Decision Theory Final Exam Last name I term: (09:50) Instructor: M. Lewandowski, PhD

Transcription

1 Problem 1 [10p] Consider the probability density functions of the following six lotteries l1 l6: PDF l 1 l 2 l 3 l 4 l 5 l 6 2,00 0,00 0,10 0,00 0,50 0,50 0,00 0,00 0,50 0,00 0,50 0,00 0,00 0,90 3,00 0,00 0,90 0,50 0,00 0,50 0,00 5,00 0,50 0,00 0,00 0,50 0,00 0,10 a) [4p] Fill in the following table (FOSD first order stochastically dominates, SOSD second order stochastically dominates) l1 FOSD l3, l4, l5, l6 l1 SOSD l3, l4, l5, l6 l2 FOSD l5 l2 SOSD l4, l5 l3 FOSD l5 l3 SOSD l4, l5, l6 l4 FOSD l5 l4 SOSD l5 l5 FOSD l5 SOSD l6 FOSD l6 SOSD CDF l1 l2 l3 l4 l5 l6 2,00 0,00 0,10 0,00 0,50 0,50 0,00 0,00 0,50 0,10 0,50 0,50 0,50 0,90 3,00 0,50 1,00 1,00 0,50 1,00 0,90 5,00 1,00 1,00 1,00 1,00 1,00 1,00 FOSD l3,l4,l5,l6 l5 l5 l5 integral CDF l1 l2 l3 l4 l5 l6 2,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,00 0,20 0,00 1,00 1,00 0,00 3,00 1,50 0,50 1,50 2,50 2,50 2,70 5,00 2,50 2,50 3,50 3,50 4,50 4,50 SOSD l3,l4,l5,l6 l4,l5 l4,l5,l6 l CDF Integral from CDF For l1 a graphical illustration: 3,5*0,5=1, CDF Integral from CDF 5*0,5+0,5*1=

2 b) [2p] Consider the orders (ranking) based on FOSD [we say that x is on a higher position than y if x FOSD y]. Fill in the following table (YES/NO): Order based on FOSD is transitive Order based on FOSD is complete Order based on SOSD is transitive Order based on SOSD is complete YES NO YES NO c) [3p] Calculate standard deviations and riskiness measures for lotteries l4 and l5 [riskiness measure R is defined as a non trivial zero of a function E log(r+li) log(r)=0, where li is a given lottery] Standard deviation 3,5 2,5 Riskiness measure 10/3 6 l4 l5 d) [1p] Suppose you have to pay a price of 2/3 for lottery l4 and you are offered lottery l5 free of charge. Which option l5 or l4 2/3 has higher riskiness? l4 2/3 l5 Riskiness measure 6,93 6 Problem 2 [8p] Consider the following problem under uncertainty: CRISIS NO CRISIS SAFE 0 2 MEDIUM 3 6 RISKY 6 10 a) [2p] What is the optimal (pure) strategy (S, M, R) according to the following criteria: Strategy Value of a criterion Maximin S 0 Maximax R 10 Hurwicz (α=0,5) R 2 Laplace R 2 Minimax regret M 4 b) [4p] Suppose now that the decision maker can choose a mixed strategy (Safe with probability p1, Medium with probability p2, Risky with probability p3). Which strategy is now optimal using the following criteria: p1 p2 p3 Value of a criterion Maximin Laplace Minimax regret option 1 0 3

3 Minimax regret option [Tip for minimax regret: min max (regret in state 1, regret in state 2); in both states regrets should be equal] [1p] Can maximin ever be better when considering mixed strategies as compared to considering only pure strategies? YES / NO In our case the possibility of a mixed strategy does not improve maximin as compared to the situation where only pure strategies are allowed, because the minima of all the strategies (SAFE, MEDIUM and RISKY) occur in one column CRISIS. It will change however if we introduce another strategy HEDGE, for which the minimum is in the other column ( NO CRISIS ) than the minima of all the other strategies: CRISIS NO CRISIS SAFE 0 2 MEDIUM 3 6 RISKY 6 10 HEDGE 5 5 Notice that the value of a maximin in pure strategies is equal to 0 and is achieved for SAFE strategy. Now consider mixing strategies SAFE and HEDGE let s take SAFE + HEDGE. Expected payoff of this mixed strategy is equal to no matter whether CRISIS or NO CRISIS occurs. CRISIS SAFE + HEDGE NO CRISIS The minimum of these two payoffs is, which is better than maximin in pure strategies (0). Hence mixed strategies may improve maximin criterion. [1p] Can Laplace ever be better when considering mixed strategies as compared to considering only pure strategies? YES / NO Consider the same problem as above with an additional strategy HEDGE: CRISIS NO CRISIS EV SAFE MEDIUM 3 6 1,5 RISKY HEDGE We have listed Expected Value of each of the strategies (probability of CRISIS and NO CRISIS is assumed to be one half in Laplace criterion due to the principle of insufficient reason we don t know anything about the probabilities so we assume symmetry) in the column on the right. Notice that RISKY strategy is the best. Consider now mixed strategies of the form p SAFE +

