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# Adv Mth Mod Macroec ECON 609

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This 38 page Class Notes was uploaded by Leopoldo Ritchie DDS on Thursday October 29, 2015. The Class Notes belongs to ECON 609 at University of Michigan taught by Miles Kimball in Fall. Since its upload, it has received 10 views. For similar materials see /class/231651/econ-609-university-of-michigan in Economcs at University of Michigan.

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Date Created: 10/29/15

Tanya Islam Econ 609 Christian Gollier s The Economics ofRisk and Time Exercise 1 7 Meyer 1977 introduces the concept of stochastic dominance with respect to a function Consider any increasing function ul and de ne the associated function set uquot z u z gt yu1 1s the set of functlons that are more r1sk averse yuu uz W than u1 a Show that yu1 is a convex set Let uA be more risk averse than ul 3 Hf Z Z LilZ 1 42 12 u22 u1quotZ anduB be more r1sk averse than ul 3 Z I 2 32 12 and ucluA1 lu3 3 age gt uquotz age u z We have to show that 122 1ulz 1 1u z ul21 u z age Auz1 Augz uz age gt 1 uzuquotz 1 A ugzu z u czuz ugzuz gt u1quotz1 ulquotz Au ltu uz u z A C uquotz uz yu1 is a convex set b Find a simple basis for yu1 and derive the integral condition for the corresponding stochastic dominance order A basis for the set of utility functions that are more concave than ul is be 2 minu1zu1t9 V2 9 and 9 6 ab Let F1 and F2 denote the cumulative distribution function for1 and E2 The integral condition for the corresponding stochastic dominance order iminu1zu16dFlz s iminu1zu16sz 2 3 fu1zdFlzju16dFlz s fu1zdF2zju16dF2z 3 u1zdFlz u19JdFlZ s iu1zsz2 Mg sz Z 3 u1zdFIz u19F1z lZS iu12sz2 16F2Z l2 3 fu1zdF1Z u19F1b 39 S ju12sz Z u19F2b 39 3 Jgu12dF1Z 191 39 S jlu1zszz 191 F2 9 3 Jul ZdF1 Z F2 2 u19F19 F2 9 S 0 Integrating by parts 3 19F19 F29JF1Z F2 2ui2dz 19F19 39 S 0 2 jEzuzdz 2 szzuzdz l 27 Examples and Kreps Portius Prefs Advanced Math Today we continue giving some examples and looking at the tricks that are helpful in solving them The first trick that we use is change of variables Begin with the equation the most simplified form of the Bellman equation that allows for variation over time A t r 2 p39YQ 39YAtrl39Yl 39Y M At 27039 which can be rewritten A t p39YQ 39YAt1 39Y r AU We see that the equation could be quite a bit simpler if we did a change of variables by which we define BlA FY 270392 Plugging this in we get ME L1Yramp Bt Bt 2yo z Multily by Bt to find M1130 pBt39YBt 7l39Yr 2y039 Move all of the B s together and multiply by l r 2 39YBU7139YrMJB pm 27039 Dividing through by 39Y and cleaning up the negative signs gives Bit ll39Ylr Bt1YpB t 2y039 and thus 1 1 2 Bt 7p 17rM1B 1 7 7 270 This equation is quite simple to integrate but there is one more trick to use Specifically we are going to allow 0 r and u to possibly change over time This change is going to be denoted with subscript t In order to solve the differential equation explicitly we are going to need to use Leibniz Rule which divides the derivative of an integral into three parts Mathematically Leibniz rule is 6 179 E M Mathematically you can see the three parts of the change in the following integral of fte maybe 11 dx b396fb66 a396fa66 lt a9 b9 The first bit is the change in the actual area under the curve the second bit is due to the beginning of the curve moving and the last bit is due to the end of the time considered moving Now notice that the differential equation is the solution to r 1 1 mz39wzw d Jim 1Wrm7zaz i e 7 7 27039 B 611 Thus the solution to the differential equation is H H 2 Iri tquotlr rNHMJdN 7 7 2 7039 Bare dt Because A is the inverse ofB 7 l H 7 2 no J jlpaquot1ilrzquot7 02 2 latquot J39 8 L7 7 7039 j dry 2 We have the fullest answer that we could possibly want Although we won t actually solve the full equation the same trick helps for solving for the value of the console When we have the console PlR you will find that M W 13 It elf dt P R R and combine this with RlPt Note also that the substitution F E and you start with rR E where R is the consol rate and r is the short rate Again moving backwards let s reconsider the last example that we had from the end of the last lecture egg dt a MaltcE0l390 e39F t st dwrwycdt dy039dz For this equation there are two state variables Thus in order to fully simplify the problem we will need to find to dimensions of symmetry The first we covered last time Let s look for the following symmetry w gtwe ccr6 This leads to Vwe t gte39quoteVwt Thus ew and we can therefore find the base for the problem VOte39 wVwt note the unfortunate part of this problem is that we must allow for negative consumption implicitly However there are still publications using this form The general method of writing this is Vwyte39 wV0yt Unfortunately there is still another important state variable there the y To simplify the problem along this dimension let s use the symmetry YYe c gtce w gtw Note the symmetry listed at the end of the last notes was correct but it would have led to a bit more algebra than the present one Because we have taken care of one of the state variables w and it doesn t change in a manner that effects 0 for this symmetry we can leave it a 0 This will generally be the case the simplest cases to deal with will leave the variable already simplified unchanged when taking care of another dimension With the symmetry tested we can look on the effect of the change on value V0yete39a9V0yt As always plug in ey And we get V0ytequot yV00t Plugging this into the Value function we get the general form Vwyte39 39 YV00t We now know exactly how V varies with all state variables This makes the equation fairly simple to solve ad will help us solve it Rather than actually solving it we just write the Bellman equation There are a few new factors in this Bellman equation The slight added level of sophistication necessary comes from the multiplicity of state variables We are just taking the Taylor expansion so we need to remember the partials with respect to all state variables VW Vy VWW wa and Vyy In this particular example these values will all wash out but we will look a bit at a version where they do not all wash out in a second era o2 pVVFmaxc 7 VWrWycVy0wa02wa0Vyy a Now suppose that the equation were a bit different but the symmetries still held Specifically suppose that we were trying to solve era MaxcE0l390 e39F t dt a st dwrwycdtlxsd5 dy039dz cor rd5 dZB Note that because you have two randomly distributed variables you need to state the covariance between them Both symmetries we used will still work just fine but we need to add that 0 gttx Now the partial derivative terms are not 0 The way to consider the covariance term