数理经济学 课件
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.。
数理经济学第一讲
p1 x1
FOC：
p2 x2 ，又由 p1 x1 p2 x2 w ，解得： 1
w
p1
x1 p, w x2 p, w
1 w
p2
x x x x2 进行比较静态分析， 发现： 1 0; 1 0; 2 0; 即实际消费量随收入增加， 0， w p1 w p2
3
第一讲 数理经济学简介
樊潇彦 复旦大学经济学院 2009
4. 从一个简单的问题开始
求解下面的消费者效用最大化问题（UMP） ：
Max U x1 , x2 ln x1 1 ln x2
x1 , x2
s.t.
G x1 , x2 w p1 x1 p2 x2 0
x1 p, w ?
x2 p, w ?
4
第一讲 数理经济学简介
樊潇彦 复旦大学经济学院 2009
思路是：写出 Lagrange 函数，求出一阶条件，进而求解。
L x1, x2 U x1, x2 G x1, x2 ln x1 1 ln x2 w p1x1 p2 x2
1.
课程目标：为准备攻读经济学类研究生课程的同学打好数理基础。 研究生课程 个人选择 Nhomakorabea2.
基本要求
相信经济学也是一门科学； 修过高等数学、线性代数、概率统计、计算机语言等相关课程； 渴望用理性的方式、科学的方法理解经济现象。
2
第一讲 数理经济学简介
樊潇彦 复旦大学经济学院 2009
Max U x; s.t. G x; 0
经济学之相关数理基础-PPT精选
製作與使用圖形
圖A1.4(c)負相關，顯示回答 問題數目與休閒時間的關
係，為一條越來越陡峭的
下降趨勢曲線
曲線呈現先緩慢、後急遽下降， 代表隨著休閒時間的增加，回答 問題的題數迅速減少。
製作與使用圖形
• 斜率
– 斜率
– 是縱軸變數的變動量除以橫軸變數的變動量， 斜率的大小可以展現一變數對另一變數的影響 程度。若以代表變動量，則y代表縱軸變數 的變動量，y代表橫軸變數的變動量，於是斜 率等於：
給初學者之建議1
• 如何在開始的時候把握一些基本的工具呢？ • 首先要把微積分讀好，經濟學會運用到一
些基本的微積分，尤其對於一階分，二階 微分的意義，全微分及偏微分更是要好好 的了解，如此對於你往後的經濟學生涯是 絕對有幫助的．
給初學者之建議2
• 在經濟學中有許多的圖形及基本的數學工具，在 圖形之中不免會有座標，有直線有曲線，座標有 橫座標及縱座標，直線有斜率，曲線有切線斜率； 數學也大概只有基本微積分而已，這些雖然都是 一些工具，但是在剛開始的時候，就要有仔細研 究的精神，不只在文字上了解經濟學，利用這些 工具所求出的結果來幫助你，更是事半功倍，了 解每一個圖形的真正意含是真的真的很重要的一 件事，不只是在初學時，在往後的每一個時期也 都是如此．
• 當一階微分大於零，其在幾何上的意義為：當ｘ 值愈大時，ｙ值也愈大，即斜率為正值．
微積分2
• （２）二階微分：二階微分由字面上看來就可以 知道就是把一階微分（切線斜率）所得的數式再 作一次一階微分，在幾何上也就是說：當二階微 分大於零時表示，ｘ值愈大則其切線斜率也愈 大．這樣子的關係當然就表示此時曲線是呈現凹 口向上的，可以自己劃圖印證一下．一個凹口向 上的曲線，表示它會有極小值產生．相反的，凹 口向下則會有極大值產生．
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。
《数理经济学》课件
数学符号在数理经济学中具有特定的意义，它们代表了经济变量、参数和函数等。理解这些符号的意义 是理解数理经济学理论的关键。
数学模型与方程
01
模型构建
数理经济学家使用数学模型来描述经济系统。这些模型通常由一组方程
式构成，用来表示不同经济变量之间的关系。
02
方程类型
在数理经济学中，常见的方程类型包括线性方程、非线性方程、微分方
数理经济学的发展历程
总结词
数理经济学的发展历程可以追溯到19世纪，其发展经 历了多个阶段，包括古典数理经济学、新古典数理经 济学和现代数理经济学等。
详细描述
数理经济学的发展历程可以追溯到19世纪，当时一些 经济学家开始尝试运用数学方法来描述和预测经济现 象。古典数理经济学阶段主要关注生产、分配和交换 等经济活动的均衡问题。新古典数理经济学阶段则强 调个体行为和市场均衡的研究，并引入了边际分析和 效用函数等概念。现代数理经济学则更加注重数学模 型的复杂性和精确性，并广泛应用于宏观和微观经济 学等领域。
在数理经济学中，证明方法多种多样 ，包括直接证明、反证法、归纳法和 演绎法等。这些方法用于证明经济定 理和推导经济关系，确保经济理论的 严谨性和准确性。
在数理经济学中，必须遵循一定的推 理原则，如公理化原则、一致性原则 和完备性原则等。这些原则确保了经 济理论的逻辑严密性和科学性。
03
