Problem Set 3 Solutions

Transcription

1 Problem Set 3 Solutions Ec 030 Feb 9, 205 Problem (3 points) Suppose that Tomasz is using the pessimistic criterion where the utility of a lottery is equal to the smallest prize it gives with a positive probability. For example, let P = (p, 0; q, 20; p q, 30). If p > 0, then U(P ) = 0, if p = 0 and q > 0, then U(P ) = 20 and if p = q = 0, then U(P ) = 30. (a) Is Tomasz risk averse? For ease of discussion, let us approach the question using the example lottery. We can show that Tomasz is risk averse by comparing his utility from participating in the lottery to receiving the EMV for sure. Given any (p, q) consistent with probability theory, we know that EMV (P ) = 0p+20q+30( p q), i.e. a probability weighted average of the three outcomes. Moreover, getting the EMV for sure is essentially the same as participating in a lottery P where we have a probability of receiving EMV (P ). In other words, in this particular case the utility of getting the EMV for sure is simply the EMV value itself: U(P ) = EMV (P ). Case : Suppose p > 0, then by inspection EMV (P ) U(P ) = 0 since the weighted average of three numbers should be at least as great as the smallest one. Furthermore, if 0 < p <, we have the strict inequality EMV (P ) > U(P ) = 0. Case 2: Suppose p = 0 and q > 0, then EMV (P ) U(P ) = 20. Furthermore if 0 < q <, EMV (P ) > U(P ) = 20. Case 3: Suppose p = 0 and q = 0, then EMV (P ) = U(P ) = 30. We see that Tomasz assigns at most the same amount of utility as EMV, but usually he would strictly prefer getting the EMV for sure. Consequently, we conclude he is risk averse. More generally, given any lottery Q with n different payoffs {x, x 2,..., x n } with respective probabilities {p, p 2,..., p n }. Without loss of generality, we can further restrict that x i < x j if i < j. Now, suppose that j is the smallest index for which p j > 0. Then by definition we can calculate the EMV as EMV (Q) = n p i x i = i= n p i x i. () Under the pessimistic criterion, Tomasz would assign U(Q) = x j to the lottery. Again, since EMV (Q) is a probability-weighted average of values x j and greater, we must have that EMV (Q) U(Q) = x j. Please huijiawu@fas.harvard.edu regarding errors/typos i=j

2 Furthermore, this inequality is strict if 0 < p j <. Hence, for all possible lotteries Q, Tomasz would (at least weakly) prefer getting EMV (Q) for sure. Thus, we conclude that he is risk averse. (b) Can Tomasz s preferences be represented by expected utility? No, Tomasz s preferences cannot be represented by expected utility. We can show this using proof by contradiction. Suppose that there exists a specification for a Bernoulli utility function u( ) such that Tomasz s prefrences can be represented by expected utility. This means that given any lottery Q in the general form described in the solution for part (a), we have EU(Q) = n p i u(x i ). (2) i= More specifically, the above should also apply for the example lottery P = (p, 0; q, 20; p q, 30). It is important to note here that u( ) does not change with whichever lotteries we are considering. Therefore, given any (p, q) consistent with probability theory we need EU(P ) = pu(0) + qu(20) + ( p q)u(30). (3) Note that since Tomasz is indifferent among all lotteries that has p (0, ], we need EU(P ) = ū across this subset of possible lotteries. Since there is an infinite number of possible p s, to be consistent with Equation 3 it must be the case that u(0) = u(20) = u(30) = ū. Then EU(P ) = ū for all possible values of (p, q), since u( ) cannot change with (p, q). However, when p = 0, we know that U(P ) = 20 > 0 or U(P ) = 30 > 0. In this case, we would need EU(P ) > ū to represent Tomasz s preference for the types of P lotteries that have p = 0 over those with p > 0. Thus, we arrive at a contradiction, and hence ends the proof. Alternatively, we can prove our claim by showing that Tomasz s preferences violate the Independence Axiom. Recall that one of the necessary conditions for the existence of an expected utility representation is the satisfaction of the Independence Axiom. Consider lotteries P i = (p i, 0; q i, 20; p i q i, 30) for i =, 2, 3, where p = q = 0, p 2 = 0, q 2 =, p 3 =, and q 3 = 0. In other words, P gives us 30 for sure, P 2 gives us 20 for sure, and P 3 gives us 0 for sure. Tomasz would strictly prefer lottery P to lottery P 2, since U(P ) = 30 > 20 = U(P 2 ). According to the Independence Axiom, since Tomasz P strictly prefers to P 2, then for any α (0, ) he must also strictly prefer αp + ( α)p 3 to αp 2 + ( α)p 3. However, these two compound lotteries both have a non-zero probability in the payout of 0. Hence, U(αP + ( α)p 3 ) = U(αP 2 + ( α)p 3 ) = 0, and Tomasz should be indifferent between these two new lotteries. This contradicts what the Independence Axiom requires. (c) Suppose that David has expected utility preferences. Is there a specification of a Bernoulli utility function u for David that would make him more risk averse than Tomasz? If yes, find one, if not show that all possible u make David less risk averse than Tomasz. No, such a specification does not exist; all possible u( ) would make David less risk averse than Tomasz. Again, we can use proof by contradiction to reason through this question. Suppose that such a Bernoulli utility function u( ) for David existed so that he is more risk averse than Tomasz. Then for every lottery Q and sure outcome x, if Tomasz prefers x to Q, then David also prefers x to Q. Let us create a lottery R that pays out y and y 2 with probabilities r and r 2 respectively. Also, let 2