4 p MEDIUM + p RISKY + p HEDGE, where p 0, and p = 1. Expected Value of such mixed strategy where CRISIS and NO CRISIS has probability one half is: EV MIXED = p EV SAFE + p EV MEDIUM + p EV RISKY + p EV HEDGE = = p 1 + p 1,5 + p 2 + p 0 And this expression will never be greater than 2 which is equal to EV(RISKY). Hence introducing mixed strategies cannot improve Laplace criterion compared to the situation where only pure strategies are allowed. Problem 3 [5p] Consider a choice function which satisfies property alpha, beta and gamma. Fill in the following tables (if you answer YES, then write by which single property you include it in the chosen set; if you answer NO, then also write by which single property you don t include it in the chosen set): Menu 1 a b d Chosen (YES/NO) YES NO NO Menu 2 a c Chosen (YES/NO) YES YES Menu 3 a b c d Chosen (YES/NO) YES NO YES NO By which property gamma alpha beta + alpha Menu 4 d c Chosen (YES/NO) NO YES By which property beta alpha + For educational purposes, we decompose properties alpha and beta into positive and negative versions (alpha +, beta +, alpha, beta, respectively). Positive version corresponds to the logical sentence p q, where p and q are positive sentences (e.g. x is chosen from set A), whereas negative version corresponds to the logical sentence q p, where p and q are positive sentences and hence p and q are negative sentences (e.g. x is not chosen from A). Both positive and negative versions are equivalent in logical sense. Property alpha + (positive version): if x is chosen from a bigger set, it must be chosen from a smaller set as well. Property alpha (negative version): if x was not chosen from a smaller set, it must not be chosen from a bigger set. Property beta + (positive version): if x and y are chosen from a smaller set and x is chosen from a bigger set, then y must be chosen from the bigger set as well. Property beta (negative version): if x was chosen and y was not chosen from a bigger set, then if x is chosen from a smaller set, then y must not be chosen from the smaller set. Problem 4 [3p] Consider a choice function which satisfies property alpha and beta. Fill in the following tables (If you answer YES, then write by which single or double (remember WARP is equivalent to

5 BOTH alpha and beta) property you include it in the chosen set. If you answer NO, then also write by which single or double property you don t include it in the choice set): Menu 1 b d Chosen (YES/NO) YES NO Menu 2 b c Chosen (YES/NO) YES NO Menu 3 a b d Chosen (YES/NO) YES YES NO By which property X alpha & beta alpha Menu 4 a b c Chosen (YES/NO) YES YES NO By which property alpha & beta X alpha WARP is equivalent to alpha & beta taken together and it says: if x and y are both in set A and set B, and if x is chosen from A and y is chosen from B, then y must be chosen from A and x must be chosen from B. Problem 5 [4p] A preference relation R defined on a finite set of objects (fruits) is complete and transitive. A utility function values u for a couple of objects is given below (function u represents preference relation R). Which of the following alternative utility functions u, u or u represent the same preferences R? apple orange banana lemon YES/NO u X u YES u YES u NO A binary preference relation is complete and transitive if and only if it may be represented by an ordinal utility function which is unique up to any increasing transformation. Hence any function that assigns numbers to alternatives in the same order as u represents the same ordinal preferences. Consider now a preference relation R defined on a finite set of lotteries. Preference relation R is complete, transitive, and additionally satisfies continuity and independence (expected utility axioms). A utility function values v for a couple of lotteries is given below (function v represents preference relation R ) Which of the following alternative utility functions v, v, v represent the same preferences R? l1 l2 l3 l4 YES/NO v X v NO v YES v NO

6 Preference relation on the set of lotteries satisfy expected utility axioms if and only if it may be represented by a cardinal von Neumann Morgenstern utility function that is unique up to increasing affine transformation. Only for v we can find numbers a>0 and b responsible for setting a unit and a zero, respectively, for which v (.)=a*v(.)+b. In our case choose a=25 and b= 3 and you get the relationship v =25*v 3. Problem 6 [6p] Suppose we are given the following two lotteries: the P bet (x,p) (meaning x with probability p and zero otherwise) and the $ bet (y,q), where y>x>0 and 1>p>q>0. Consider an individual who exhibits the following preference pattern: Prefers the P bet in a direct choice But assigns higher certainty equivalent to the $ bet. [Certainty equivalent or selling price for a lottery l is defined as the minimal amount of money which the decision maker is willing to accept to forgo the right to play lottery l] Can you construct a sequence of trades (a Dutch Book) which exploits this person? 1 take the $ bet 2 exchange for the P bet and pay me a penny 3 exchange for a certainty equivalent of the P bet 4 exchange for a certainty equivalent of the $ bet and pay me a penny 5 exchange for the $ bet 6 Repeat If YES, write the sequence and explain which axiom of consistency is violated here. transitivity Problem 7 [6p] [Extra] Consider selling price S(W,x) for a lottery x at wealth level W, where E U[W+x]=U[W+S(W,x)] and buying price B(W,x) for a lottery x at wealth level W, where E U[W+x B(W,x)]=U[W] Assume that U is a DARA utility function (Absolute Risk Aversion decreases with wealth). Prove the following statement: S W, x B(W, x), for all W (and x) for which S(W,x) and B(W,x) exists Tip: You can use the fact that for DARA utility functions S(W,x) is increasing in wealth.

7 1 First notice that S[W B(W,x),x] = B(W,x) 2 If B(W,x)>0, then by the fact that S is increasing in wealth we have 0 <= B(W,x) <= S[W B(W,x),x] <= S(W,x) 3 If B(W,x)<0, then by the fact that S is increasing in wealth we have 4 Hence S W, x B(W, x), QED S(W,x) <= S[W B(W,x),x] <= B(W,x) < 0