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Lecture 5

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Option Value

Value relates to the willingness to pay to guarantee the availability of the services of a good for future use by the individual.

Concept was introduced by Weisbrod (1964) in considering a national park and the prospect of its closure. He argued the benefit of keeping the park open would be understated by just measuring current consumer surplus for visitors and that there should be added to that a measure of the benefit of future availability -> called option value.

U

U(Y)

U(A)

A

C

YA-Y*: option price, OP, the maximum amount that the individual would be willing to pay for the option which would guarantee access to an open park.

YA-Y**: expected value of the individual’s compensating surplus1), E[CS].

Option value, OV: OP - E[CS]=Y**-Y*.

“Option value is a risk aversion premium” (Cicchetti and Freeman, 1971, p.536)

N

U(N)

YN

Y*

Y**

YA

Y

Figure 13.2 Risk aversion, option price and option value (Perman et al.: page 448)

1) Compensating surplus: WTP for improvement of the environment to happen.

Irreversibilities

Examples:

loss in biodiversity

landscape changes

GHG emissions

several forms of pollution (pesticides, SO2)

sunk investment costs

Irreversibility with future known.

A: amenity value of wilderness area

MC: marginal costs

MB: marginal benefits

NB: net benefits

MNB: marginal net benefits

Irreversibility with future known.

A1: amenity value now, period 1

A2: amenity value in the future, period 2

MNB2 > MNB1

A1NI: MNB1 = 0.

A2NI: MNB2 = 0.

A2NI > A1NI

Irreversibility with future known.

- Considering irreversibility:
- Optimal preservation at AI1, AI2
- Cost of irreversibility:
- area abc in period 1
- area def in period 2
- Cost of ignoring irreversibility: edhi – abc.

Irreversibility with future unknown.

Figure 13.5 Irreversibility and development with imperfect future knowledge

(Perman et al.: page 454)

MNB known in period 1, unknown in period 2

E[MNB2]=p*MNB21 + (1-p) MNB22

Irreversibility implies: A2 A1

AI1, AI2: outcome where there is irreversibility but no risk.

If MNB21, A1NI is chosen.

If MNB22, A2I is chosen

If uncertainty about MNB2, outcome AI1, AI2, if p = 0.5.

- Notation and assumptions:
- D: development, P: preservation, Ri: return from i-th option
- Bpt: preservation benefits, Bdt: development benefits
- Cdt: development costs, only in the period development project is undertaken
- two periods t, 1 now and 2 the future,
- benefits, B, and costs, C, in period 2 are in present values (discounted)
- DM has complete knowledge of all period 1 conditions
- at the start of period 1, period 2 outcomes can be listed and probabilities attached to them
- at the end of period 1, complete knowledge about period 2 will become available to the DM
- decision to be taken at the start of period 1 is whether to permit development

Two-period development/preservation options.

return immediate development: R1 = (Bd1 – Cd1) + Bd2

return preservation in period 1: either R2 or R3

Two-period development/preservation options.

- return preservation in period 1: either R2 or R3 (Option 2 or 3)
- => if Bp2 > (Bd2 - Cd2) => choose R3, preservation period 2
- returns from preservation in the first period, Rp :
Rp = Bp1 + max{Bp2,(Bd2 - Cd2)}

Two-period development/preservation options.

assume complete knowledge over future circumstances

=> develop if Rd > Rp -> Rd - Rp > 0

=> (Bd1 - Cd1) + Bd2 - Bp1 - max{Bp2,(Bd2 - Cd2)} > 0

(Bd1 - Cd1) - Bp1 actually known to DM, other terms not, that’s why it is not an operational decision rule

- assume complete knowledge over future circumstances
- => develop if Rd > Rp -> Rd - Rp > 0
- => (Bd1 - Cd1) + Bd2 - Bp1 - max{Bp2,(Bd2 - Cd2)} > 0
- (Bd1 - Cd1) - Bp1 actually known to DM, other terms not, that’s why it is not an operational decision rule
- Now, assume possibleoutcomes of Bd2, Bp2,(Bd2 - Cd2) are known and DM can assign probabilities to the mutually exclusive outcomes.
- => (Bd1 - Cd1) - Bp1 +E[ Bd2] - max{E[Bp2],E[(Bd2 - Cd2)]} > 0
- ignores, more information are available at the start of period 2.

- Now, assume possible outcomes of Bd2, Bp2,(Bd2 - Cd2) are known and DM can assign probabilities to the mutually exclusive outcomes.
- => (Bd1 - Cd1) - Bp1 +E[ Bd2] - max{E[Bp2],E[(Bd2 - Cd2)]} > 0
- ignores, more information are available at the start of period 2.
- if area is developed in period 1, information cannot be used
- if area is preserved in period 1, information can be used whether or not to develop in period 2
- but decision has to be made in period 1, but DM also knows the outcomes and the probabilities

- Now, assume possible outcomes of Bd2, Bp2,(Bd2 - Cd2) are known and DM can assign probabilities to the mutually exclusive outcomes.
- => (Bd1 - Cd1) - Bp1 +E[ Bd2] - max{E[Bp2],E[(Bd2 - Cd2)]} > 0
- but decision has to be made in period 1, but DM also knows the outcomes and the probabilities
- this leads to the following decision rule, develop if:
- Rd = (Bd1 - Cd1) - Bp1 +E[ Bd2] - E[max{ Bp2,(Bd2 - Cd2)}] > 0

Develop if, (considering availability of future information):

(Bd1 - Cd1) - Bp1 +E[ Bd2] - E[max{ Bp2,(Bd2 - Cd2)}] > 0

Develop if, (ignoring availability of future information):

(Bd1 - Cd1) - Bp1 +E[ Bd2] - max{E[Bp2],E[(Bd2 - Cd2)]} > 0

Quasi option value:

E[max{ Bp2,(Bd2 - Cd2)}] - max{E[Bp2],E[(Bd2 - Cd2)]} > 0

The amount by which the net benefits form development project

that includes irreversible costs have to be reduced.

=> reflects the benefits of keeping the option alive for future preservation

Simple numerical example:

max{E[Bp2],E[(Bd2 - Cd2)]} =

max{E[0.5*10 + 0.5 * 5], E[0.5*6 + 0.5 * 6]} = max {7.5, 6}= 7.5

develop if : (Bd1 - Cd1) - Bp1 +E[ Bd2] - 7.5 > 0

result: 7.75 - 7.5 = 0.25 > 0.

Simple numerical example:

- Now consider E[max{ Bp2,(Bd2 - Cd2)}], there are two outcomes
- A where Bp2 > (Bd2 - Cd2), Bp2 =10, pA = 0.5
- B where Bp2 < (Bd2 - Cd2), (Bd2 - Cd2)= 6, pB = 0.5
- Hence: E[max{ Bp2,(Bd2 - Cd2)}] = (0.5 * 10) + (0.5 * 6) = 8
- develop if : (Bd1 - Cd1) - Bp1 +E[ Bd2] - 8 > 0
- result: 7.75 - 8.0 = - 0.25 < 0 ! => don’t develop first period

Simple numerical example:

QOV= E[max{ Bp2,(Bd2 - Cd2)}] - max{E[Bp2],E[(Bd2 - Cd2)]}

= 8 - 7.5 = 0.5.

The QOV is always positive, as it allows to reduce losses compared to the situation where the arrival of information is ignored.