell you what we knows on the status of the negotiations.. This analyst is known to be very successful in similar situations. Every time an agreement was reached he had predicted it 90% of the times. However, he hasn't been so reliable when the negotiations were not successful. In such cases, he only called it right 60% of the time.

How much would AIC pay now for such information ?

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

Buy

Buy

Don't buy

Don't buy

Agreement

(??)

Disagreement(??)

Agreement

(??)

Disagreement(??)

(??)

(??)

$0

$0

$3

$3

-$1

-$1

Full Information

Disagreement(0.5)

This is what we know . . .

Predicts Success

Predicts Failure

(0.9)

(0.1)

(0.3)

(0.45)

(0.05)

(0.2)

Path probabilites

Agreement(0.5)

Predicts Success

Predicts Failure

(0.4)

(0.6)

Predicts Failure(B)

This is what we need to know . . .

Agreement

Disagreement

(C)

(D)

(0.3)

(0.45)

(0.05)

(0.2)

Path probabilites

Predicts Success(A)

Agreement

Disagreement

(E)

(F)

”Flipped Probabilites”

A = 0.45 + 0.2 = 0.65B = 0.05 + 0.3 = 0.35

A X C = 0.45 0.65 X C = 0.45 C = 0.69A X D = 0.20 0.65 X D = 0.20 D = 0.31 B X E = 0.05 0.35 X E = 0.05 E = 0.14 B X F = 0.30 0.35 X F = 0.30 F = 0.86

Computation of ”Flipped Probabilites”

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

Buy

Buy

Don't buy

Don't buy

Agreement

(C)

Disagreement(D)

Agreement

(E)

Disagreement(F)

(A)

(B)

$0

$0

$3

$3

-$1

-$1

Full Information

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

Buy

Buy

Don't buy

Don't buy

Agreement

(0.69)

Disagreement(0.31)

Agreement

(0.14)

Disagreement(0.86)

(0.65)

(0.35)

$0

$0

$3

$3

-$1

-$1

Full Information

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

Buy

Buy

Don't buy

Don't buy

Agreement

(0.69)

Disagreement(0.31)

Agreement

(0.14)

Disagreement(0.86)

(0.65)

(0.35)

$0

$0

$3

$3

-$1

-$1

Full Information E(Buy) = 0.69 x 3 + 0.31 x -1 = $1 .76

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

Buy

Buy

Don't buy

Don't buy

Agreement

(0.14)

Disagreement(0.86)

(0.65)

(0.35)

$0

$0

$3

-$1

Full Information$1 .76

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

Buy

Buy

Don't buy

Don't buy

Agreement

(0.14)

Disagreement(0.86)

(0.65)

(0.35)

$0

$0

$3

-$1

Full Information$1 .76

E(Buy) = 0.14 x 3 + 0.86 x -1 = -$0.44

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

Buy

Buy

Don't buy

Don't buy

(0.65)

(0.35)

$0

$0

Full Information$1 .76

-$0.44

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

Buy

Buy

Don't buy

Don't buy

(0.65)

(0.35)

$0

$0

Full Information$1 .76

-$0.44

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

(0.65)

(0.35)$0

Full Information

$1 .76

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

PredictsSuccess

PredictsFailure

(0.65)

(0.35)$0

Full Information

$1 .76

E(Partial Information) = = 0.65 x 1.76 + 0.35 x 0 = $1 . 144

Buy

Don't buy$0

$1

Wait$0.45

$1 .5

PartialInformation

Full Information

$1 . 144

How much would AIC pay for such partial information ?

Expected gain without infomation E(Buy) = $100,000

Expected gain with partial information

E(Partial Information) = $114,400

Thus, the information has a value of

$14,400

Summary

Complex decision scenarios usually involve both the choice of an alternative (action) and the influence of uncertain events (lotteries) on the final outcome of the decisions

Information on these uncertain events is, hence, valuable Full Information is seldom available Partial information, though, can be useful to obtain a

more accurate measure (updated probabilities) of the uncertain events

Decision Trees are useful diagrammatic representations of decision problems that help in the search for the best decision using Backward Induction