数理经济学solutions3

Ec2030:Mathematics for EconomistsHandout 6Harvard University 2January 2003Solutions to Problem Set 3Problem 1.Parts (a)-(h)are standard and are discussed carefully in the notes.At the steady state we have:f (k ∗)−c ∗=k ∗(1)Plugging this into the FOC gives us:=u (c ∗)−δV (k ∗)V (k ∗)=δ[V (k ∗)f (k ∗)](2)From the last equation we get f (k ∗)=1δ.This pins down k ∗.The Inadaconditions ensure that there exists some k ∗>0which satisfies this equation.Intuitively,the slope of f is infinite at 0and then converges to 0for k →∞.Therefore,it has to be 1δfor some k ∗between 0and ∞.When you linearize you have to very careful.Always linearize the original setof equations before you plug in the steady state condition k ∗=f (k ∗)−c ∗.Otherwise you get non-sensical results.Start with the first equation:0=u (c ∗)+u (c ∗)(c −c ∗)−δV (f (k ∗)−c ∗)−δV (f (k ∗)−c ∗)[f (k ∗)(k −k ∗)−(c −c ∗)](3)Now use the steady state conditions and set ∆c =c −c ∗and ∆k =k −k ∗:u (c ∗)∆c =δV (k ∗) 1δ∆k −∆c (4)The second equation gives us:V (k ∗)+V (k ∗)∆k =δ[V (f (k ∗)−c ∗)+V (f (k ∗)−c ∗)(f (k ∗)∆k −∆c )][f (k ∗)+f (k ∗)∆k ]](5)Ignoring second order terms and using the steady state conditions we get:V (k ∗)∆k=δV (k ∗)(1δ∆k −∆c )1δ+δV (k ∗)f (k ∗)∆k V (k ∗)∆k =V (k ∗)(1δ∆k −∆c )+u (c ∗)f (k ∗)∆k V (k ∗) ∆c −1−δδ∆k =u (c ∗)f (k ∗)∆k (6)We have to linear equations in ∆k and ∆c which both contain V (k ∗)aboutwhich we know nothing.But we can eliminate this expression by dividing oneequation through the other one:V (k ∗) ∆c −1−δδ∆k δV (k ∗) 1δ∆k −∆c=u (c ∗)f (k ∗)∆k u (c ∗)∆c (7)Ec2030Handout 6:Solutions to Problem Set 32Let’s set β=u (c ∗)f (k ∗)u (c ∗)which is a positive parameter:∆c −1−δδ∆k ∆k −δ∆c =β∆k ∆c (8)The easiest way to solve this equation is to realize that ∆c =γ∆k for some γwhich we have to find (this is the linearized solution):∆k ∆c ∆k −1−δδ ∆c ∆k ∆c −δ=β∆k ∆c γ−1−δδ1γ=β(9)This gives us a quadratic equation in γ:γ2−γ1−δδ=β−βδγγ2−γ 1−δδ−βδ =β(10)Solve this equation and get the linearization.The interesting root is the positive one (there is exactly one positive and one negative root):having more capital than in steady state means higher consumption.Therefore,we have:γ=1−δδ−βδ+ 1−δδ−βδ 2+4β2(11)Problem 2.Let S >0be a real number.Consider the problem:Maximize ni =1x i =x 1x 2..x nsubject to x 1+x 2+..+x n =S and x 1≥0,x 2≥0,..,x n ≥0.2(a)This is obvious -otherwise the maximized value would be 0and you can certainly always do better by setting x i =S n for example.2(b)We want to maximizeL =ni =1ln x i −λ n i =1x i −S (12)Ec2030Handout6:Solutions to Problem Set33 Note,that we maximize the log of the objective function instead of the objective function itself.This is possible because anything which maximizes the log max-imizes the original function as well.This is often very convenient when dealing with maximization of products.The FOC conditions are:1x i=λ(13)But this implies that all x i are identical and hence x i=Sn.2(c)The log-functions are concave and hence the objective function is concave as the sum of concave functions.The constraint is linear and hence convex.This shows that we have a concave programming problem which has a unique max.2(d)Take those numbers and define y i=x ini=1x isuch thatni=1y i=1.Wethen have:n i=1y i≤ni=11n=1n n(14)But this gives us:n i=1x i≤=(ni=1x i)nn n(15)Now take the n th root on both sides and we get the claim.Problem3.There was of course a mistake here-you can only minimize the problem instead of maximizing it:the objective function is a sum of parabolas which are opened to the top,i.e.are convex.So the sum is convex rather than concave.So let’s minimize the problem instead.To get back a concave maximization problem we maximize the negative of the objective function:L=2x1−x21−1−x22−λ[x1+x2]−µx21−4(16)Note that the constraints are convex so that we have a concave problem.There-fore,the KT conditions are sufficient in this case.The FOC are:2−2x1=λ+2µx1−2x2=λ(17) We now discuss the various comparative slackness cases.If both conditions bind then x1=2and x2=−2or x1=−2and x2=2.Only thefirst