数理经济学课件精品共152页文档
数理经济学lecture10
Lecture XI:Constrained Optimization IIMarkus M.MobiusOctober29,20021Inequality ConstraintsIn economics,we oftenfind maximization problems with inequality constraints rather than equality constraints:maxx∈R nf(x)such thatg1(x)=b1,..,g m(x)=b m andh1(x)≤c1,..,h p(x)≤c p(1) We assume n>m+p(otherwise the system is overdetermined).Kuhn-Tucker gives thefirst order condition with this problem.Constraint Qualification The Jacobian matrix of the equality constraints and the bind-ing inequality constraints has full rank.E.g.constraints1,..,p0are binding.The vectors{∇g1(x),..,∇g m(x),∇h1(x),..,∇h p(x)} are linearly independent.Theorem.Let x be a local maximizer of the problem.Assume that the constraint qualification is satisfied.Then there are multipliersλ1,...,λm,µ1,...,µp such that:∂f ∂x i (x)=mk=1λk∂g k∂x i(x)+pl=1µl∂h l∂x i(x)1for all i.Furthermore,µl≥0for all l=1,..,p,andµl[h l(x)−c l]=0 for all l=1,..,pRemarks:•concise notation in term of gradients:∇f=mk=1λk∇g k+pl=1µl∇h l(2)•The last condition simply means thatµl=0for any constraint that doesn’t bind.•The positivity condition has to be interpreted in term of arbitrage.Consider the case of only one constraint/inequality.Draw picture in an indifferencef diagram.Going in opposite direction to the gradient of the constraint,∇h,is allowed(the inequality is preserved),so it must decrease f,which implies that we are going in opposite direction to∇f.in conclusion,∇f and∇h have same directions.•Explain what happens when the inequality constraint does not bind.Draw picture in dim2.You can get rid of the constraints which you know won’t bind at the optimum.E.g positivity constraints etc.Short theory of gradient vectors and iso-surfaces.Take f:R n→R.Consider the iso-surface S c={x∈R n|f(x)=c}Property1Let x∈S c then the vector f (x)=grad x f is orthogonal to the affine hyper-plane tangent to S c at x.2Proof:f(x+h)=f(x)+f (x)h+o(h).If f(x+h)=f(x)then f (x)·h=0.f (x)points in the direction where f increases.Take h going the same direction of f (x)then f(x+h)>f(x).Explain why multiplier is positive or zero.LagrangianL(x,λ,µ)=f(x)− m k=1λk g k(x)− p l=1µl h l(x)Write FOC as∂L ∂x i (x,λ,µ)=∂f∂x i(x)−mk=1λk∂g k∂x i(x)−pl=1µl∂h l∂x i(x)=0GeometryConsider only inequality constraints.When only one constraint is binding,simple tangency condition.Otherwise,we can have a kink.The condition∇f(¯x)=µ1∇h1(¯x)+µ2∇h2(¯x)implies something about the geometry at the kink.Example1Max U(x1,x2)st p1x1+p2x2≤I,x1≥0,x2≥0with U(x1,x2)strictly increasing in x1and x2.The budget constraint always binds.L=U(x1,x2)−λ(p1x1+p2x2−I)+µ1x1+µ2x2Three possibilities:1)µ1=µ2=ual FOC:U1=λp1U2=λp22)µ1>0,µ2=0.x1=0,x2=I/p2and U1=λp1−µ1.Explain in words.3)µ1=0,µ2<0.Example23Maximize f(x,y)=x2+x+4y2st2x+2y≤1and x≥0and y≥0.The matrix of constraints always has full rank.The constraint qualification holds. Draw graphLagrangianL=x2+x+4y2−λ1(2x+2y−1)+λ2x+λ3yFOC:∂L ∂x =2x+1−2λ1+λ2∂L∂y=8y−2λ1+λ3We see that f(x,y)is increasing in x and y,therefore the constraint2x+2y=1is binding.A.Suppose thatλ2>0.Then(x,y)=(0,1/2)and f(0,1/2)=1.B.Supposeλ3>0.Then(x,y)=(1/2,0)and f(1/2,0)=3/4.C.Suppose thatλ2=λ3=0.Then2x+1=8y,implying that y=(2x+1)/8.We then infer that2x+(2x+1)/4=1or equivalently x=3/10,y=1/5.We then check f(3/10,1/5)=11/20.2The Envelope’s TheoremConsider an optimization problem with a parameter a.How does the optimal value of f depend on the parameter?We start with the simplest case:unconstrained optimization:Envelope Theorem(Unconstrained Case).Let U denote an open subset of R n and I an open subset of R.Consider the C1mapping f:U×I→R,(x,a)−→f(x,a). Suppose that for each a,there is a solution x(a)of class C1.Let V(a)=f[x(a),a]=Max x∈R n f(x,a)Then:V (a)=∂f∂a[x(a),a](3) 4Proof:First order condition:∂f∂x[x(a),a]=0.Now,V (a)=∂f∂a[x(a),a]+x (a)∂f∂x[x(a),a]=∂f∂a[x(a),a].QEDThe function V(a)is called the value function.Graphic interpretation.We can generalize this:Envelope Theorem(Constrained Case).Let f,g1,...,g k:R n×R→R be C1func-tions.Let x(a)denote the solution of problemMax x∈R n f(x,a)st g1(x,a)=0,..,g m(x,a)=0.for anyfixed choice of the parameter a(Some of these can be inequality constraints)Suppose that the solution x(a)and the Lagrange multipliersλ1(a),..,λm(a)are C1func-tions of a and that the qualification constraint holds.Then,V (a)=∂L∂a[x(a),λ1(a),..,λm(a)](4)=∂f∂a[x(a),a]−mk=1λk(a)∂g k∂a[x(a),a](5)Proof:At the optimumV(a)≡L[x(a),λ(a),a]≡f[x(a),a]−mk=1λk(a)g k[x(a),a]since either g k[x(a),a]orλk(a)=0. Differentiate with respect to a:V (a)=ni=1∂f∂x i[x(a),a]dx ida+∂f∂a[x(a),a]−mk=1λ k(a)g k[x(a),a]−mk=1λk(a)ni=1∂g k∂x i[x(a),a]dx ida+∂g k∂a[x(a),a]5and thusV (a)=∂L∂a[x(a),λ1(a),..,λm(a)]−mk=1λ k(a)g k[x(a),a] +ni=1dx ida∂f∂x i[x(a),a]−mk=1λk(a)∂g k∂x i[x(a),a]We note that the FOCs imply that∂f ∂x i [x(a),a]−mk=1λk(a)∂g k∂x i[x(a),a]=0.We then show thatλ k(a)g k[x(a),a]=0for all k.Either g k[x(a),a]=0,in which case we are done.Or g k[x(a),a]<0.Thenλk(a)=0on a neighborhood of a andλ k(a)=0.This theorem easily generalizes to vector parameters(a1,...,a q).Application:g1(x,a)=h1(x)−a.L=f(x)−λ[h1(x)−a]Then V (a)=∂L/∂a=λ.Example:utility maximization problemλquantifies the sensitivity of optimal utility to changes in the initial wealth level. For this reason,it is often called the marginal utility of income.6。
数理经济学课件.docx
Mathematical Preliminaries1.1.Sets•A set is a well-defined collection of elements・Example: A = {a, b, c} is a set with three elements・a G A means that the element a is in the set Ad g A means that the element d is not in the setA.•0 denotes the null set or empty set, i.e., the set with no element.•set A is a subset of set B 讦a w B for all a w A. This is denoted by A u B, or B A・•union of sets: A u B = {x: x G A or X G B}.•intersection of sets: A n B = {x: x G A and x G B}.•A and B are disjoint sets if A n B = 0.•the complement of set A in set B is the set B\A = {x: x G B and x G A}・•Cartesian product: Let x G X and y e Y, and let (x, y) be an ordered pair. Then (x, y) 6 XxY, where XxY is the Cartesian product of X and Y・1.2.LogicConsider two propositions P and Q・If P implies Q、then P is a sufficient coixlition for Q, and Q is a necessary condition for P. This is denoted by P => Q.If P implies Q and Q implies P, then “P holds if and only 讦Q holds，： P and Q are equivalent, and P is a necessary and sufficient condition for Q. This is denoted by P <=> Q.Let {not P} denote the statement that P is not true.CorHrapositive: If P implies Q, then {not Q} implies {not P}.Example 1: Let x be a real number, P = {x > 0) and Q = {x2 > 0). Then, P => Q and {not Q}=> {not P} hold, but {Q => P} is false.Example 2: Let P = {x > 0) and Q = {x3 > 0}・ Then P o Q, and {not P) <=> {not Q) hold.1.3.Numbers•natural numbers: N = {1, 2, 3,…}・integers: Z = {-3, -2, -1,(), L 2, 3,...}•rational numbers: Q = {a/b: a G Z, be Z, bHO}・irrational numbers: 2 ： 3 …•real numbers: R = {x: x is rational or irrational}.complex numbers: C = {a + b i: a e R, b G R, i = (-1)12), where a is the real part, and b is the imaginary part・the n-fold Cartesian product of R: R n = Rx...xR = {(x h x2, •••, x n): Xj G R, i = 1,n), where Xj is the i-th coordinate of x = (x b x2,x n)1.4.FunctionsLet X and Y be sets.Definition: f: X t Y is a mapping associating each element of X with an element of Y.X is the domain of ff(x) is the image of x under ff(X) = {f(x): x G X} is the image of X under f•a function: if only one point in Y is associated with each point in X・ a coirespondence: if more than one point in Y can be associated with each point in X •inverse function: x = f *(y) if and only if y = f(x).a function f: X T Y is onto 讦f(X) = Y.It means that the equation f(x) = y has at least one solution for each y.•if f(x) and f ly) are both single・valued, then f is on—to-one.It means that the equation f(x) = y has at most one solution for each y.•composite function: h = g(f(x)) = g • f is the composition of f with g satisfying f: A t B u C, g:C t D, g • f: A t D.L5e BoundsLet S c R.•S is bounded from above (from below)讦there exists a G R (b G R) such that x < a (x > b) for all x G S. Then, a is an upper bound of S, and b is a lower bound of S・•The least upper bound (lub) or suDremum (sup) of S is the upperbound of S such that there does not exist a smaller upper bound. It is denoted by sup(S).•The supremum of S is called a maximum (max) of S if sup(S) G S. It is denoted by max(S). •The greatest lower bound (gib) or infimum (inf) of S is the lower bound of S such that there does not exist a larger lower bound. 1( is denoted by inf(S).•The infimum of S is called a minimum (min) of S if inf(S) e S. It is denoted by min(S).