3 y be the lesser of the two outcomes. Considering the infinite number of lotteries with r (0, ], then under the pessimistic criterion Tomasz would prefer getting y or more for sure over participating in any of these lotteries. Since David is more risk averse, then he must also prefer getting y or more for sure over lottery R. Additionally y 2 > y, so we also must have that u(y 2 ) u(y ). Mathematically, we have that u(y ) r u(y ) + r 2 u(y 2 ) u(y ) u(y ) = r u(y ) + r 2 u(y 2 ) ( r )u(y ) = r 2 u(y 2 ) u(y ) = u(y 2 ), (4) where the last line comes from the fact that r 2 = r. The above then must holds for all y 2 y. But since y can be arbitrarily small, we can take y. The only function u( ) consistent with this is if u( ) = ū for some constant ū. However, if u( ) were constant, then the David is indifferent between any lottery and its EMV, whereas Tomasz prefers getting the EMV for sure. Thus even in this corner case, we cannot make David more risk averse than Tomasz so we reach a contradiction. 3

4 Problem 2 (2 points) A corporation must decide between three mutually exclusive projects: Project P has an equal probability of four gross payoffs: $80 million, $00 million, $20 million, and $40 million, Project Q has an equal probability of three gross payoffs: $90 million, $0 million, and $30 million, Project R has an equal probability of two gross payoffs: $90 million and $30 million. Suppose that you know that the CEO of the corporation has expected utility preferences and is risk averse. Can you determine his preferences over P, Q, and R? Similar to what we did in lecture, we can use the following tree diagrams to illustrate each lottery: $40 mil $30 mil $30 mil /4 $20 mil /3 /2 P /4 /4 $00 mil Q /3 $0 mil R /4 /3 /2 $80 mil $90 mil $90 mil Let u( ) denote the Bernoulli utility function associated with the expected utility preferences of the CEO. Since the CEO is risk averse, we know that his utility function u( ) is concave. In other words, for any (x, y) and any p (0, ) we have u(px + ( p)y) pu(x) + ( p)u(y). (5) Using this concavity feature, we can determine the CEO s preferences over P, Q, and R. More precisely, when the CEO has expected utility preferences and is risk averse, he will (at least weakly) prefer project Q over project R over project P. To prove this we need to show that given any concave u( ) we have EU(Q) EU(R) EU(P ). For ease of reading, define M =, 000, 000. EU(Q) = 3 u(90m) + 3 u(0m) + 3 u(30m) = 3 u(90m) + ( 3 u 2 (90M) + ) 2 (30M) + 3 u(30m) 3 u(90m) + [ 3 2 u(90m) + ] 2 u(30m) + 3 u(30m) = 2 u(90m) + 2 u(30m) = EU(R) 4