is to view it as stating that dzd5Bdt First dw with respect to dt As always dwdt is an order larger than dwd5 Thus we get the 52022 square it to make the correct orderid z The same method works for the dydz term from this comes the 012 Last we consider the dwdy term This problem is solved by taking a close look at the stochastic optimal control section of the book chapter The authors consider the general format Maximize yFtx st dxgtxdtZk1mU ktxdzk for il n zaosd6dz As the text tells us replace d5dz with and we have the full Bellman equation 2 a2 s 2 039 wa BatxsVyy Y egg pVVmaxc 7 VwrwycVy0VWW a The negative aspect of the utility function is that tx gttx In other words with this utility function stock holding is independent of wealth This is a result from CARA At this point we look at the one type of symmetry that is particularly useful for all utility functions changing time Write the general maximization problem MaxcytEt I we39Pti39tuct dt st dwrwclxu r dtlX039dZ Consider the following symmetry t gtt0 Ct gtcte At gttxttx Wt gtwte As long as U u and r are independent of t we get an interesting result This is because we also implicitly use rt gtrte which we don t know much about Only if these variables are time independent do we know that they do not change Plugging this in we see that VwteVwt In other words if the characteristic parameters of the equation are not changingiand time is going to 00 then the value of the problem is a constant The problem with T other than 00 is the same as with riyou would need to consider what happens at other times You increase T by e and when it is 00 nothing changes Plugging in as usual we see that 64 and thus VwtVw0 17 again This problem will C Now let s consider speeding up time using the special utility function u 1 not be generally solvable for other values The rest of the problem is as in the general case above Consider the following symmetry Wt gtwet Note that this is not a big deal as w0 gtwo the starting point doesn t change txt gttx0t Zt gtZetltgtOl gteol and thus 039 gt0quote p90 ueu raer C t gtOC 0t Note that this is not the simplest transformation It is relatively straightforward to show that this symmetry leads to a fully feasible set How about the resulting change in the utility function 9177 V gt 7 V 9 You could come up with what looks like a general symmetry for all utility functions but then you would also need to change u gteu 9 which doesn t leave you with much meaning Let s consider 9177 t9 V gt V In order to truly account what is going on we need to write V as a function of all of the items we changed for the symemtry VW909u9rU e 939Ww p u r 0 You can look at this and consider exactly what need to go on when time moves faster Note that this is simply another symmetry of the same equation that we began the class with Furthermore we also know that all symmetries need to hold for all simplified forms of algebra Therefore if you are worried that you have done the algebra incorrectly it is always useful to find other symmetries then check that the symmetries hold in the forms after you have mauled them with algebra In other words you could check your algebra in the reduced form of the V equation er7 2 1 y lYpl l39Yr 7r2 Y M7 Now that we have gone several lectures into looking at symmetry we consider whether this tool is really beneficial The fact that it is used in a lot of published papers is not enough Furthermore in order to fully simplify the problem we need to have a symmetry for all of the state variables This type ofmega symmetry does not exist in a lot of problems Even so there is often at least some symmetryiand we earlier noted that each symmetry gets rid of one degree of difficulty Let s face it solving two freaky differential equations is much harder than solving one In general it is a good idea to try and deal with the trend variables using symmetry ie things that can go towards infinity as say wealth goes to infinity Once you have done away with such trend variables then you can generally use Taylor expansion to deal with the remaining ones like number of working hours which are limited to 24 at maximum We will see the exact type of trick that we use for such problems beginning in the next lecture Let s look at the following twostate problem that will have to be approached just this way C is consumption N is labor supply ck MachEo I 0 e39F t 1 3 elsUr Udt st dKZNf 6Kcdt ZN dZZuZod6 Note that we are implicitly assuming independence of the error terms Note that there is likely a typo on the second exponent of the maximized function First check for symmetry of the form KaeK c gtec Z gtez N gtN You see that it is consistent by plugging into the constraints including that the d terms are also changed by the same amount and everything cancels Plugging into the Value function leads to vael39Bv We use this to get some more information Namely that VezeKel39BVzK Plug in 91Z gives you 1 H V1KZE VKZ And thus VZKV1KZZ1quot5 Let s consider the economic meaning We can not tell the exact form of the change in valuation when technology changes as we have only taken care of one of the dimensions of uncertainty However this is enough to do a partial analysis of technology improvement In fact you can consider any surprise increase in technology as having to elements 1 Z115 means that the economy is really a lot better off 2 The VlKZ shows that in order to really be Z115 better off you need to have the same ratio of capital stock increase as you had Z increase Thus it is as if your capital stock is falling behind your technology and this is a bad thing A common pattern will be one or two easy dimensions with one final difficult dimension that you are unable to solve using symmetry However with only one dimension it is possible to chop it to pieces with a computer You can graph it and consider it stability etc For example if you remember 602604 you can graph the following type of equation Where the dotted line is the saddle Krepsa Portius Preferences The relevant articles are the Kimball article and the Puffer and Epstein article We introduce Kreps Portius preferences in order primarily to show that the symmetry methods we have been learning about are not limited to the simple type of stochastic control problem we have been dealing with Yes even the simple stochastic control problems can get difficult but we will show that even adding extra sophistication like special preferences does not make thing that much harder An additional reason for introducing these preferences is that they have been used fairly commonly throughout the literature The reason that these preferences were created was that essentially all changes to utility functions that affect time preference also affect risk aversion However you often want