• Property:If S c R and S has an upperbound, then S has a supremum.If S u R and S has a lowerbound, then S has an infimum.16 Vector SpaceConsider a set V.LI- associative law: x + (y + z) = (x + y) + z, for all x, y, z, w VL2- identity: there exists O G V such that x + 0 = x for all x G VL3- inverse: there exists (-x) e V such that x + (-x) = 0 for all x e VL4- commutative law: x + y = y + x for all x, y G VL5- associative law: a・([3・x) = (a-p) x for all a, p e R, and for all x G VL6- identity: there exists 1 G V such that l x = x for all x e VL7- distributive law: a・(x + y) = a x + a y for all ae R, and for all x, y w VL8- distributive law: (a + p)-x = a x + p x for all a,卩G R, and for all x G V L9- closure: x e V and y G V implies that (x + y) G V LIO- closure: x G V and ae R implies that (a・x) e V. Definition: A set V is vector space (or linear space) if it satisfies Ll-Ll0. Then x G V is called a vector. Examples: R l\ or C n is each a vector space.1.7. Norms and distancesConsider a function d(x, y) satisfying:M1: d(x, y) = 0 if and only if x = y M2: d(x, y) + d(y, z) > d(z, x) M3: d(x, y) > 0 for all x, y M4: d(x, y) = d(y, x).Definition: For a given set X,讦a function d: XxX —> R satisfies M1-M4, then: X is a metric space, denoted by (X, d)d is a metricd(x, y) is the distance between points x and y.Examples:di(x，y) = [Zi (xj - yi)2!12 = Euclidian distance，denoted by ||x - y||d2(x, y) = maxi 风-y s|d3(x, y) = Zi lx, - y.lNote: Topology consists in studying the properties of sets that are independent of the distance measure chosen.Definition: Let V be a vector space・ A real value function N: V t R is called a norm on V if: N(x) > 0 for all x G V N(x) = 0 if and only if x = 0N(r x) = |r| N(x) for all re R and x e V, andN(x + y) < N(x) + N(y) for all x, y G V.Example: N(x) = d|(x, 0) = [Xi (xj)2]12 = ||x|| is the Euclidian noim of x in R.R n, with Euclidian norm and Euclidian metric, is a normed vector space.Every normed vector space is a metric space with respect to the induced metric defined by d](x, y) = llx - yll.Convex SetsLet X be a vector space (e.g・，X = R n).Definition: A set S c: X is convex if any x, y G S implies that(0x + (1-0) y) e S, for all 0 e R, 0 < 0 < 1.Note: (0 x + (1-0) y) is called a linear combination of x and y.Properties:・Any intersection of convex sets is convex.•Let Si, i = 1,m, be convex sets in vector space X. Then:•(Ejei (Xj Si) = {x: x = Z i=i■•…m oti Xi, XjG Sj, otf R, i = 1,…，Hi} is a convexset.•(SixS2x...xS m) = x i=1■•…m (Si) is a convex set.L9. Compact SetsLet S u R n.Definition: An open ball about x0 e R n with radius r e R, r > 0, is defined as:B r(x0) = {x: x G S, d(x, x0) < r}, where d(x, x0) is the Euclidian distance betweenpoints x and x0.Definition: An open set S u R n is a set S such that, for each x e S, there exists an open ball B r(x) completely contained in S.• The union of open sets is open..A finite intersection of open sets is open.Definition: The inierior of a set S, denoted by int(S), is the union of all open sets contained in S..A set S is open 讦and only 讦S 二int(S).Definition: A set S c R n is closed 讦the set (R n\S) is open.・The intersection of closed sets is closed・.A finite union of closed sets is closed・Definition: The closure of a set S, denoted by cl(S), is the intersection of all closed sets containing S..A set S is closed if and only if S = cl(S).Definition: The boundary of a set S u R" is the set cl(S)ncl(R n/S).Definition: A set S is bounded if there exists an open ball with a finite radius which contains S・Definition: A collection of open sets (S a)a€A in a metric space X is said to be an open cover of a given set S u R n if S u 低入 S a.The open cover (S a)ae A of S is said to admit a finite subcover if there exists a finitesubcollection (SQ^F such that S u S®Definition 1: A set S c R n is compact if and only if it is closed and bounded・Definition 2: A subset S of a metric space X is compact if and only if every open cover of S has a finite subcover.Note: The definition 2 of compactness applies to sets in any metric space, while definition 1 applies only to sets in R n.LIO. SequencesLet (X, d) be a metric space (e.g., X = R n), and let S c X.Definition: A sequence {xj： j = 1,g} in S converges to y if, for any e > 0, there exists a positive integer j" such that j hj，implies d(y, xj) < e.This is denoted by y = limj* {xj, where y is the limit of {Xj}.Note: It does not follow that y = lirOjT® {xj} G S・Definition: A sequenee {Xj： j = 1，…，g} in S is a Cauchy sequence if for any e > 0, there exists a positive integer such that, for any ij > j\ d(x i9 Xj) < £.Definition: If every Cauchy sequence in a metric space is also a convergent sequence, then the* A sequence {xp j = 1,…严} in R" is a Cauchy sequence if and only if it is a convergent sequence, i.e. if and only if there is y G R n such that linij》{Xj} —> y.metric space is said to be complete.By the above definition, this implies that R n is complete (although not all metric spaces are complete)・Definition: Let m(j) be an increasing function: m: {12 3,…} T {1, 2, 3,such that m(k+l) > m(k).Given a sequence {xj： j = 1, 2,…，g}, {m = 1,…，g} is a subsequence of {xj： j =1,2,…，oo}..A set S c R n is closed if and only if every convergent sequenee of points in S converges to a point in S・・ A set S e R n is compact if and only if every sequence in S has a convergent subsequence whoselimit is in S.