5 Thus we have shown that EU(Q) EU(R) for any concave u( ) function. Similar to the reasoning above, we can also demonstrate that EU(R) EU(P ). EU(R) = 2 u(90m) + 2 u(30m) = 2 u ( 2 (80M) + 2 (00M) ) + 2 u ( 2 (20M) + 2 (40M) ) 2 ( 2 u(80m) + ) 2 u(00m) + 2 ( 2 u(20m) + ) 2 u(40m) = 4 u(80m) + 4 u(00m) + 4 u(20m) + 4 u(40m) = EU(P ) Putting these results together, we get EU(Q) EU(R) EU(P ) for all concave Bernoulli utility functions u( ). To further illustrate our results, let us consider the case when u(x) = x. Then EU(P ) = 4 ( 80M + 00M + 20M + 40M) = , EU(Q) = 3 ( 90M + 0M + 30M) = , and EU(R) = 3 ( 90M + 30M) = So under this particular specification we have shown that the CEO prefers project Q, then project R, then project P, consistent with our general results. 5

6 Problem 3 (3 points) Consider the insurance example we studied in Lecture 8. We derived the demand for insurance when the agent was using the probability weighting formula and the price γ was actuarially fair (γ = p). Suppose now that the price is some arbitrary number γ, not necessarily equal to p. Derive the range of γ for which the agent is willing to buy full insurance and the range where he wants zero insurance. The agent wants to choose the optimal λ dollars of insurance to maximize the following. Without insurance, the agent receives V (P λ ) = π(p)[w L + λ( γ)] + ( π(p))[w λγ] = w π(p)l + π(p)λ( γ) λγ + π(p)λγ = w π(p)l + λ[π(p) γ] (6) V (P 0 ) = π(p)[w L] + ( π(p))w = w π(p)l. (7) Thus we see that the agent s decision depends on the sign of the coefficient on λ, π(p) γ. In other words, his/her choice boils down to the relation between π(p) and γ. Case : π(p) > γ, agent is willing to buy full insurance. Case 2: π(p) < γ, agent wants zero insurance. Case 3: π(p) = γ, agent is indifferent about getting insurance. 6

7 Problem 4 (2 points) A state lottery sells tickets for a cost of $ each. The ticket has a probability of /(2,400,000) of winning $,000,000 and otherwise nothing. (a) What is the EMV of buying one ticket? EMV = 2, 399, 999 ($, 000, 000 $) + 2, 400, 000 2, 400, 000 ( $) = ($, 000, 000) $ 2, 400, 000 = $ (8) (b) In the hope of increasing profits, the state considers increasing the award to $2,000,000 and reducing the probability to /(4,800,000). A statistician said that it s not worth the trouble, because the expected profit remains precisely the same. What do you think? The EMV of this alternative lottery is EMV = 4, 799, 999 ($2, 000, 000 $) + 4, 800, 000 4, 800, 000 ( $) = ($2, 000, 000) $ 4, 800, 000 = $ (9) Although the statistician is correct in the sense that the expected profit remains the same, there is still some merit in switching to the new lottery. First of all, potential lottery buyers do not assess lotteries using EMV; otherwise, they would not participate in either lottery since the EMV is negative. Potential buyers might overweight the probability of winning due to their hopes of receiving large gains (i.e. possibility effect). Additionally, the relative degree of overweighting tends to increase as the winning probabilty approach zero. If this were the case, it is possible for potential buyers to value the alternative lottery more than the original one. Furthermore, buyers may also underestimate the probability of not winning the lottery. (c) Use prospect theory (with u(x) = x) to formalize your argument. What features of the probability weighting function does your argument rely on? Assuming u(x) = x for potential buyers and let π( ) be the probability weighting function, under prospect theory they evaluate each lottery using π(p)u(x ) + π( p)u( ) = π(p)(x ) π( p) (0) where (p, x) are the probability of and amount gained in winning the lottery, respectively. Furthermore, π(p) is consistent with the fourfold pattern we saw in lecture. Note that this means agents would overweight small probabilities and underweight high probabilities. Due to the possibility effect, π(p) is steep at p = 0 (steeper than 45 degree line). However, π(p) flattens (becomes less steep than 45 degree line) at intermediate values of p ( meh effect) and steepens at high values of p (certainty effect). Let the subscripts a and b represent the two lotteries in parts a and b respectively. We want to compare 7