to study the effect of changing risk aversion without changing the time preferences The method of doing this study is to use a special form of preferencesiKreps Portius Preferences Some of the characteristics of Von Neuman Morganstern preferences will hold f Kreps Portius preferences but others will not Specifically timing of when you find out about the result of a stochastic process t ou have no control over will effect preferences This is a bad thing VNM and KrepsPorteus preferences only overlap if the result of all gambles is immediately observable We are going to solve for these preferences using the limit as distinct time intervals goes to zero The psi function is the one that does the job of taking away the risk aversion The first psi term makes the person more risk averse increases the concavity of their preferences Then the expectation of this is taken and last the psiinverse term takes effect This psiinverse takes away any changes in time preferences but leaving the risk aversion note that sub t does not mean partial with respect to t here KrepsPortius Preferences are the following type of valuation VthUXtte39Ph Vlr E V thl Here s how it works You take a more concave value of all of the possible future outcomes By Jensen s theorem this leads to lower value of the future paths when the expectation is taken over it Submission to the Journal of Answers to Gollier James M Sallee Submitted January 27 2005 Exercise 82 Consider the following nonseparable interternporal preferences Uc1c2 lnc1 lnc2 7 hcl o Interpret h depending on whether it is positive or negative 0 Show that the optimal consumptions c1 and cz are respectively decreasing and increasing with h Solution 82 Part 1 If h is positive then consumption is habit forming and h represents the degree of habit formation This means that in the second period you care about the growth in consumption lntuitively this would cause a consumer to consume less in the rst period since they suffer a penalty from that consumption in the second period If h is negative then consumption is durable and h represents the degree of durability This means that you enjoy a boost in the second period from the rst period consumption lntuitively in the two period model this will cause an increase in rst period consumption since you enjoy part of that consumption again in the second period Part 2 The solution requires explicating a budget constraint Let c1 c2 3 5 It is clear that the constraint will bind Substituting c2 5 7 c1 we can consider the unconstrained problem maxlnc1 Midi 7 c1 7 hcl 01 This yields a basic rst order condition in terms of c1 We know that this will de ne a global maximum by the strict concavity of the objective function 1 61 k FOO 7 0 i c1 071hc1 5 7 k01 k01 gt O 01 1 7 6X1 k Denote this optimal rst period consumption as cf We can determine the marginal change in cf from a change in h by simple differentiation 3c 5 71 17 1k2lt039 1 l 25 Boyd s symmetry examples Advanced Math The example we begin with is actually the example that we used all last lecture to illustrate Boyd s symmetry theorem We will find several things to be true I With just Boyd s symmetry we can get a lot of information regarding comparative statics We can also get a lot of economically pertinent information I The economically pertinent information can be used for quite a bit of data testing I The economic information combined with the comparative statics will be useful in solving for the control rule However the control rule will still not be a breeze In the last lecture we showed that the problem MaxcychOIo e39PtU Ct s t dWrWCtxurdt lXO39dZ C 7 1 where Uc We also showed that this plugs into the Bellman equation Q K t pVKxt VtKxt maXUK x t VK K tAK x t VKK K I And becomes 6177 a 2 a 2 7 Vwwtrwctxur wawt pVwtVwtmaxoyc 1 y 2 The LHS of the Bellman equation is always the same and the second term the Vt only matters if the parameters are not time invariant andor T is finite In other words quite a bit of simplification can be done by just making the tim ehorizon infinite and everything timeindependent ie that r is constant as is u etc The other term of particular importance is the one after the VW This term is simply the element before dZ in the constraint squared and divided by two We showed last lecture that if the following changes are made w gtew c gtec tx gt61x then the budget constraint is satisfied and thus all plays are feasible In short the problem shows symmetry for that transformation What s more we can easily consider what happens for all possible paths when we multiply consumption by 9 then we increase value by 91quot This transformation is true for all possible paths Since it is true for all paths then Vewtel39YVwtfor all e and W This one relationship written just above gives you a huge amount of information about the functional form Consider the case where 91w Then we get Vltlw139YVwt Solving this for Vwt leads to VwtVltw139Y In other words we have a very nice method of comparing what happens when we multiply wealth by a constant From here it is quite straightforward to solve for the decision rule in the time invariant case as long is it can be characterized by Boyd symmetry The hope for anyone doing research is that the solution of the timeinvariant case will help get the insights to solve the general nontime invariant case We are going to go as many steps as we can in the general time variant case then veer off allowing all tim evariant terms to be zero The general method is to consider the first order conditions Then take the information and plug that back into the Bellman equation and with luck finish solving the resulting differential equation Let s see we are dealing with maximization so we need a FOC We are going to look specifically for elements to substitute in for Vwt Vwwt and wawt Write out the Bellman equation and note that the control variables are all on the RHS 6177 202 pVwtVtwtmaXoc 1 Vwwtrwclxur wawt 2 7 Take the FOCC 6V1tw1 7 c Yvwwt V1t1 Yw Y W 1 1 CV1t 7139Y 7 And therefore V1tAt7V 1 7 Doing even this much information allows us to consider some deeper meaning Specifically we originally noticed that At was the average propensity to consume cw Therefore we have 1 1 AtVlt 7 l39Y 7 Substituting in VwtVltw leads to AU 7 W 1 7 Vwt 1quot Wow We have V as a function of solely W and t Taking the derivative with respect to W once and twice gives VWWtAt39YW39Y wawt39YAt39YW39Y391 Write out the Bellman equation again for consideration of FOCM 1202 2 pVwtVtWtmathc 161 VWWtrwctxur wawt FOCM leads to VWWturwawto020 Which leads to a very economically meaningful relationship u r VW w t 0392 VWW w t Where the second fraction on the RHS is the absolute risk tolerance The absolute risk tolerance is the inverse of the