・ A sequence {xj： j = 1, 2,…，g} in R n converges to y if and only if every subsequence of {xj： j =1,…，g} converges to y.・Every bounded sequence contains a convergent subsequence・Definition: A sequence {xj： j = 1,2,…，oo} is (strictly) increasing if, for all m > n, x m > (>) x n for all n.A sequence {xj： j = 1,2,…，<»} is (strictly) decreasing if, for all m > n, x m < (<) x n forall n・Let X u R, X H 0. If X is bounded from above (below), there exists an increasing (decreasing) sequence in X converging to sup(X) (inf(X)).Definition: Assume that ±oo are allowed as limits of a sequence.The lim sup of the sequence {xj： j = 1,in R is defined as limj^oo {可:j =1,2 •••}, where aj = sup{xj, Xj+b Xj+2. ...}・ It is denoted by linij^oo supk>j x k, or simply by lim SUpjTco Xj・The lim inf of the sequence {Xj： j = 1,in R is defined as limj^ {bj： j = 1,2 …}, where bj = inf{x jt Xj+i, x i+2, ...}. It is denoted by inf k>j x k, or simply by lim infj†Xj・† A sequence xj in R converges to a limit y G R if and only if y = lim supj* Xj = lim in与* Xj.。
数理经济学ppt课件
主讲 ：刘照德 博士 副教授 QQ: 912804610 Email:lzhaode@
广东财经大学经济贸易学院 2013.9.2
1
本课程简介及基本要求
➢ 1.本课程的开设背景及教学计划 ➢ 2.基本要求
➢ 每人需要有一本教材，可以复印或打印 ➢ 认真听课，自学相关基础知识 ➢ 上课不许接听电话； ➢ 随机点名，不得无故旷课，有事需请假 ➢ 上课可以随时提问问题 ➢ 课下邮件或QQ群内讨论 ➢ 按时交作业
10
§1.2 数理经济学的产生与发展
数理经济学在它的历史进程中经历了3个重 要阶段:
➢边际分析阶段 ➢集论与线性分析阶段 ➢汇合阶段
11
§1.2 数理经济学的产生与发展
➢边际分析阶段(1838一1947) 1838年到1947年，是经济学向数学借用武
器的历史发展阶段，借用的基本工具：微 积分，尤其是偏导数、全微分和拉格朗日 乘数法。 边际分析法是这一时期产生的一种经济分 析方法，同时形成了边际效用学派，代表 人物有walras，Jevollss，Gossen，Menger， Edgeworth，Marshall，Fisher，Clark等人。
均衡的稳定性理论、资源最优配置、一般交易论
13
§1.2 数理经济学的产生与发展
➢集论与线性分析阶段(1948一1960) 1838年到1947年，是经济学向数学借用武
器的历史发展阶段，借用的基本工具：微 积分，尤其是偏导数、全微分和拉格朗日 乘数法。 边际分析法是这一时期产生的一种经济分 析方法，同时形成了边际效用学派，代表 人物有walras，Jevollss，Gossen，Menger， Edgeworth，Marshall，Fisher，Clark等人。
数理经济学的基本方法第三章(经济学中的均衡分析)
3.3局部市场均衡--非线性模型
解法三：另一图解法
图3.3
无忧PPT整理发布
3.3局部市场均衡--非线性模型
*离次多项式方程
定理一：给定一个多项式方程
xn+an-1xn-1+……+a1x+a0=0 这里所有系数都是整数，并且xn的系数是1。如 果存在整数根，那么每一个都必须是a0的除数。
不适用情况：x4+5/2x3-11/2x2-10x+6=0
3.3局部市场均衡--非线性模型
P*1=1 P*2=-5(舍） 无忧PPT整理发布
3.3局部市场均衡--非线性模型
解法二：代数法（利用二次公式）
无忧PPT整理发布
3.3局部市场均衡--非线性模型
或
同理应用于.3.7可得 P*1= 1, P*2= -5(舍） 再利用3.6求出Q*=3
无忧PPT整理发布
无忧PPT整理发布
3.2局部市场均衡—线性模型
模型的构建
假设条件：当且仅当超额需求为零（Qd-Qs=0），
即市场出清时，市场实现均衡。
建模：Qd=Qs,
Qd=a-bP, (a,b>0) Qs=-c+dP, (c,d>0)
(3.1)
无忧PPT整理发布
3.2局部市场均衡—线性模型
图3.1
无忧PPT整理发布
无忧PPT整理发布
3.1均衡的含义
缺乏依据的结论：均衡是事物的一种理想的或合意的状态， 因为只有理想状态才会缺乏变化动力。（利润最大化、非充 分就业的国民收入均衡水平）
唯一合理的解释是：均衡是这样一种状态，其一旦达到且外 力不发生变化时，就有维持不变的倾向。
本章--非目标均衡：不是由于对特定目标的刻意追求，而是 由于非个人的或超个人的经济力量相互作用与调节所致。 （给定供求条件下的市场均衡、给定消费与投资方式下的国 民收入均衡）
数理经济学课件
＞0 i=1,2…n 图（a） 图（a ＜0 i=1,2…n 图（b） 图（b 图（c） 图（c
⑶追求享受品种多样化假设：
U ( x1 , x2 )
图（d 图（d）
得到的都是向下弯曲 的截线，故效用函数的整 个曲面是向下弯曲的，数 学上称为凹函数，重要任 务就是寻找一种简洁的凹 函数作为效用函数的数学 表达式。
k (σ −1)⋅δ
σ ⋅( δ −1)
<k
(σ −1) (σ −1)
倍，符合效用函数凹的假设，实际中常用δ=σ， 倍，符合效用函数凹的假设，实际中常用δ=σ，因其推导出的需求 函数表达式一致。 故， 其中：
U ( x1 ⋯ xn ) = A[α1 x1
σ
1
σ
+ ⋯ + α n xn
σ
1
σ
σ
]
(σ −1)
数理经济学
——理论与应用 ——理论与应用
（研究生用）
说
• • • • • •
明
1、课堂学时：40，课外与课堂学习比例为3︰1 2、教 材： 数理经济学——理论与应用 清华大学出版社，张金水著 。 3、参 考 书： （1） 可计算非线性动态投入产出模型，清华大学出版社，张金水著 。 （2) 一般均衡理论，上海财经出版社，罗斯·M.斯塔尔著。 (3) 数理经济学导论，中国统计出版社，伍超标著 （4）数理经济分析入门，中国科学技术大学出版社，候定丕。 （5）价值理论及数理经济学的20篇论文，首都经济贸易大学出版社，吉 拉德·德布鲁著。
2.1 生产过程中投入量与产出量之间定量关系；生产函数的数 生产过程中投入量与产出量之间定量关系； 学表达式
茹少峰数量经济学课程PPT第一章
定义方程 定义方程实质上是数学恒等式，常用符号“ =” 表 示。定义方程一般用于描述经济学概念或前提假设。 行为方程 行为方程描述经济现象的规律，由所研究问题内 含的经济学规律决定。行为方程在数学上是两个或两 个以上变量的一种函数关系，而在经济学上，是两个 或两个以上经济学变量的行为关系。 均衡条件 均衡条件仅出现在均衡模型中，它是联结行为方 程和方程组的桥梁和纽带。在均衡模型中，通常通过 均衡条件方程来求得模型的均衡解。
第三节 数理经济学的研究方法和基本问题
1.研究方法 数理经济学通常是从一定的假设条件出发，将经济活 动量转化为一个或一组变量，继而写出函数式或方程组， 从而得到相应的经济现象或经济系统的数学描述，然后运 用数学推理方法得出结论，这是数理经济学的一般研究方 法，简言之，数理经济学研究方法就是建立经济问题的数 学模型与求解模型。
第一章 数理经济学概述
本章主要学习的内容： 1、数理经济学的定义 2、数理经济学的诞生和发展 3、数理经济学的研究方法和基本问题 4、数理经济学研究的内容与地位
第一节 数理经济学的定义
目前对于数理经济学尚无统一的定义，以下是几种 有代表性的定义： 阿罗(Kenneth J. Arrow)：数理经济学是包括数学概念 和方法在经济学，特别是在经济理论中的各种应用。 蒋中一(Alpha C. Chiang )：数理经济学是一种经济分析 方法，是经济学家利用数学符号描述经济问题，运用已 知的数学定理进行推理的一种方法。
总结
由以上定义可以看出：数理经济学主要是介绍数学 方法如何应用到经济分析中，如经济问题如何用数学模 型表示，一个变量的变化如何影响另一变量的变化等问 题。因此，数理经济学与其说是一门经济学分支学科， 不如说它是一种经济学分析方法。
经济数学课件完整版
fprintf语句
fprintf 为 输 出 命 令 , 其 格 式 为 :fprintf('text
format',val),
其中,text为需要输出的文本内容,val 为需要输
出的变量值,format是对变量值val的显示格式说
明.说明val的值为整数时用%d;说明val的值为以
科学记数法显示时用%e;说明val的值以浮点数
1.0 学习任务1 等额本金还款法还房贷
等额本金还款法是在还款期内把贷款总额按还款期数(贷款分几次还清就是几期)均分,每期偿
还同等数额的本金和剩余贷款在该期所产生的利息.
若贷款总额为b,银行月利率(年利率的1/12)为r,每月一期,总还款期数为n,第k期的还款额记为
