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Course: ECON 206, Spring 2007
School: Brown
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Theory Microeconomic II Assignment 2: Risk and Information Solutions February 15, 2007 Problem 1. Concerning Measures of Risk Aversion: Let ui : , i = 1, 2, be two strictly increasing, strictly concave, twice dierentiable Bernoulli utility functions; and, let F : [0, 1] be a cumulative distribution function for some real-valued random variable, y , which is the (positive or negative) reward from accruing some...

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Theory Microeconomic II Assignment 2: Risk and Information Solutions February 15, 2007 Problem 1. Concerning Measures of Risk Aversion: Let ui : , i = 1, 2, be two strictly increasing, strictly concave, twice dierentiable Bernoulli utility functions; and, let F : [0, 1] be a cumulative distribution function for some real-valued random variable, y , which is the (positive or negative) reward from accruing some risky situation. So, agent is preferences over gambles can be represented as follows: Ui (F ) = ui (y)dF (y), i = 1, 2. Dene c(F ; ui ) to be the certainty equivalent of the gamble represented by F for agent i: c(F ; ui ) u1 [ ui (y)dF (y)]. We say that agent 1 is more risk i averse than agent 2 if and only if: F : c(F ; u1 ) c(F ; u2 ). Prove that the following three claims are equivalent: (i.) Agent 1 is more risk averse than agent 2. (ii.) There exists an increasing concave function v : v(u2 (y)). (iii.) y Solution . (ii.) (iii.): Notice that since both u1 (y) and u2 (y) are increasing we can always nd an increasing function v() such that u1 (y) = v(u2 (y)). Taking the derivative in in this equation and using the chain rule we have that u1 (y) = v (u2 (y))u2 (y). Prepared by Omer Ozak (ozak@brown.edu), Department of Economics, Brown University. If you nd any typos or mistakes please let me know, so that it can be xed. such that, y : u1 (y) = : u1 (y) 1 u (y) u2 (y) 2 u (y) 1 Microeconomic Theory II Taking logarithms on both sides and dierentiating again we get u1 (y) u (y) v = 2 + u1 (y) u2 (y) v so that v 0 if, and only if, u1 (y) u (y) 2 . u1 (y) u2 (y) Solutions (ii.)(i.): By denition u2 (c(F ; u2 )) = u2 (y)dF (y), the hypothesis implies that u1 (c(F ; u1 )) = v(u2 (y))dF (y). Since v() is concave, then v(u2 (y))dF (y) v Putting the pieces together we get u1 (c(F ; u1 )) = v(u2 (y))dF (y) v u2 (y)dF (y) = v(u2 (c(F ; u2 ))) = u1 (c(F ; u2 )) c(F ; u2 ). u2 (y)dF (y) . and since ui () is strictly increasing, we have that c(F ; u1 ) (i.)(ii.): Let v() again be the increasing function that makes u1 (y) = v(u2 (y)). By hypothesis c(F ; u1 ) c(F ; u2 ) so that u1 (c(F ; u1 )) u1 (c(F ; u2 )). But this implies that v(u2 (y))dF (y) = u1 (c(F ; u1 )) u1 (c(F ; u2 )) = v i.e. v() is concave. Problem 2. Concerning Blackwells Theorem: Let C ={c1 , ...cN } be a nite set of consequences, and let p N 1 be an agents prior probability distribution on C. (So, pn =Prob{cn }, n = 1, ..., N .) Let S = {s1 , ..., sK } be a nite set of signals with arbitrary element sk , k = 1, ..., K. And, dene an experiment to be some conditional likelihood function l : SC [0, 1] such that: cn C : K l(sk | cn ) = 1. k=1 Let be the set of all possible experiments (a compact, convex subset of a nite-dimensional Euclidean space.) Given an experiment, l , dene q l : N 1 S N 1 to be the function which associates with each prior belief and signal pair a posterior beliefs l for the agent, derived using Bayess