absolute risk aversion divided by W WOW You see that as long as preferences are of the form we assumed it is possible to sum the absolute risk tolerance across individuals to get a composite risk Substitute in for the VW and wa terms and we see another clear relationship Vwwt w w VWW W t W 7 Take the inverse and multiply by w to get the relative risk aversion Y In other words the risk aversion of the value function of this problem is the same as the risk aversion of the utility function of the underlying individual While many papers suggest that this is true in general it is only true due to the symmetry that we are taking advantage of Note however that we have not assumed time invariance thus far The equation that we solved for above for D will be useful in some circumstances but there is more that we can say In particular let s consider the information from the FOCM in addition to FOCC Specifically plug in AU 7 w 1 7 VwwtAt39Yw39Y wawt39YAt39Yw39Y391 Vwt 1quot Into the above equation for D u r VW w t 0392 VWW wt and we find that it s nice when so many terms drop out This equation also states that if we double all initial elements w then we double the amount of 0 Once againinote that this is solely the symmetry talking although it can also get your name published Now we really get working on the differential equation For this we need to substitute all of the information we have from the FOC s into the original Bellman equation In other words we combine 1 7 VwwtAt39Yw39Y wawt39YAt39YwT1 Vw t W 7039 With 6177 202 pVwtVwtmaX xc 1 Vwwtrwctxur wawt 2 7 using cAtW which leads to c1 Arw1 7 Aay W 1 y 1 y to get 1 7 o 1 7 w pA t v39YA tAt39V1 w 1 y 1 177 2 2 At1Y w At39Yw39YrwAtw 2r wur YAtYwY1 wzi 1 y ya y o 2 This form of the Bellman equation is quite a mess yet there are a number of straightforward simplifications It would have been possible to write this as a function of Vlt instead of At but the At s have economic meaning to them Thus by writing it this way we have the intuition that comes with economic logic to help us construct a solution 7 The first Simplification is to notice that there are a lot of 1 terms Taking these and adding only terms that do not have corresponding values leads us to Zr llTH 739v1YAt39i 7039 0A t WA mmVI At139Yl39YAt39YrAt i 7 O 2 1 u r2 02 Simplify Just the last term 39Yl 39Y At39Yw39Y39 Y by canceling a 39Y and 02 in numerator and 7 Oquot denominator to get u r pA t Y39Y1i tAt39V1 At139Yl39YAt39YrAt 2 7039 llTH 71 39YAt39Y 27039 u r Now spread out the terms l39YAt39YrAt 702 ur to get pAt Y39Y21 tAt39V1 At139Yrl 39YAt39Y1 39YAt139Yl 39Y w 7 1 39Y At39Y 02 ya 2y0392 But we note that the last two terms are half one another pA t 39Y39Y tAt39V1 At139Yrl 39YAt39Y1 39YAt139Y l 39Y At39Y gt 7 7039 Multiply through by A39Y and we have Air 71 r 270392 p39Y m Atrl 39Y1 39YAt1 39Y which simply becomes p39Y 39YAtrl 39Y1 39Y girl This is as far as we will go with the tim edependent valuation It is a differential equation in A and it is solvable For time independency we assume that Tooand we assume that all parameters are independent of time notably r u and 039 With time independence we know that A 0 and so we can write 02 p YAtrlYl Y 2 27039 which is rewritten 2 39YA t pr1 39Y1 39Y M 27039 No differential equation to solve at all All we have is that 1 Z AtpY 1 7 r M 7 270 Plug in numbers such as 39Y4 to get an idea of what is happening and we seel Z Iquot A14 34 r M 27039 You could use this to estimate a fairly realistic simulation Say by plugging in 002yr u06yr u2Jyr and r02yr You could also go ahead and try to solve the differential equation We are going to do one last substitution on this problem We look at the valuation function and plug in our results to see that wH Vw1A39Y 1 p11w rgtZH W 7 7 2702 17 You see that this form of the valuation is just not all that complicated It is particularly useful if you want to do comparative statics If r increases by itself you see that there is a positive and negative effect The positive effect is higher returns while the negative effect is higher cost of capital for investment in the tock market On the other hand you see that 1 increasing leads to the person being strictly better off In addition while we don t know a base 0 we do know how it varies with widirectly from the symmetry I r Xi 72 7039 In other wordsistockholding is proportional to risk toleranceias you would expect It also has the W implication that despite the present high stock price people must be willing to still hold itiwhich therefore means that they expect the price to continue to rise One special case that will often be useful is the limit as 39Y gtl39 this leads to the log utility case In that situation Ap This result is frequently cited in the literatureialthough it is not robust39 it is only true for log utility We now spend a bit of time considering the log utility case One method for considering it would be to take the limit as 91 but we consider it directly in order to get more practice with the mathematical rules The general problem is the same as before note that we want to consider what is occurring at all times t MachthtftT e39PtquottU Ct dt s t dWrWCtxurdt lXO39dZ where uctlnct Consider the following for symmetry c gtec w gtew tx gt61x This just happens to be the same as the last case If we used the term B where lxBw then you would have BB Anyway with this symmetry it is straightforward to show that vvytTe39o 01nedt Doing the integration leads to 7p e ItTe39P quot1n0dt ln6ItTe39P quotdt lne 1 er T l and thus 1 er T l VewtVwt p Plugging in 91w gives 1 enema V1tVwtln1w And soving for Vwt gives 1 er 07 VwtVlt lnw p This result once again is a commonly used one in the literature but is only true for this particular utility funciton We now consider one more problem This particular problem is particularly useful for considering safety motive for saving For this problem we use the utility form egg 1K00444 a We are not interested in the portfolio of this person but concentrate on their spending habits The random variable is y the amount of work that this person is able to receive Thus we are looking specifically for precautionary savings The maximization problem is 7a e MaxcE0l390 e39F t dt a st dwrwycdt dy039dz We only have time to consider the time inconsistent case so we assume that r is constant Symmetry when we have the power in there has got to look a bit different Let s look for the following symmetry w gtwe ccr6 This leads to Vwe t gte39quoteVwt Thus ew and we can therefore find the base for the problem VOte39 wVwt note the unfortunate part of this problem is that we must allow for negative consumption implicitly However there are still publications using this form The general method of writing this is Vwyte39 quot V0yt 2 