f(k),请完成如下任务:
的定义域是各部分的自变量取值集合的并集.求分段函数
的函数值f(x0)时,要根据x0所在的范围选用相应的解析式,
其图形要在同一坐标系中分段作出.
1.1 函数及其性质
显示时用%f,如果该语句的输出完成后需要换行
的话用\n说明.
0.2 数学软件MATLAB的基本用法
0.2.7
平面图形
在MATLB系统中,用plot(x,y)绘制平面曲线y=f(x)的图形,
其中x是自变量的取值范围;y是对应于自变量x函数值.
自变量x的取值常用如下两种形式给出:
(1)x = a∶d∶b,表示自变量x从a开始,以d为间距,在闭区
Out[3]=1.74755
(*这里的1.74755是系统给出的运算结果*)
更一般地,用N [exp,n]得到表达式具有n位有效数字的数值结果.
0.1 数学软件Mathematica的基本用法
数理经济学 chapter1 课件
.Chapter1Basic Probability Theory•A random experiment is a process that generates well-defined outcomes. One and only one of the possible experimental outcomes will occur,but there is uncertainty associated with which one will occur.•Fundamental axioms of modern econometrics:–An economic system can be viewed as a random experiment governed by some probability distribution or probability law.–Any economic phenomena(often in form of data)can be viewed as an outcome of this random experiment.•Two essential elements of a random experiment:–The set of all possible outcomes—sample space.–The likelihood with which each outcome will occur—probability func-tion.•Definition:Sample Space.The possible outcomes of the random experiment are called basic outcomes,and the set of all basic outcomes is called the sample space,denoted by S.When an experiment is performed, the realization of the experiment is one outcome in the sample space.•Examples:Finite Sample Space.Experiment Sample SpaceToss a coin{Head,Tail}Roll a die{1,2,3,4,5,6}Play a football game{Win,Lose,Tie}•Example:Infinite Discrete Sample Space.Consider the number of accidents that occur at a given intersection within a month.The sample space is the set of all nonnegative integers{0,1,2,···}.•Example:Continuous Sample Space.When recording the lifetime of a light bulb,the outcome is the time until the bulb burns out.Therefore the sample space is the set of all nonnegative real number{t:t∈R,t≥0}.•Definition:Event.An event A is a subset of basic outcomes from the sample space S.The event A is said to occur if the random experiment gives rise to one of the constituent basic outcomes in A.That is,an event occurs if any of its basic outcomes has occurred.•Example:A die is rolled.Event A is defined as“number resulting is even”. Event B is”number resulting is4”.Then A={2,4,6}and B={4}.•Remarks:–The words”set”and”event”are interchangeable.–Basic outcome∈sample space,event⊂sample space.•Definition:Containment.The event A is contained in the event B, or B contains A,if every sample point of A is also a sample point of B. Whenever this is true,we will write A⊂B,or equivalently,B⊃A.•Definition:Equality.Two events A and B are said to be equal,A=B, if A⊂B and B⊂A.•Definition:Empty Set.The set containing no elements is called the empty set and is denoted by∅.The event corresponding to∅is called a null(impossible)event.•Definition:Complement.The complement of A,denoted by A c,is the set of basic outcomes of a random experiment belonging to S but not to A.•Definition:Union.The union of A and B,A∪B,is the set of all basic outcomes in S that belong to either A or B.The union of A and B occurs if and only if either A or B(or both)occurs.•Definition:Intersection.The intersection of A and B,denoted by A∩B or AB,is the set of basic outcomes in S that belong to both A and B.The intersection occurs if and only if both events A and B occur.Venn Diagram:Use circles(or other shapes)to denote sets(events).The inte-rior of the circle represents the elements of the set,while the exterior represents elements which are not in the set.Figure1:Venn Diagrams:(a)Complement,(b)Union,and(c)Intersection•Definition:Difference.The difference of A and B,denoted by A\B or A−B,is the set of basic outcomes in S that belong to A but not to B, i.e.A\B=A∩B c.The symmetric difference,A÷B,contains basic outcomes that belong to A or to B,but not to both of them.Figure2:Venn Diagrams:Symmetric Difference.1.3Review of Set Theory•Definition:Exclusiveness.If A and B have no common basic out-comes,they are called mutually exclusive(or disjoint).Their intersection is empty set,i.e.,A∩B=∅.•Definition:Collectively Exhaustive.Suppose A1,A2,···,A n are n events in the sample space S,where n is any positive integer.If∪n i=1A i=S, then these n events are said to be collectively exhaustive.•Definition:Partition.A class of events H={A1,A2,···,A n}formsa partition of the sample space S if these events satisfy(1)A i∩A j=∅for all i=j(mutually exclusive).(2)A1∪A2∪···∪A n=S(collectively exhaustive).Laws of Set Operations•Complementation(A c)c=A,∅C=S •CommutativityA∪B=B∪A,A∩B=B∩A.•AssociativityA∪(B∪C)=(A∪B)∪C,A∩(B∩C)=(A∩B)∩C.