case works as thenλ=4>0.But thenµ<0which is impossible.Ec2030Handout6:Solutions to Problem Set34 If no condition binds thenλ=µ=0.Then x1=1and x2=0.But thisviolates x1+x2≤0.If thefirst condition only binds thenµ=0and x1+x2=0.This gives x1=12 and x2=−12such thatλ>0.Hence this is a solution.If the second condition only binds we getλ=0and x1=2or x1=−2.In both cases we getµ<0.Hence the unique solution to the problem is x1=12and x2=−12.Problem4.4(a)Consumers which face a two-part tariffchoose the optimal quantity q such thatθu(q)−qp(18) is maximized,i.e.p=u (q).Note,that they purchase more as prices are lower. Also note,that thefixed fee does not affect consumer behavior as long as they prefer buying some output to no output at all(the monopolist would never set the fee that high).The monopolist sets the fee at F=θu(q)−qp and thus extracts all the remaining consumer surplus from consumers.His total profit is therefore:π=θu(q)−qp+q(p−c)=θu(q)−qc(19) His profit is maximized if u (q)=c,i.e.p=c.He therefore prices at marginal cost.4(b)The monopolist would like to set a higherfixed cost to high-value con-sumers in order to extract more surplus from them.But if there is a second-hand market,high value consumers can buy units from low-value consumers and avoid the higherfixed cost.In fact,even low-value consumers can pool their purchases and therefore divide up thefixed cost.Effectively,the monopolist can only charge a standard price in this case.4(c)In this case the monopolist will not necessarily set prices to marginal cost. Instead he sets F i and p i to induce type i to buy from him.Both consumer types will do so if two conditions are fulfilled for them:(1)Each consumer does not want to imitate the other consumer and choose the bundle intended for him instead of the other consumer type.(2)Each consumer wants to buy his bundle instead of not buying at all.Thefirst condition is called the incentive com-patibility condition and the second one is the individual rationality condition. It will be easier to change variables.Instead of working withfixed costs and prices we assume that the monopolist offers a bundle(q i,T i)to both consumersEc2030Handout6:Solutions to Problem Set35 where T i=F i+p i q i is the total payment made by consumer type i.The four conditions can then be written as:θ1u(q1)−T1≥θ1u(q2)−T2θ2u(q2)−T2≥θ2u(q1)−T1θ1u(q1)−T1≥0θ2u(q2)−T2≥0(20)4(d)The monopolist maximizes his total profitsπ=T1+T2−(q1+q2)c subject to the four conditions above.This is a standard Kuhn-Tucker type problem.It’s concave and hence has a unique solution.Rather than work through all cases we take an informed guess.(a)The low type consumer is just indifferent between buying and not buying.(b)The high-type consumer is indifferent between buying his bundle and choosing the low-type bundle instead.Therefore we have:θ2u(q2)−T2=0θ1u(q2)−T2=θ1u(q1)−T1(21)This gives us T1=θ2u(q1)and T1=θ1u(q1)−(θ1−θ2)u(q2).Plugging this into the objective function gives us(we could also do constrained optimization with a Lagrangian but this is easier):π=θ1u(q1)−(θ1−θ2)u(q2)+θ2u(q2)−c(q1+q2)(22)The FOC become:θ1u (q1)=cθ2u (q2)−(θ1−θ2)u (q2)=c(23)From thefirst condition we see that q2is the same as in(a).Hence the monop-olist uses marginal cost pricing for the high type.However,the lower-type faces a higher price.4(e)Intuitively,the monopolist generates most surplus from the high-type consumers.He starts from a setup where he only sells to those consumers: he then prices at marginal cost and sets F to extract all surplus from them. However,no low-type consumer would buy.So he offers a new tarifffor those consumers with a lowerfixed cost but a higher price.He increases the price only for the low-type consumers because he wants to distort the decision of the high types as little as possible.Ec2030Handout6:Solutions to Problem Set36 Problem5.Thefirst function is concave.The second is quasi-concave be-cause it’s a monotonic transformation of a concave function.The other two are quasi-concave for positive values because they are monotonic transformations as well.。