8 Extremals and Convex cones Advanced Math Today we discuss some method for economics of uncertainty The tools that we are going to use are extremals and convex cones The idea is the following suppose that something is true at all of the extremes of some convex cone In addition suppose that there is a linear relationship between all points in the convex cone Then it must be true that it is true for all points in the convex cone Of course we spent the end of the last lecture showing how expected utility is really linear despite utility being concave Thus we have the linear relationship that is used in the last paragraph For probability the convex cone is the set of possible combinations of outcomes 2P where P is the probability that outcome i occurs For iel2 this graphically looks like P2 P1 The extremals are the points on the triangle the points where one of the outcomes occurs for certain The convex cone is of course the shaded area of possible probability combinations All possibilities in the convex cone are convex combinations of the extremals We are going to use these methods to check for characteristics utility in an uncertain universe We are specifically going to use it to check for characteristics that uncertain choices must have in order to be strictly preferred and then will use the same tools to see how people with different types of utility functions react differently to gambles For gamble comparison we use the method to find first order and second order stochastic dominance For utility rather than using points on a plane we use functions of a certain class Example one we derive the meaning of first order stochastic dominance Suppose that we have two gambles that we can take X 1 and X2 Furthermore assume only that valuation of the gambles is done by some function that is increasing in the outcome ie a utility function that is increasing in income We want to know what the necessary and sufficient conditions are to guarantee EuOCzKEuOCz We want the extremals for the utility function the function that we use to evaluate the gamble The extremal of an increasing function is called a Heavy sidefimcn39an H which is defined by lif g S x H X 01f g gt x In other words the heavy side function is a very simple step function It equals one if its characterizing variable xi is less than x the realization of the random variable in this instance Graphically it looks like HA I gt E Of course the derivative of the heavy side function is what is called a Dyvac delta and looks like the following with the spike really thin 3H A gt E Although we won t prove it it is simple to see that it is possible to find the utility of any outcome X from heavy side functions using the following formula UXI w HX U d It will not be much of a problem if U does not exist as it could be replaced by dU However for the example we assume continuity in the u function The important aspect is that we can create any increasing utility function with a convex combination the integral over heavy side functions Thus the heavy side functions are in fact the extrem als of increasing utility functions To do the actual proof you would need to do use Leibniz rule to solve the integral and show that it is in fact equal to the utility function Now that we have the extremals we can now see what characteristics of the random variable lead to the characteristic we desire EUX 22EUX 1 If this characteristics is true for all of the extremals then it is also true for the function that is the convex combination of the extrem als Thus the next step is to see what needs to be true about the random variables in order for the above comparison to be true for all of the extremals If EHXE with E from X 2S EHXE with E from Xul for all E Note that EHXE with E from Xuz can be written EHXEX 2 ls true for all of the extremals ie the heavyside functions then it is true for all increasing utility functions To see this just note that we have utility simply from I w EHX X 2U dES I EHX X 1U d This is a necessary characteristic as we have made up the utility function from the H functions This is a necessary condition as we showed that all increasing functions are convex combinations of heavyside functions If you can t make the condition true for all heavyside functions then there will be at least some increasing utility function for which the above inequality does not hold Denote F as the distribution function for variable X Thus we can rewrite EHX X 2S EHX X 1 As EHX 2 S EHX 1 Where EHX aEI HX dFaX FiXl 1Fi The algebra from here is straightforward to see that EUX 2SEUX 1 iff EHX 2ESEHX 1E for all E Which is equivalent to lF2ESlF1E for all E In other words the above is true iff F2 first order stochastically dominates F1 F2ZF1 Example two Second order stochastic dominance We are going to derive the characteristics of two gambles that are necessary and sufficient conditions for one gamble to be preferred to another for all increasing and concave utility functions In other words we are now setting not just the first derivative but also the second derivative of the utility function Just to get a sense of the pattern this will follow we note that the extremal for this type of function has two items to worry about the first and second derivative Well the second derivative we saw we could deal with by a heavy side function actually since we want the derivative to be negative it is the inverse of a heavy side function but same idea We drew out the heavy side function earlier Increasing and concave comes from some function which is the integral of heavy side functionsithese are l the straight line fxx 2 the min function fxminxE0 where E describes the min function Graphically min functions look like H Once again we do not prove that this is the set of extremals but you can intuitively see that we can create Ux from UXU I u minX ld The first derivative is a negative heavy side function an the second derivative is dyvac delta Although E we don t do it you could prove that this is the case by integrating by partsior you could just believe it because it seems reasonable Just as before we want to find the expected value of each min function given each random variable Algebraically we do I w 1 nX 0dFaXI X dFaX Integrate by parts F1XX l I 1F1Xdx The first term is zero One method of proving it is zero is to take the limit as x goes to negative infinity noticing that the value of the low probability of a low value must go to zero by Chebyshev s inequality However for economics problems it is generally enough to simply state that the CDF of X has a finite lower bound Anyway we have I w 1 nX 0dFaX I w F1XdX which means that EUX 2SEUX 1 iff I wEF2xdXSI w F1xdx we also need to make sure that it is also true for the final extremal linear utility Thus