•Distributivity:A∩(B∪C)=(A∩B)∪(A∩C),A∪(B∩C)=(A∪B)∩(A∪C). More Generally,for n≥1,B∩(∪n i=1A i)=∪n i=1(B∩A i),B∪(∩n i=1A i)=∩n i=1(B∪A i).•De Morgan’s Laws:(A∪B)c=A c∩B c,(A∩B)c=A c∪B c. More Generally,for n≥1,(∪n i=1A i)c=∩n i=1(A c i),(∩n i=1A i)c=∪n i=1(A c i).Use Venn diagrams to check De Morgan’s Laws for the case of two events(A∪B)c=A c∩B c.Figure3:Validation of(A∪B)c=A c∩B c.Left panel:(A∪B)c;right panel:A c∩B c.•How to prove the general case of De Morgan’s Laws(n>2)(∪n i=1A i)c=∩n i=1(A c i).•Proof.x∈(∪n i=1A i)c⇔x/∈A i for any i=1,···,n⇔x∈A c i for all i=1,···,n⇔x∈∩n i=1(A c i).Hence,the equality holds.•Definition:Probability Function.Suppose a random experiment has a sample space S .The probability function P maps an event to a real number between 0and 1.It satisfies the following properties:(1)0≤P (A )≤1for any event A in B .(2)P (S )=1.(3)Countable Additivity :If countable number of events A 1,A 2,···∈B are mutually exclusive (pairwise disjoint),then P (∪∞i =1A i )=∑∞i =1P (A i ).•Definition:Countable Set.A set S is called countable if the set can be put into1-1correspondence with a subset of the natural numbers.•Remarks:–We can use a(finite or infinite)sequence to list all elements in a countable set.–Finite sets are countable.–The set of natural numbers,the set of integers,and the set of rational numbers are countable sets.–The set of all real numbers in interval(a,b),b>a,is uncountable.Properties of Probability Function:•P (Φ)=0.•P (A c )=1−P (A ).•If C 1,C 2,···are mutually exclusive and collectively exhaustive,thenP (A )=∞∑i =1P (A ∩C i ).•If A ⊂B ,P (A )≤P (B ).•Subadditivity :For events A i ,i =1,2,···,P (∪∞i =1A i )=∞∑i =1P (A i \∪i −1j =1A j )≤∞∑i =1P (A i ).•Theorem:For any n events A1,···,A n,P(∪n i=1A i)=∑i P(A i)−∑i1<i2P(A i1A i2)+∑i1<i2<i3P(A i1A i2A i3)+···+(−1)n+1P(A1A2···A n).•Proof by induction.–When n=2,the theorem is true.–Assume for n=k events,the theorem is true.–For n=k+1events,we haveP(∪k+1i=1A i)=P(A k+1)+P(∪k i=1A i)−P(∪k i=1(A i A k+1)).Apply the formula of n=k events to P(∪k i=1A i)and P(∪k i=1(A i A k+1)), we obtain the formula for n=k+1events.1.5Methods of Counting•For the so-called classical or logical interpretation of probability,we will assume that the sample space S contains afinite number N of outcomes and all of these outcomes are equally probable.•For every event A,P(A)=number of outcomes in AN.•How to determine the number of total outcomes in the space S and in various events in S?1.5Methods of Counting•We consider two important counting methods:permutation and combi-nation.•Fundamental Theorem of Counting.If a random experiment con-sists of k separate tasks,the i-th of which can be done in n i ways,i=1,···,k,then the entire job can be done in n1×n2×···×n k ways.•Example:Permutation.Suppose we will choose two letters from fourletters{A,B,C,D}in different orders,with each letter being used at mostonce each time.How many possible orders could we obtain?There are12ways:{AB,BA,AC,CA,AD,DA,BC,CB,BD,DB,CD,DC}.The word“ordered”means that AB and BA are distinct outcomes.Permutations•Problem:Suppose that there are k boxes arranged in row and there are n objects,where k≤n.We are going to choose k from the n objects tofill in the k boxes.How many possible different ordered sequences could you obtain?–First,one object is selected tofill in box1,there are n ways.–A second object is selected from the remaining n−1objects.Therefore, there are n−1ways tofill box2.–..–The last box(box k),there are n−(k−1)ways tofill it.•The total number of different ways tofill box1,2,···,k isn(n−1)···(n−(k−1))=n! (n−k)!.•The experiment is equal to selecting k objects out of the n objectsfirst, then arrange the selected k objects in a sequence.•Each different arrangement of the sequence is called a permutation.•The number of permutations of choosing k out of n,denoted by P k n,isP k n=n! (n−k)!.•Convention:0!=1.•Example:The Birthday Problem.What is the probability that at least two people in a group of k people(2<k≤365)will have the same birthday?–Let S={(x1,x2,···,x k)},x i represents the birthday of person i.How many outcomes in S?365k.–How many ways the k people can have different birthdays?P k365.–The probability that all k people will have different birthday is P k365/365k.–The probability that at least two people will have the same birthday is p=1−P k365/365k.–k=10,p=0.1169482;k=20,p=0.4114384,k=30,p=0.7063162, k=40,p=0.8912318,k=50,p=0.9703736;k=60,p=0.9941227.Combinations•Choosing a subset of k elements from a set of n distinct elements.