Rule. In particular: qn (p, sk ) =Prob{cn | sk }. Dene B : S S [0, 1] to be a garbling of signals, meaning that B(sj | sk ) is the probability that signal sj gets reported to a decision-maker when signal sk has actually been realized. (So: sk S : K B(sj | sk ) = 1.) Finally, let l (l; B) be the experiment which results from j=1 the original experiment, l, when its realizations have been garbled by B. 2 u2 (y)dF (y) Microeconomic Theory II (a) Using Bayess Rule, write an explicit expression for the function q l . (b) Write an explicit expression showing how l (l; B) can be derived from l and B. Solutions (c) (Not Trivial) Derive an explicit expression showing how q l and q l (l;B) are related to one another. (Hint: q l are the agents beliefs after observing signals from the garbled version of l so, intuitively, they should be some kind of garbled version of what the agents beliefs would be if he were observing l directly.) Solution . (a) From Bayess Rule we have that l qn (p, sk ) = P rob(cn | sl ) = k l(sk | cn )pn N m=1 l(sk | cm )pm . (2.1) (b) Let B = (Bkj )K j,k=1 be the matrix with element Bkj = B(sj | sk ) (this is a transition probability matrix) and let L(l) = ((Ll )K )N be the matrix with element typical nk k=1 n=1 Ll = l(sk | cn ). Then, nk Ll (l;B) = Ll B i.e. l (l; B)(sj | cn , sk) = j=1 K B(sk | sj )l(sj | cn ). (2.2) (c) Now, we have that if we see the signal sk from the garbled signal, then the probability that the original signal was sj is (j, k) = P rob(sj | sk ) = This implies, that K l qn (l;B) (p, sk ) K B(sk | sj ) K i=1 B(sk | N m=1 l(sj | cm )pm si ) N l(si | cm )pm m=1 = P rob(cn | sk ) = j=1 P rob(cn | sj )P rob(sj | sk ) = j=1 l (j, k)qn (p, sk ). Problem 3. (The value of information.) Suppose some agent can make an even-money bet on the outcome of the next ip of a coin. Let this agents initial wealth be W . The agent is an expected utility maximizer with Bernoulli utility function: u(y) = ln(y). Initially the agent knows the coin is 3 Microeconomic Theory II Solutions biased, such that either it has probability 1/3 of coming up heads and 2/3 of coming up tails; or, its the other way around, with the probabilities being 2/3 heads, and 1/3 tails. Initially, the agent thinks either of these events is equally likely. For a fee, this agent can to get see a trial ip of the coin before placing his bet. (a) How much would this agent be willing to pay to see the trial ip, and how would his betting after seeing the trial ip depend on its outcome? (b) Suppose now that the agent could see two trial ips instead of one if he wanted to, but would have to pay more to see the two ips. How much more would this agent be willing to pay? (c) Obviously, seeing two trial ips should always better than seeing one for any decision maker who has to choose some action, the return from which depends on the outcome of the next ip of this coin. Prove this, using Blackwells Theorem on the Comparison of Experiments. Solution . (a) Let b be the bet the agent does and assume he wins if heads comes up. Then his expected utility before doing the experiment is U0 (b) = 1 1 2 1 ln(w + b) + ln(w b) + 2 3 3 2 1 = (ln(w + b) + ln(w b)) 2 2 1 ln(w + b) + ln(w b) 3 3 and his optimal bet would be b0 = 0 and his optimal utility is U0 = ln w. If he is able to see one ip before deciding, then the updated probabilities of coin Ci , i = 1, 2 are given by 11 23 11 23 12 23 P (C1 | H) = P (C2 | H) = + 12 23 12 23 11 23 1 = 3 = 2 3 P (C1 | T ) = P (C2 | T ) = 12 23 12 23 11 23 + 11 23 11 23 12 23 = = 2 3 1 3 + + so that his expected utility if he sees H in the trial is U1 (b | H) = 1 1 2 2 ln(w + b) + ln(w b) + 3 3 3 3 5 4 = ln(w + b) + ln(w b) 9 9 4 1 2 ln(w + b) + ln(w b) 3 3 Microeconomic Theory II Solutions 1 H in which case he would bet bH = 9 w which gives him utility U1 = ln w+ 5 ln 10+ 4 ln 8ln 9. 