we also require that EX 2SEX 1 This last condition is not important for this second order stochastic dominance However it would prove to be important for the third order Let s consider third order stochastic dominance What does it take for one gamble to be preferred to another one when utility is increasing concave and has a positive third derivative Of course the key is once again that the lowest derivative with assigned sign is characterized by a heavy side function the next derivative up is characterized by a min actually here it will be a max function and the next one is more curvedia quadratic Graphically you see that the quadratic will look like H In other words the extrem als will be 81 gOi MinOlt ZOl b UxX2 this is for the limit case as E goes to 00 We need to find the expected value of this for X being a random variable and thus I w 1 nX 0ZldFaXI ElmEzldFaOO for the a type extremals Integrate by parts 2F1XXE I 2XEF1xdx Of course the first term is zero again This leaves us I w 2XF1XdXI wEEF1XdX Integrate the first by parts 2XIN F1xdxN 2I MEI w5F1xdxd5 I NEF1xdx I m sure there is an algebra mistake but so be it I believe that is the correct approach However that s a start Furthermore notice that for part b you also need to guarantee that EX 2 2 ie the variance Moving on to risk aversion signing Suppose that we want to consider only the situation in which we have increasing risk aversion We will check utility in logs as it simplifies matters so we write this as the following inequality risk aversion is decreasing in X dlnr ml K Xx lt 0 dx uquot39x uquotx uquotx u x which is rewritten uquot39x gt uquotx uquotx u39x Where we know the RHS is positive risk averse and u lt0 Therefore u gt0 in order for the equality to lt0 work out Thus when you assume decreasing relative risk aversion you assume positive third derivative This positive third derivative really does seem reasonable as wealthier people do tend to invest more in riskier assets Let s move on39 we have three derivatives let s go for four Suppose that we think one of two items 1 wealthier will respond less than poor peoplelesser precautionary savings for a risk of a given size39 ie the possibility of a loss of 10000 2 People increase holding of the risky asset less for a given increase in income as their wealth builds If we buy into either of those conditions then we are assuming that d 25 lt 0 dx The math is the same and thus taking the actual derivative leads to uquotquotx gt uquot39x uquot39x uquotx which means that the fourth derivative needs be gt0 We see that the four derivatives that we have signed thus far are alternating in sign It turns out that there is a particular function that is the extremal as the limit of the number of alternating sign derivatives goes to infinity That function is Uxe39ax While there certainly is no reason to go a heck of a lot further then fourth derivative it is nice to note that such a simple function has the signs of all derivatives we might want correct The infinite alternating sign aspect is not useful for economics However the fact that it gets the first four signs correct means that it may be a reasonable check for some things le If you think some result may be true you could plug in this utility function and see if it holds Because it has the correct signs for the important derivatives we may be inclined to think we are on to something Next class s problems During the next lecture we are going to consider problems in which the following must be true If EfX SfO then EgX Sg0 Note that 0 is the outside option which we simply normalize to Oiwe do not mean the null set In other words we are going to be interested in ordering people or their preferences their things etc Let s consider some of the interesting possibilities 1 If Fxx then we can consider how strongly people like given things 2 If fxXU wx then as we found in the last lecture we are thinking about how much risky asset people ofa type like Le it is coming from maxx EUwtxx 3 If we want to consider how much precautionary savings one type of person is going to do then fx is uquotwx u39w x uquotW WW If fx is greater than 0 then we know x uquotw x gt u w x quot00 WW Multiply through by u and do a bit of simple algebra note that the sign of the inequality changes when we multiply a negative number from one side to the other and we this transforms into this inequality Euquotw x gt uquotw uquotwx u39w The X here is background risk le suppose that xD and Y are two statistically independent risks We want to know how the existence of the one risk effects reaction to the other risk Consume one of the risks xu into a new utility function UwEUwX We therefore have ExEyUWX Y EyUwY EUwX Y E UwY In other words we are now able to sign the risk aversion for one riskigiven the effect on another risk Specifically we have quot w EuquotwX gt uquotw 39w Eu wX u39w 2 24 Monotone comparative statics Advanced Math Rather than a complete discussion of monotone comparative statics today s lecture should be considered an introduction to lattices monotone comparative statics and a reader s guide to Milgrom and Shannon 1994 Before defining a lattice graphically it is useful to get an idea of what one is graphically The following two graphs show lattices in two dimensions and one dimension in turn X2 More carefully a lattice is a set in which the meet and join are defined and in which the meet and joint of all pairs of points within the set are also within the set A complete lattice is a set in which the meet and join of all as many as all points within the set is both defined and an element of the lattice The function meet of A and B is denoted AAB And the joing of A and B is denoted AVB It is useful to consider the idea of meet and join by considering sets which do satisfy the definition of a lattice For sets the meet is defined by the intersection AnB And the join is played defined by the union AUB Similarly the real numbers R is a lattice as is R2 For R meet is the min of two numbers andjoint is the max For R2 the simplest definition ofmeet of AA1A2 and BB1B2 AB is NinA1gtB1gt MinA1gtB1 and the simplest definition of join of A and B AVB is MaXA1gtB1gt M3XA1gtBr In other words a lattice is an order relationship Not all points in the lattice are explicitly ordered see X and X in the following diagram However for all points in the lattice there is a point within the lattice that is the max of two points the join and the min of the two the meet Thus while not a complete ordering a lattice does give a partial ordering X is the notation for vector X2 X1 The use of the lattice will be for comparative statics Similar to calculus comparative statics monotone comparative statics will ask the question if we change this parameter how much