•The order of the elements is irrelevant.For example,the subsets{a,b}and {b,a}are identical.•Each subset is called a combination.•The number of combinations of choosing k out of n is denoted by C k n.We haveC k n=the number of choosing k out of n with ordering the number of ordering k elements=P k n/k!=n!/k!(n−k)!.•Example:Combination.Suppose we will choose two letters from four letters{A,B,C,D}.Each letter is used at most once in each arrangement but now we are not concerned with their ordering.How many possible pairs could we have?There are six pairs:{A,B},{A,C},{A,D},{B,C},{B,D},{C,D}.•Example:A class contains15boys and30girls,and10students are to be selected at random for a special assignment.What is the probability that exactly3boys will be selected?–The number of combinations of10students out of45students is C1045.–The number of combinations of3boys out of15boys is C315.–The number of combinations of7girls out of30girls is C730.–Thus,p=C315C730/C1045.•C k n is also denoted by (nk).This is also called a binomial coefficientbecause of its appearance in the binomial theorem(x+y)n=n∑k=0(nk)x k y n−k.•Properties of the binomial coefficients–∑nk=0C k n=2n,∑nk=0(−1)k C k n=0.–∑ik=0C k n C i−kn=C i2n.–C k n+C k−1n=C k n+1.•Example:The Matching Problem.A person types n letters,types the corresponding addresses on n envelopes,and then places the n letters in the n envelopes in a random manner.What is the probability p n that at least one letter will be placed in the correct envelope?•ANS:–Let A i be the event that letter i,i=1,···,n,is placed in the correct envelope.We need to determine the value of P(∪n i=1A i).–Use formulaP(∪n i=1A i)=∑i P(A i)−∑i1<i2P(A i1A i2)+∑i1<i2<i3P(A i1A i2A i3)+···+(−1)n+1P(A1A2···A n).–∑i1<···<i kP(A i1···A ik)=C k n×(n−k)!/n!=1/k!.–P(∪n i=1A i)=1−12!+13!−···+(−1)n+11n!=n∑i=1(−1)i+11i!=−[n∑i=0(−1)i1i!−1]≈1−e−1≈0.632.1.6Conditional Probability•Different economic events are generally related to each other.Because of the connection,the occurrence of event B may affect or contain the information about the probability that event A will occur.•Example:Financial Contagion.A large drop of the price in one market can cause a large drop of the price in another market,given the speculations and reactions of market participants.•Definition:Conditional Probability.Let A and B be two events in(S,B,P).Then the conditional probability of event A given event B, denoted as P(A|B),is defined asP(A|B)=P(A∩B) P(B)provide that P(B)>0.•Properties of Conditional Probability:–P(A)=P(A|S).–P(A|B)=1−P(A c|B).–Multiplication RulesP(A∩B)=P(B)P(A|B)=P(A)P(B|A).•Example:Suppose two balls are to be selected,without replacement,from a box containing r red balls and b blue balls.What is the probability that thefirst is red and the second is blue?ANS:Let A={thefirst ball is red},B={the second ball is blue}.ThenP(A∩B)=P(A)P(B|A)=rr+b·br+b−1.•Theorem:Chain Rule.For any events A1,A2,···,A n,we haveP(A1A2···A n)=P(A1)P(A2|A1)P(A3|A1A2)···P(A n|A1···A n−1) provided P(A1···A n−1)>0.•Remark–P(A1···A n−1)>0implies P(A1)>0,P(A1A2)>0,···.•Theorem:Rule of Total Probability.Let{A i,i=1,2,···}be a partition(i.e.,mutually exclusive and collectively exhaustive)of S,P(A i)> 0for i≥1.For any event B in S,P(B)=∞∑i=1P(B|A i)P(A i).•Example:Suppose B1,B2,and B3are mutually exclusive.If P(B i)=1/3 and P(A|B i)=i/6for i=1,2,3.What is P(A)?(Hint:B1,B2,B3are also collectively exhaustive.)•Bayes’Theorem:P(B|A)=P(A|B)P(B)P(A).•Remarks:–We consider P(B)as the prior probability about the event B.–P(B|A)is posterior probability given that A has occurred.•Alternative Statement of Bayes’Theorem:Suppose E1,···,E n are n mutually exclusive and collectively exhaustive events in the sample space S.ThenP(E i|A)=P(A|E i)P(E i)P(A)=P(A|E i)P(E i)∑ni=1P(A|E i)P(E i).•Example:Auto-insurance Suppose an insurance company has three types of customers:high risk,medium risk and low risk.From the com-pany’s consumer database,it is known that25%of its customers are high risk,25%are medium risk,and50%are low risk.Also,the database shows that the probability that a customer has at least one accident in the current year is0.25for high risk,0.16for medium risk,and0.10for low risk.What is the probability that a new customer is high risk,given that he has had one accident during the current year?•ANS:Let H,M,L are the events that the customer is a high risk,medium risk or low risk customer.Let A be the event that the customer has had one accident during the current year.ThenP(H|A)=P(A|H)P(H)P(A|H)P(H)+P(A|M)P(M)+P(A|L)P(L).Given P(H)=0.25,P(M)=0.25,P(L)=0.50,P(A|H)=0.25, P(A|M)=0.16,P(A|L)=0.10,we haveP(H|A)=0.410.