Topics in Economic Analysis 1 (TEA1)Problem Set 1 - SKETCH of Answers

1. (a) Owing an uninsured car leads to uncertain wealth. Greg’s expected utility in theabsence of insurance when parked in the street is

EUstreetuninsured = 0.2√

90, 000− 50, 000 + 0.8√

90, 000 = 280

Greg’s expected utility when uninsured and parked in a garage:

EUgarageuninsured = 0.1

√90, 000− 50, 000− 599 + 0.9

√90, 000− 599 = 288.95

> 280 = EUstreetuninsured

Thus, in the absence of insurance, Greg will park in the garage.

(b) Full insurance leads to certain wealth. Greg’s utility when fully insured andparked in a garage:

EUgaragefull =

√90, 000− 5, 000− 599 = 290.52

and his utility when fully insured and parked in the street:

EU streetfull =√

90, 000− 5, 000 = 291.5

Thus, with full insurance Greg will park in the street.

Once Greg parks in the street which leads to the chance of loss being 0.2, whilebeing fully insured at 0.1 per unit of coverage, the insurer makes losses of (0.1−0.2)50, 000 = −5, 000.

(c) This is a moral hazard problem (hidden action is present).

2. (a) With the partial coverage, Greg’s utility of this coverage with garage parking is:

EUgaragepartial = 0.9

√90, 000− 599− 0.1X + 0.1

√90, 000− 599− 0.1X − 50, 000 +X =

= 0.9√

89, 401− 0.1X + 0.1√

39, 401 + 0.9X

His utility of this coverage with street parking is:

EU streetpartial = 0.8√

90, 000− 0.1X + 0.2√

90, 000− 0.1X − 50, 000 +X =

= 0.8√

90, 000− 0.1X + 0.2√

40, 000 + 0.9X

We are looking for the partial coverage X that would make Greg indifferent be-tween garage and street parking, i.e. EUgaragepartial = EU streetpartial. Thus, one needs to

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solve this equation for X. This is a formidable task, and you are not expected tosolve it. (One can solve this equation numerically, for example, using Mathemat-ica software, or graphical calculator, or even playing around with Excel.) Thus,for this course, it is sufficient to write the above equation and substitute in it thevalue X equal to $44,083.5:

EUgaragepartial = 0.9

√89, 401− 0.1× 44, 083.5 + 0.1

√39, 401 + 0.9× 44, 083.5 = 290.502

EUstreetpartial = 0.8√

90, 000− 0.1× 44, 083.5 + 0.2√

40, 000 + 0.9× 44, 083.5 = 290.502

We always assume in this type of problems that, when indifferent, a risk-averseindividual will take the action because this would reduce risk. Thus the insurerwill make zero profit here as probability of loss when the action is taken equalsto the per-unit premium 0.1.

To see that the condition EU garagepartial = EUstreetpartial) indeed determines the maximum

coverage, let us consider two alternatives. First, suppose instead that the insureroffers a higher partial coverage of X1 > X, at the premium per unit of 0.1.Consider, for example, X1 = 44, 100. It is easy to show that in this case Gregwould buy the partial coverage and park in the street, as EUgaragepartial < EU streetpartial,so that the insurer incurs losses. Now consider instead that the insurer offers alower partial coverage of X2 < X, at the premium per unit of 0.1. Consider, forexample, X2 = 44, 000. It is easy to show that in this case Greg would buy thepartial coverage and park in the street, as EUgaragepartial > EU

streetpartial - just what the

insurer wanted, as parking in the garage would lead to zero expected profits. Butgiven the free entry into insurance market, another insurer could come over andoffer Greg a slightly higher partial coverage of X3, at the premium per unit of0.1, such that X2 < X3 < X. Consider, for example, X3 = 44, 050. One can showthat Greg would prefer partial coverage X3 to partial coverage X2, so that Gregwould sign insurance contract with the new entrant, rather than with our originalinsurer.

Thus, the only way an insurer can make business with Greg and not to lose moneyis to offer him a partial contract such that Greg is indifferent between parking inthe street and in the garage.

(b) Since the partial coverage is $44,083.5, the co-insurance value α = 44,083.550,000

100% =

88.17%.

3. (a) The actuarially fair “on average” insurance premium is

Premium = (.8× .5 + .2× .5) = 0.5

so that the total premium is 0.5×1, 000 = 500. Note that the actuarially fair “onaverage” premium, in effect, involves “subsidization” of rock-lovers by jazz-lovers.