8 the values that agents assign for each lottery using Equation 0, noting that In particular, we want to show that it is possible for agents to prefer lottery b over lottery a, and that some agents who might not participate in lottery a may start participating in lottery b. [π(p b )(x b ) π( p b )] [π(p a )(x a ) π( p a )] = [π(p b )x b π(p a )x a ] + [π(p a ) π(p b )] + [π( p a ) π( p b )] }{{}}{{}}{{} (), + (2), + (3), () We see that the difference in the anticipated value between lottery a and b can be broken down into three parts that we can sign. Part (): The weighting function flattens as p increases from 0 to intermediate values (i.e. locally concave between 0 and intermediate values). For any small p we should have 2 π(0) + 2π(2p) π(p) (since π(0) = 0, we can rewrite this inequality as π(2p) 2π(p)). The fact that p a = 2p b implies π(p a ) = π(2p b ) 2π(p b ). So π(p b )x b π(p a )x a 2 π(p a)(2x a ) π(p a )x a = 0. Parts (2) and (3): The weighting function is monotonically increasing (i.e. p < p implies π(p) π(p )). Since p a > p b, we know π(p a ) π(p b ) 0. Similarly, p a < p b implies π( p a ) π( p b ) 0. So when the probability weighting function π( ) is shaped in a way that () and (2) together dominate (3), the agent would prefer lottery b over a. In other words, let ɛ 0 denote the wedge between π(p b ) and 2 π(p a), so that π(p b ) = 2 π(p a) + ɛ. Similarly, we define ɛ 2 = π(p a ) π(p b ) 0 and ɛ 3 = π( p b ) π( p a ) 0. Then Equation is nonnegative if 2ɛ x a + ɛ 2 ɛ 3 0. Since 2x a = x b is very large, we see that () and (2) together would tend to dominate (3) for nonextreme values of ɛ 3. (Note that we cannot claim this for all possible weighting functions π( ), unless we restrict the relative degree of overweighting in small probabilities to underweighting in large probabilities.) We now have shown that it is possible for agents to prefer lottery b over lottery a. Additionally, since there is a wedge between the anticipated value of lotteries a and b, we can construct instances where π(p b )x b π( p b ) > 0 and π(p a )x a π( p a ) < 0. In this case, by switching from lottery a to lottery b, agents would switch from non-participation to participation. Lastly, note that the overweighting of low probabilities (i.e. π(p b ) > p b ) and underweighting of high probabilities (i.e. π( p b ) < p b ) make it possible for agents to view lotteries favorably, whereas the EMV approach would lead to nonparticipation in both lotteries. Alternative solution of part (c) Imagine that we drive the winning probability to zero and keep the EMV of the lottery constant. That is, P = (p, x ; p, ) is a lottery such that EMV (P ) = c, where c is constant. We want to show that for very small p the anticipated value of the lottery is bigger than zero. The constant EMV condition looks like this: p(x ) + ( p)( ) = c, which means that x = c+ p. The anticipated value of the lottery is π(p)(x ) + π( p)( ). We want to show this is positive for small p. 8

9 Let s write this condition down using the formula for x π(p)( c + p ) > π( p) π(p)(c + p) > pπ( p) π(p) p > π( p) π(p) c + Now as p goes to zero, we have the right hand side numerator going to one and the denumerator is just =0.47, so the whole right hand side goes to 2.4. On the other hand the left hand side is approximately equal to π (0), so as long as π (0) > 2.5 the agent will like a lottery with small p (and large x). 9