will that change However unlike comparative statics through calculus monotone comparative statics will be able to answer the question for even discrete changes A sublattice is defined as a portion of a lattice in which the meet and join of all points within the portion is also in the portion A sublattice is defined by the lattice and the divisors between the elements within the portion from the members of the lattice that is outside the lattice The paper includes the proof that any portion of a lattice that is separated from the rest by lines that are increasing is a sublattice Let s draw a few examples to get the idea A cross of two sets on the axis is a lattice39 this is used in the following examples In the following example the points are the elements of the lattice while the dashed lines are the separator It is easy to see that all meets and joins of points within the sublattice between the lines are also within the sublattice X2 Theorem 1 any portion of a lattice that is separated from the rest of the lattice by boundaries that have positive slope is a sublattice Note how the following is not a sublattice the arrows point to the join of two points within the portion cut off by the dotted boundaries X2 X1 We now move on to theorem 2 Strong set ordering All sets can be ordered The notation for the set S being strongly set ordered greater than S is the folloowing 25 5 The definition of strongly ordered is as follows SZSS iff for all seS and s eS sAs eS and sVs eS If the two sets S and S are nonoverlapping then this relationship is plain to see the arrow draws a line between the meet and join of two points which are the points themselves X2 When the sets S and S are overlapping it is a bit more subtle The set S is the set of points within the sublattice defined by the dashed lines but above the heavy dotted line The set S is the set of points within the sublattice defined by the dashed lines but below the heavy solid line X2 The portions of S and S that are not elements of the union of S and S are follows exactly as the example above The portions of S and S that are elements of SnS are only slightly more sophisticated The meet and join of many of the points within the union are in both S and S The important element is that they are in the S or S necessary for the condition to hold You can create a similar graph of strongly ordered sets in one dimension as well The set below the solid line is S while the set above the dashed line is S Note that all points in S below other points in S must also be in S Otherwise there will be no ordering Theorem 3 Spence Mirlese condition This was discussed in the last lecture Theorem 4 Monotonicig theorem this is the denoument of the paper The meaning of this theorem is a bit more sophisticated than the first three and thus we cover it in a bit more detail We begin with the exact words of the theorem I Let f be a function from XxT gtReals where I X is a lattice I T is a partially ordered set and I SCX I Then argmaxxesfxt is monotone nondecreasing in ts Iff I f is quasi supermodular in x I and f satisfies the single crossing property in xt xeX teT There are two elements of the theorem that bear special consideration The first is fxt is monotone nondecreasing in ts This phrase means two things 1 Increasing t increases the argm ax value of f 2 Increasing the set of possible X s chosen over does not decrease the argmax Quasi supermodularity is a weaker form of supermodularity Supermodularity is defined as fxVyfxyzfxf which leads to fXVYfO ZfXfXAY This condition implies quasi supermodularity Definition f is quasisupermodular if 1 fXZfX Y fXVYZfO ii fxgtfxy gt fxVygtf I As you can see the comparison of f over the points is the same for both supermodularity and quasi supermodularity See the points in R2 below Xquot O O X AXquot X 0 O The definition of single crossing in xt follows the same general pattern we considered in the last lecture Definition fxt is single crossing in xt if I For all x which are both I xeS where 5255 where S is the set containing only y ie x is bigger than y I x is in the preferred set compared to y fxtZfQt I if t gtt then fxt Zfyt I For all x which are both I xeS where SZSS where S is the set containing only y ie x is bigger than y I x is in the strictly preferred set compared to y fxtgtf 1t I if t gtt I then fxt gtfyt We now consider an outline of the proof First we show sufficiency Define txEargmaxxesfxt Suppose xeutx Note that we allow for multiple elements that maximize f over S given t x eut s We want to show that t s zsuts The definition of strong ordering gives us that we want to show 1 xVx eut S 2 xAx euts Step 1 The meet ofx and x XAX is an element of S as SZSS by assumption We have defined x to be the argmax of fxt Therefore fxtzfxx t Step 2 We are assuming fxt is quasisupermodular and part of the definition of quasisupermodularity is fxtZfxx t gt fxVx tZfx t Therefore fxVx tZfx t Step 3 x is an element of an S2 the set in which x lies It is thus a member of a set strongly ordered greater than the set containing XVX Therefore it is possible to simply plug in the definition of single crossing to get fxVx tZfx t but fx t is an argmax m39 Essentially the same proof is done to show that xAx euts Step 1 by definition of argmax we have fx t ZfxVx t Step 2 by quassupermodularity we have fxVx t fx t Step 3 by single crossing we have FxVx tZfxt but fxt is an argmax m39 The proof for necessity is a bit messier This also is a twopart proof but the method for the two parts is essentially the same The first part which we look at is showing that f must be quasisupermodular We begin with 5255 and we write possible combinations that lead to the argmanx in each set S S S S S S x xVy x xVy x xVy my y x y y x y y x xVy x xVy x xVy amy y X y y X y y x xVy x xVy x xVy amy y 95A y 95A y The larger versions of the letters are the proposed argmaxes Each set of four corresponds to the four possible argmax combinations that are within S and 5 However three of these options do not satisfy the ordering of the sets S S 39 S S 39 S S 39 x xVy x xVy x xVy my y xA y y xA y y x xV x xVy x xVy M xA y y xA y y Kxxyy However you will note that all of the remaining possibilities satisfy the definition of quasi K superm odularity The method is the same for showing that the f must be characterized by single crossing However the definitions of the rows are t t rather than S S as in the table above Each box contains X X Y Y The pattern of smaller versus larger elements proposed combinations of maxima is the same After crossing out the combinations that do not satisfy the ordering of the set you find the definition of single crossing must be satisfied Advanced Math 1 6 Dynamic Programming Before beginning with dynamic programming let