(b) For either type, if fully insured then:

EUfull,0.5 = ln(10, 000− 0.5× 1, 000) = 9.1590

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The rock lovers will buy full coverage insurance at this premium because:

EU rockuninsured = .8 ln(10, 000− 1, 000) + .2 ln 10, 000 = 9.1261 < 9.1590 = EUfull,0.5

But the jazz lovers will be better off without insurance:

EUjazzuninsured = .2 ln(10, 000−1, 000)+ .8 ln 10, 000 = 9.18927 > 9.1590 = EUfull,0.5

(c) Since only the rock lovers buy insurance, the insurance company’s expected profitper unit of coverage is:

Eπ = 0.2× 0.5 + 0.8(0.5− 1) = 0.1− 0.4 = −0.3

Since the actual total premium per average contract is .8 × 1, 000 = 800 > 500,so the company is losing money.

(d) To make sure that the company is not loosing money, the premium should becloser to the probability of loss by the rock-lovers. Recommend the companyinstead to offer the insurance at a premium per unit of coverage of 0.8 at a totalpremium of $800.

If there is only one type of insurance contract offering full coverage at 0.8 per unitof coverage, the rock lovers still fully insure - even though this is more expensiveinsurance:

EU rockfull,0.8 = ln(10, 000− 0.8× 1, 000) = 9.1269 > EUrockuninsured = 9.1261

But the jazz lovers still opt for no insurance as

EUjazzfull,0.8 = ln(10, 000− 0.8× 1, 000) = 9.1269 < EU jazzuninsured = 9.18927

Yet the company will break even, as this is an actuarially fair policy for the rocklovers.

(e) This is an adverse selection problem (hidden characteristics are present).

4. If the company can not see the music taste, then the company should offer two differentpolicies: one offering full coverage at $800, and the other offering partial coverage ofX at total premium 0.2X, where X is such that the rock-lovers are indifferent betweenfull and partial coverage, and the jazz-lovers are better off buying the partial coverage.

EU rockpartial = .8 ln(10, 000− 1, 000 +X − 0.2X) + .2 ln(10, 000− 0.2X) =

= .8 ln(9, 000 + 0.8X) + .2 ln(10, 000− 0.2X) = EU rockfull = 9.1269

This is a difficult equation to solve (so just writing the above equation would be suffi-cient). If one solves the above numerically, the solutions are $12.644 and $49,942.64.Obviously, the second value is too high for the company to offer. As usual for this

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type of problems, we assume that the rock lovers will take the full insurance whenindifferent, so that the insurer will make zero profits on their contracts.

Check that the value of $12.644 also works for the jazz lovers:

EUjazzpartial = .2 ln(9, 000 + 0.8× 12.644) + .8 ln(10, 000− 0.2× 12.644) = 9.18929

which is just a bit better than being uninsured.

Again, to understand why we use the condition EU rockpartial = EUrockfull , let us consider two

alternatives. First, suppose instead that the insurer offers a higher partial coverage ofX1 > X, at the premium per unit of 0.2. Consider, for example, X1 = 13. It is easy toshow that in this case rock lovers would buy the partial coverage as EUrockpartial > EU

rockfull ,

so that the insurer incurs losses. Now consider instead that the insurer offers a lowerpartial coverage of X2 < X, at the premium per unit of 0.2. Consider, for example,X2 = 12. It is easy to show that in this case rock lovers would buy the full coverage, asEU rockpartial < EU

rockfull - just what the insurer wanted, as buying full coverage at premium

of 0.8 by rock lovers would lead to zero expected profits. But given the free entry intoinsurance market, another insurer could come over and offer a slightly higher partialcoverage of X3, at the premium per unit of 0.2, such that X2 < X3 < X. Consider, forexample, X3 = 12.6. But that would attract jazz lovers as EU jazzpartial,X3

> EUjazzpartial,X2

,so that jazz lovers would sign insurance contract with the new entrant, rather thanwith our original insurer.

Thus, the only way an insurer can make business with both jazz and rock loversand not to lose money is to offer one full insurance designed for rock lovers and onepartial insurance designed to make the rock lovers to be indifferent, and which is moreattractive than no insurance by the jazz lovers.

5. * This problem is based on the material of Topic 2a “Moral Hazard in the InsuranceMarket: Continuous Case”.

(a) For any given level of care e, Barbara’s total expected utility is:

EU(e) =

(1−

1

e

)√L+

1

e

√0− 0.5e2 =

(1−

1

e

)√L− 0.5e2

First-order condition: √L

e2− e = 0⇒ e3NI =

√L

Thus, Barbara’s optimal choice in the absence of insurance is eNI = L1/6 =2.92402.