•Example:In a certain group of people the ratio of the number of men to the number of women is r.It is known that the incidence of color blindness among men is p,and the incidence of color blindness among women is p2. Suppose that the person that we randomly selected is color blind,what is the probability that the person is a man?•ANS:Let M and W denote the events“man selected”and“woman se-lected”,and D be the event“the selected person is color blind”.ThenP(M|D)=P(D|M)P(M)P(D|M)P(M)+P(D|W)P(W)=pr1+rpr1+r+p21+r=rr+p.•Example:In a TV game there are three curtains A,B,and C,of which two hide nothing while behind the third there is a Big Prize.The Big Prize is won if it is guessed correctly which curtain hides it.You choose one of the curtains,say A.Before curtain A is pulled to reveal what is behind it,the game host pulls one of the two other curtains,say B, and shows that there is nothing behind it.He then offers you the option to change your decision(from curtain A to curtain C).Should you stick to your original choice or change to C?•ANS:Let A,B,and C be the events“Big Prize is behind curtain A”(respectively,B and C).We can assume P(A)=P(B)=P(C)=1/3. Let B∗be the event“host shows that there is nothing behind curtain B”. ThenP(A|B∗)=P(B∗|A)P(A)P(B∗|A)P(A)+P(B∗|B)P(B)+P(B∗|C)P(C).–If the prize is behind curtain A,the host randomly pulls curtain B or curtain C,so P(B∗|A)=1/2.–If the prize is behind curtain B,P(B∗|B)=0.–If the prize is behind curtain C,P(B∗|C)=1.Therefore,P(A|B∗)=12×1312×13+0×13+1×13=1/3.Similarly,P(C|B∗)=2/3.•If two events A and B are unrelated.Then we expect that the information of B is irrelevant to predicting P(A).In other words,we expect that P(A|B)=P(A).•Definition:Independence.Two events A and B are said to be statis-tically independent if P(A∩B)=P(A)P(B).•Remarks:–By this definition,P(A|B)=P(A∩B)/P(B)=P(A)P(B)/P(B)=P(A).Similarly,we have P(B|A)=P(B).Therefore,the knowledge of B does not help in predicting A.–If P(A)=0,then any event B is independent of A.•Example:Random Walk Hypothesis(Fama1970).If a stock market is fully efficient,then the stock price P t will follow a random walk; that is,P t=P t−1+X t,where the stock price change{X t=P t−P t−1}is independent across different periods.•Example:Geometric Random Walk Hypothesis.The stock price {P t}is called a geometric random walk if X t=ln P t−ln P t−1is independent across different time periods.Note that X t≈P t−P t−1approximates theP t−1relative stock price change.•Example:Suppose two events A and B are mutually exclusive.If P(A)> 0and P(B)>0,can A and B be independent?•ANS:A and B are not independent becauseP(A∩B)=0=P(A)P(B).•Theorem:Let A and B be two independent events.Then(a)A and B c;(b)A c and B;(c)A c and B c are all independent.•Proof.(a)P(AB c)=P(A)−P(AB)=P(A)−P(A)P(B)=P(A)[1−P(B)]=P(A)P(B c),A andB c are independent.•Remark:Intuitively,A and B c should be independent.Because if not, we would be able to predict B c from A,and thus predict B.•Definition:Independence Among Several Events.Events A1,···,A nare(jointly)independent if,for every possible collection of events A i1,···,A iK,where K=2,···,n and i1<i2<···<i K∈{1,2,···,n},P(A i1∩···∩A iK)=P(A i1)···P(A iK).•Remark:We need to verify2n−1−n conditions.For example,three events A,B,and C are independent ifP(A∩B)=P(A)P(B),P(A∩C)=P(A)P(C),P(B∩C)=P(B)P(C),P(A∩B∩C)=P(A)P(B)P(C).•Remark:It is possible tofind that three events are mutually(pairwise) independent but not jointly independent.It is also possible tofind three events A,B,C that satisfy P(A∩B∩C)=P(A)P(B)P(C)but not independent.•Example:SupposeS={a,b,c,d}and each basic outcome is equally likely to occur.Let A1={a,b},A2= {b,c},and A3={a,c}.Then we haveP(A1)=P(A2)=P(A3)=12,P(A1A2)=P(A1A3)=P(A2A3)=14,butP(A1A2A3)=0=1 8 .•Example:Suppose S={a,b,c,d,e,f,g,h}and each basic outcome is equally likely to occur.Let A1={a,b,c,d},A2={a,b,c,d},and A3={a,e,f,g}.ThenP(A1)=P(A2)=P(A3)=1 2 ,P(A1A2A3)=1 8 ,butP(A1A2)=1/2=P(A1)P(A2).•Theorem:If the events A1,A2,···,A n are independent,the same is true for events B1,B2,···,B n,where for each i,the event B i stands for either A i or its complement A c i.•Theorem:If the events A1,A2,···,A n are independent,thenP(A1∪···∪A n)=1−{[1−P(A1)]···[1−P(A n)]}.•Example:Consider experiments1,2,···,n such that in each of them an event D may or not occur.Let P(D)=p for every experiment,and let A k be the event“D occurs at the k-th experiment”.•ANS:A1∪···∪A n is the event“D occurs at least once in the n experi-ments”.ThenP(A1∪···∪A n)=1−{[1−P(A1)]···[1−P(A n)]}=1−(1−p)n.。