s do a little discussion of three characteristics of maximization functions that are often discussed 1 Dynamic consistency If look at the function to be maximized at some time and then at some time in the future the tail for the initial time that corresponds to the future actions must be the same as when you look at the weighting of future actions from the later point in time A difference of a scale factor is allowed Another way of stating this is that if you maximize over the entire function you get the same optimal path as you would if you set the first element and then maximize over the rest of the choices Even more the ordering of the future paths will be the same independent of whether you have not yet maximized the first element or not Note that this is different than the definition of dynamic consistency that is often used in macro re optimal monetary policy When this does not hold you have a fairly odd game It is as if the person playing the game today is different from the person playing the game tomorrow One example of this is that of a hyperbolic discount factor utility In hyperbolic utility discounting is at greater than an exponential form Exponential discount may be Hyperbolic discounting is steeper time time An explicit example of hyperbolic utility is Uuc0 A uc1 18 uc2l16uc3 Which discounts the second period hugely then has a discount rate of 2 after that From t1 the hyperbolic utility looks like uc1 l uc2l8uc3 which can not be transform ed into the tail from the point of the first period by multiplying the entire thing by any scale factor The type of thinking in the hyperbolic utility function does seem to play a role for example in the benefits of the snooze button At that point in the morning the extra five minutes seems very worthwhile For modeling purposes it is generally easiest to model this type of situation with a different player at each point in time The optimal from the initial time perspective is to somehow promise not to do what you are tempted to do in the future 2 Stationarity From some F1 on the future looks the same as it did from t0 on How things in the future look relative to now is the same for a bunch infinite number of nows While the first element of dynamic consistency holds for most normal situations this is generally assumed simply because it simplifies the math For example it is perfectly reasonable to assume that discount rates are lower when you are young than when you are old39 however as long as you can take that into account when you are young then dynamic consistency will still hold One example in which stationarity does not hold is going to the front During time zero you are at the front and may die Therefore you discount accordingly Uuc0 A uc118 uc2 116 uc3 This may look hyperbolic but it s not Assuming you re not dead from time one on it looks like uc112 uc2 14 uc3 which can be transformed into from the initial time simply by multiplying by A 3 Recursiveness Past elements do not enter into the utility function For example Uctmc0c1 is very difficult to solve This could be reasonable if habit formation is an important element However this is not a big deal39 you can solve for nonrecursive elements by including them in a state variable Uuctxt where xthabit1ext1c1 For other types of problems this could be the history explicitly and be directly in the utility function There is a discussion in micro theory about whether this is important it doesn t seem to be no A utility function that satisfies all of the above properties is the simple 2 uct 20 The idea behind dynamic programming is backward recursion This is the same principle of optimality as that used in subgame perfection This same concept is used in as simple an analysis as Stephen Covey Consider what you would like them to say at your funeral and then figure out how to make it all true The solution plan will be optimal from all points on For a mathematical discussion of dynamic programming we turn to Kamien and Schwartz The only element of the three discussed above that will be inherent in these problems is dynamic consistency it is hard wired in through the setup of the J equation Kamien and Schwartz do a problem with a salvage value the last term in what is maximized but we ignore this for this discussion X is the state variable eg capital stock t is time up to time T u is control variable The general problem is as follows T o Maxjfa xudt st x gtxu xt0 x0 a 0 Defining the value function of this as follows T o Jt0xo Maxjftxudt st x gt xu xt0 x0 a 0 Keeping the two constraints in the background we can split the maximization into two parts tuAt T Jt0xo Max Iftxudt Max Iftxudt zezu1um tn tEtnuAlTtnAt Which can be rewritten as 2UAt Jt0xo Max j ftxudtJt0 Atx0 At a ring tuAt in From that point many people use contraction mapping theorem from here This is simple if you set Al you don t need to integrate We will not do it this way Rather we will begin by rewriting this and calling it equation 1 2UAt 1 Jt0 Atx0 At Jt0xo Max jftxuat zeumznmz in For continuous time we will use the Taylor expansion using the term hot for the higher order terms of minimal interest The Taylor expansion of the value function looks as follows Jt0Atx0AxJtox0 Lt0x0At Jxt0x0Ax hot In the stochastic case the second order terms as well as the first order terms will play a role Moving the first item on the right hand side of the equation over leads to Jt0Atx0AxJtox0 Lt0x0At Jxt0x0Ax hot Plug in from equation 1 2UAt 0 Max J39 fa x udt Jt0x0At Jxt0x0Ax hot zeunjinmz in Treat the At as the dt that it is divide through by it and we get 0Max ftxu Lt0x0 Jxt0x0AxAt a 0Max ftxu Lt0xo J xtoxo x and thus we get the Bellman Equation 0Max ftxu Lt0xo Jxt0xogtgtxgtul This we can solve using the cookbook Hamiltonian From math class remember that the Hamiltonian is the method used to solve the following eqation HMax ftxugtxu Of course A is the marginal valuation and thus is the Jx in the problem we are trying to solve The following three equations solve the Hamiltonian FOC Hu0 Euler Equation 1 Hx Transversality Condition xT0 We use this as above when there is a finite horizon Before discussing the Hamiltonian we had the above equation in the form equation 2 2 Max f txugtxul tx Therefore we are going to need to slightly modify to get the RHS to look like the Hamiltonian Ltx Max ftxugtxu and the RHS is the Hamiltonian To solve Ex Mz39nJo e39quotax2 bu2 dt st x u x0 x0 gt 0 a gt 0b gt 0 1 Write down in the form of equation 2 Of course fe39 ax2bu2 and gu so J t Mi 7 e39 ax2bu2JXu 19 Do the FOC Take the partial derivative of the RHS wrt u Solve that equation for u and plug back in for u in equation 2 2e39 buJx0 solves to uJxe39 2b J e39 ax2 szezn4b szenZb

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