(b) If the level of care was observable, and if an insurer offered an actuarially fair fullcoverage based on the level of care, Barbara’s utility is:

EU(e) =

(1−

1

e

)√

L

(1−

1

e

)+1

e

√

0 + L

(1−

1

e

)−0.5e2 =

√

L

(1−

1

e

)−0.5e2

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First-order condition:√L

2e2√

1− 1

e

− e = 0⇒ 2e3FI

√1−

1

eFI=√L

Thus, the implicit equation for eFI is

4e6FI − 4e5FI − L = 0⇒ 4 · 2.524346 − 4 · 2.524345 − 625 ≈ 0

(c) To show that eFI < eNI , note that

e3NI =√L and 2e3FI

√1−

1

eFI=√L

so that

e3NI = 2e3FI

√1−

1

eFI⇒(eNI

eFI

)3= 2

√1−

1

eFI

But since e ∈ [2, 3], we have that, for any e

1.41 < 2

√1−

1

e< 1.63

Thus, (eNI

eFI

)3> 1⇒ eNI > eFI

An alternative way involves using the general method presented in Topic 2a.

Notice that our functional forms satisfy the requirements that v′ > 0, v′′ < 0,d′ > 0, d′′ > 0, p′ < 0, p′′ > 0, as well as that Lv′(L(1− p(emin)) < v(L). Equatethe two first-order conditions (as both are equal to zero):

−p′(eNI)v(L)− d′(eNI) = −Lp′(eFI)v′(1− p(eFI)− d′(eFI)

Rearrange:

d′(eNI)− d′(eFI) = Lp′(eFI)v′(1− p(eFI)− p′(eNI)v(L) (1)

Since eFI > emin, we have that 1− eFI < 1− emin, and since p′ < 0, we have thatp(1− eFI) > p(1− emin), i.e.

eFI > emin ⇒ 1− eFI < 1− emin ⇒ p(1− eFI) > p(1− emin)

Now recall that v′′ < 0. That means that v′ is a decreasing function. Thus sincep(1− eFI) > p(1− emin) we have that v′(p(1− eFI)) < v′(p(1− emin)), i.e.

p(1− eFI) > p(1− emin) ⇒ v′(p(1− eFI)) < v′(p(1− emin))⇒⇒ Lv′(p(1− eFI)) < Lv′(p(1− emin))

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That leads us to

Lv′(p(1− eFI)) < Lv′(p(1− emin)) (2)

Now the technical condition Lv′(L(1−p(emin)) < v(L) means that the right-hand-side of (2) is less than v(L). In turn, that means that the left-hand-side, also isless than v(L). In other words:

Lv′(p(1− eFI)) < Lv′(p(1− emin)) < v(L)⇒Lv′(p(1− eFI)) < v(L)

Now because p′ < 0, we have that

p′(eFI)Lv′(1− p(eFI)) > p′(eFI)v(L)

We can now use the above to evaluate the right-hand-side of the equation (1):

p′(eFI)Lv′(1− p(eFI))− p′(eNI)v(L) > p′(eFI)v(L)− p′(eNI)v(L) = v(L)(p′(eFI)− p′(eNI))

Combining this with the left-hand-side of the equation (1):

d′(eNI)− d′(eFI) > v(L)(p′(eFI)− p′(eNI)) (3)

This is our main inequality. We need to consider three cases:

• Case 1: eFI = eNI . This cannot be the case as rather than inequality (3) wewould have equality:

d′(eNI)− d′(eFI) = v(L)(p′(eFI)− p′(eNI))

• Case 2: eFI < eNI . Since d′′ > 0 we have than d′(eFI) > d′(eNI) so that the

left-hand side is negative. Since p′′ > 0 we have than p′(eFI) > p′(eNI) sothat the right-hand side is positive:

d′(eNI)− d′(eFI)︸ ︷︷ ︸<0

< v(L)(p′(eFI)− p′(eNI))︸ ︷︷ ︸>0

so that instead of inequality (3) we got the opposite inequality. Thus, wecannot have eFI < eNI .

• Case 3: eFI > eNI . Since d′′ > 0 we have than d′(eFI) < d′(eNI) so that the

left-hand side is positive. Since p′′ > 0 we have than p′(eFI) < p′(eNI) so that

the right-hand side is negative:

d′(eNI)− d′(eFI)︸ ︷︷ ︸>0

> v(L)(p′(eFI)− p′(eNI))︸ ︷︷ ︸<0

so that we got inequality (3). Thus, the only case we can have is eFI > eNI .

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(d) Notice that with any full coverage, the monetary utility is the same, but thedisutility of care is increasing in care e. Thus, with unobservable care, Barbarashould choose the lowest possible level, e = 2, which leads to losses for the insurer:

Eπ = L(p(eFI)− p(emin)) = L(

1

eFI−

1

emin

)= L

emin − eFIemineFI

< 0

which is always negative because eFI > emin. To calculate profit numerically, get:

Eπ = 625

(1

2.52434−

1

2

)= −64.91

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