1 9 9 If he sees T in the trial his expected utility is U1 (b | T ) = 2 1 2 1 ln(w + b) + ln(w b) + 3 3 3 3 4 5 = ln(w + b) + ln(w b) 9 9 2 1 ln(w + b) + ln(w b) 3 3 T so that his optimal bet would be bT = 1 w which gives him utility U1 = ln w + 5 ln 10 + 1 9 9 4 ln 8 ln 9. Since before observing the trial the probability of observing heads and tails 9 are P (H) = 1 = P (T ) 2 the expected utility of observing the experiment is U1 = ln w + 5 4 ln 10 + ln 8 ln 9 > ln w = U0 9 9 so that he would be willing to pay p1 U1 U0 . (b) Suppose now that he can observe two trials, then his updated probabilities are P (C1 | HH) = P (C1 | HT ) = P (C1 | T H) = P (C1 | T T ) = 1 1 2 2 3 1 2 +1 3 2 112 233 2 1 + 123 3 23 112 233 2 1 + 123 3 23 1 2 2 2 3 2 2 +1 3 2 1 2 11 23 11 23 2 2 3 1 = 5 P (C2 | HH) = P (C2 | HT ) = P (C2 | T H) = 1 2 1 2 1 2 = = 1 2 1 2 = 4 5 1 2 1 2 3 P (C2 | T T ) = 1 2 1 2 2 2 3 1 2 +1 3 2 112 233 12 2 + 131 33 2 3 112 233 12 2 + 131 33 2 3 1 1 2 2 3 2 2 +1 3 2 2 2 3 = 4 5 = = 1 2 1 2 1 = . 5 1 2 3 If he sees HH then his expected utility is U2 (b | HH) = 2 4 1 1 ln(w + b) + ln(w b) + 5 3 3 5 9 6 = ln(w + b) + ln(w b) 15 15 1 2 ln(w + b) + ln(w b) 3 3 9 6 HH and his bet bHH = 1 w which gives him utility U2 = ln w + 15 ln 6 + 15 ln 4 ln 5. If he 2 5 T HT sees HT or T H he is as uninformed as before the trials, so that U2 = U2 H = U0 = ln w 5 Microeconomic Theory II and bHT = bT H = b0 = 0. Finally, if he sees T T then his expected utility is 2 2 U2 (b | T T ) = 4 1 1 2 ln(w + b) + ln(w b) + 5 3 3 5 6 9 = ln(w + b) + ln(w b) 15 15 Solutions 2 1 ln(w + b) + ln(w b) 3 3 1 9 6 T so that he will bet bT T = 5 w and U2 T = ln w + 15 ln 6 + 15 ln 4 ln 5. Before seeing the 2 trials the probability of observing each combination is given by P (HH) = 1 2 5 = 2 3 18 112 121 2 P (HT ) = P (T H) = + = 233 233 9 2 2 1 2 1 1 5 P (T T ) = + = 2 3 2 3 18 + 1 2 1 3 2 2 so that the expected utility of two trials is U2 = ln w + 5 9 9 6 ln 6 + ln 4 ln 5 15 15 > ln w = U0 and it can be seen that U2 > U1 so that he would be willing to pay p2 U2 U0 , but he would have to pay at least p1 . The maximum amount he would be willing to pay for the chance to get two trials instead of one is U2 U1 > 0. Notice that we have assumed that he has to decide the number of trials he wants to see before he sees the rst trial. Otherwise l we would have to do a similar procedure, but using the values of Uk and the value of the next ip would depend on the history of the trials until the moment of choice. (c) To see this we need to nd numbers bk1 ,k2 [0, 1], k1 {H, T }, k2 {H, T }2 such that P (k1 | Ci ) = k2 bk1 ,k2 P (k2 | Ci ). such that P (H P (T P (H P (T k1 bk1 ,k2 = 1. Verify that | C1 ) + 1 P (HT | C1 ) + 0 P (HT | C2 ) + 1 P (HT | C2 ) + 0 P (HT | C1 ) + 0 P (T H | C1 ) + 1 P (T H | C2 ) + 0 P (T H | C2 ) + 1 P (T H | C1 ) + 0 P (T T | C1 ) + 1 P (T T | C2 ) + 0 P (T T | C2 ) + 1 P (T T | C1 ) | C1 ) | C2 ) | C2 ) | C1 ) = 1 P (HH | C1 ) = 0 P (HH | C2 ) = 1 P (HH | C2 ) = 0 P (HH so that the experiment with one ip is a garbling of the experiment with two ips. 6 Microeconomic Theory II Solutions As an example of what we proved in problem 2, you can verify that the following holds: 5 2 2 P (C1 | H) = P (C1 | HH) + P (C1 | HT ) + P (C1 | T H) + 0 P (C1 9 9 9 2 2 5 P (C1 | T ) = 0 P (C1 | HH) + P (C1 | HT ) + P (C1 | T H) + P (C1 9 9 9 5 2 2 P (C2 | H) = P (C2 | HH) + P (C2 | HT ) + P (C2 | T H) + 0 P (C2 9 9 9 2 2 5 P (C2 | T ) = 0 P (C2 | HH) + P (C2 | HT ) + P (C2 | T H) + P (C2 9 9 9 i.e. that seeing one trial is a garbling of seeing two trials. | TT) | TT) | TT) | TT) 7
