Q-4 eaitil, a constant risk Study Problem 4.1 (Kreps 1988, Problem 7, text p.70)

Question

Q-4

eaitil, a constant risk Study Problem 4.1 (Kreps 1988, Problem 7, text p.70) 3. Suppose a decision maker has constant absolute risk aversion over the range -$100 to $1000. We ask her for her certainty equivalent for gamble with prize $0 and $1000, each with probability one-half, and she says that her certainty equivalent for this gamble is $488. What should she choose, if faced with the choice of a: a gamble with prizes -$100, $300, and $1000, each with probability 1/3; a': a gamble with prize $530 with probability ¾ and SO with probability ¼; or a": a gamble with a sure thing payment of $385? (Hints: constant absolute risk aversion, we have utilityfunction (z)s-ae-uth zE[-100, 10001;assume u (0) = 0, u(1 000) = 1; and we have u(488) = 2) + u(100); and then find U(a), U(a')and U(a")) 4. An investor has $1000 to invest in speculative stocks. The investor is considering investing $a dollars in stock A and $(1000 - a) in stock B. An investment in stock A has a 0.6 chance of doubling in value, and 0.4 chance of being lost. An investment in stock B has 0.7 chance of doubling in value and a 0.3 chance of being lost. The investor's utility function for a change in fortune (outcome) z, is utz) log (0.0007z +1) z E [-1000, 1000]. a) What is Z (the outcome set for a fixed a)? b) What is the optimal value of a in terms of expected utility? c) What is the optimal value of a in terms of expected utility when ulz) - log (z+ 3000).

Explanation / Answer

a)

The outcome for stock A is,

Za = 2 * $a * 0.6 + 0 * 0.4 = $1.2a

The outcome for stock B is,

Zb = 2 * ($1000 - a) * 0.7 + 0 * 0.3 = $(1400 - 1.4a)

The outome set is {$1.2a, $(1400 - 1.4a)}

b)

The Expected utility for stock A is,

Ua = 0.6 log(0.0007 * 2a + 1) + 0.4 log(0.0007 * 0 + 1) = 0.6 log(0.0014a + 1)

The maximum value of a can be $1000. So, the maximum value of Ua is,

Ua = 0.6 log(0.0014 * 1000 + 1) = 0.5253

The outcome for stock B is,

Ub = 0.7 log(0.0007 * 2(1000-a) + 1) + 0.3 log(0.0007 * 0 + 1) = 0.7 log(2.4 - 0.0014a)

The minimum value of a is 0. So, the maximum value of Ub is,

Ub =  0.7 log(2.4 - 0.0014 * 0) = 0.6128

The maximum expected utility is 0.6128 for a = 0. So the optimal value of a = 0

c)

The Expected utility for stock A is,

Ua = 0.6 log( 2a + 3000) + 0.4 log( 0 + 3000) = 0.6 log(2a + 3000) + 3.203

The maximum value of a can be $1000. So, the maximum value of Ua is,

Ua = 0.6 log(2*1000 + 3000) + 3.203 = 8.313

The outcome for stock B is,

Ub = 0.7 log(2(1000-a) + 3000) + 0.3 log( 0 + 3000) = 0.7 log(5000-2a) + 2.402

The minimum value of a is 0. So, the maximum value of Ub is,

Ub = 0.7 log(5000-2*0) + 2.402 = 8.364

The maximum expected utility is 8.364 for a = 0. So the optimal value of a = 0

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