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LSU - CE - 2460
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LSU - CE - 2460
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LSU - CE - 2460
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LSU - CE - 2460
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LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 156.For the sun, andg = 274 m/s 2 R= 1 1 D = 1.39 109 = 0.695 109 m 2 2()Given that an =gR 2 v2 and that for a circular orbit: an = r r2 r= gR 2 v2El
LSU - CE - 2460
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LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 168.Change to rectangular coordinates.cos =r=x randsin =y rEquation of the path:3 3 3r = = yx sin cos yx rr or y = x + 3.from whichAlso,yx=3
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 143.(a)v A = 420 km/h v B = v A + v B/ Aor,v B = 520 km/h60v B/ A = v B v A = v B + ( v A )Sketch the vector addition as shown.2 2 2 vB/ A = v A + vB
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 167.Differentiate the expressions for r and with respect to time.r = 6t 1 + 4t 2 ,&amp; r = 6 1 + 4t 2 + 24t 2 1 + 4t 2 &amp; = 72t 1 + 4t 2 r()1 2()1 2 9
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 151.Let be the slope angle of the trajectory at an arbitrary point C. Then,( aC )n = gcos=C2 vC,orC =2 vC gcosBut, the horizontal component of vel
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 160.Radius of Earth Radius of orbit Normal accelerationR = ( 3960 mi )( 5280 ft/mi ) = 20.908 106 ft r = ( 3960 + 10900 )( 5280 ) = 78.4608 106 ft an = gR 2 r2 a
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 170.From geometry, Differentiating with respect to time, Transverse component of acceleration&amp; &amp; a = r + 2r =r=b cos&amp; r= &amp; b sin 2 cos &amp; &amp; b 2b sin 2 + 2 c
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 182.From problem 11.97, the position vector isr = ( Rt cos nt ) i + ctj + ( Rt sin nt ) k.Differentiating to obtain v and a, dr v= = R ( cos nt nt sin nt )
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 183.For A = 3 and B = 1, r = ( 3t cos t ) i + 3 t 2 + 1 j + ( t sin t ) k Differentiating to obtain v and a.v= a= dr t = 3 ( cos t t sin t ) i + 3 j + ( sin t + t
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 188.(a)Construction of the curves. Construct the at curve.0 &lt; t &lt; 10 s:10 s &lt; t &lt; 26 s: 26 s &lt; t &lt; 41 s: 41 s &lt; t &lt; 46 s: 46 s &lt; t &lt; 50 s:a = slope of v t cur
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 191.The horizontal and vertical components of velocity arevx = v0 sin15 v y = v0 cos15 gt At point B,vx v0 sin15 = = tan12 v y v0 cos15 gtorv0 sin15 + v0 co
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 192.First determine the velocity vC of the coal at the point where the coal impacts on the belt. Horizontal motion:( vC ) x = ( vC ) x 0 = 1.8cos 50 = 1.1570 m/s
LSU - CE - 2460
COSMOS: Complete Online Solutions Manual Organization SystemChapter 11, Solution 193.Given: Then, (a)t = 0,( vA )0 = 0,( a A )t =dv A = 0.8 in./s 2 dtv A = ( v A )0 + ( a A )t t = 0.8 tv A = 0,( a A )n =2 vA=0a A = ( a A )t (b