数理经济学的应用
宏观经济学中的应用
经济增长与经济发展
数理经济学在研究经济增长、经济发展等方面发挥了重要作用，通 过建立数学模型来解释国家或地区的经济增长和发展趋势。
财政政策与货币政策
利用数理经济学方法分析财政政策和货币政策的效果，为政府制定 经济政策提供科学依据。
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.。
数理经济学-数理经济学本083jok-PPT精品文档
A
1
21
练习4：利润率=1时，求供给函数及要素需求函数。
L bp Y w
p r w 要素 需求函数 L/Y
1 1
1 A
22
练习4：利润率=1时，求供给函数及要素需求函数。
1 1 1 1 1 1 1 p a r b w A
练习：请写出生产函数表达式
27
N1原材料面粉消耗 N2原材料调料消耗 K固定资本 生 产 函 数 Q产出数量
生 产 L 函 数
V增加值
28
N1原材料面粉消耗 N2原材料调料消耗 K固定资本 生 产 函 数 Q产出数量
生 产 L 函 数
V增加值
N 1 N 2 Q min , ,V 1 b 2 b
数理经济学丶课间休息
1
第3 讲
第2章:生产函数与供给函数 及要素需求函数
2
了解生产函数与供给 函数及要素需求函 数在实际中的应用
3
r
利润最大决策
K
生 产 函 数 Y
w
Max Y s.t rK+wL=C
L
C
4
练习1：
写出生产函数的数学表 达式并画出等产量线。
5
练习1：写出生产函数的数学表达式
Max S.t
Y Y (K, L) r K w L C
Y 1 Y 1 K r L w
13
练习3：边际利润率=利润率?
Max Y Y(K, L) S.t r K wL C Y p Y p pY ? K r L w C
1 1
A A
1
1 1
数理经济学讲义
1.0.3 经济分析中使用数理方法的好处ice between literary logic and mathematical logic, again, is a matter of little import, but mathematics has the advantage of forcing analysts to make their assumptions explicit at every stage of reasoning. This is because mathematical theorems are usually stated in the “if-then” form, so that in order to tap the “then” (result) part of the theorem for their use, they must first make sure that the “if” (condition) part does conform to the explicit assumptions adopted.
• P.Samuelson提出了新古典综合体系。 • 20世纪70年代以来，经济动态分析逐渐兴起
和快速发展。
数理经济学讲义
1.0.3 经济分析中使用数理方法的好处
• 数理经济学与非数理经济学 • Since mathematical economics is merely an approach to economic
数理经济学讲义
1.0.2 数理经济学的历史发展
• 数理经济学的先驱是法国学者A.A.Cournot，他于 1838年发表了《以数学原理研究财富的理论》，提 出了需求函数理论，把人们熟视无睹的商品需求量 与价格之间的关系写成了函数形式。
数理经济学第五讲
Fy T 0, y T ymin , Fy T y T ymin 0
F T Fy T y T 0, T Tmax ,
截断水平终结线
F T F T y T T T 0
V V 0 ，但此时的欧拉方 0和
Fy
程为：
dFy
dt ，对所有 t 0, T 。 dF Fz z Fzt Fzy y Fzy y Fzz z Fzz z dt
T 0
Fyt Fyy y Fyy y Fyz z Fyz z
F t , y, y dt ， 给定广义积分 0
目标泛函收敛（两个充分条件） ： (1) 若 F 在积分区间上有限，且如果 F 在某个有限时刻 t0 之后均为零，则此积分收敛。
8
第五讲
古典变分法
樊潇彦 复旦大学经济学院经济系
(2) 若 F 具有形式 G t , y, y e t ，其中 是正的折现率，且函数 G 有界，则积分收敛。 今后如果函数 F 以通常的形式给出，我们将总是假设积分收敛，否则当采用具 体函数时，则要明确地检查收敛性。
为 了 去 掉 任 意 扰 动 曲 线 的 影 响 ， 我 们 需 要 进 一 步 分 解 p t 。 记 终 结 时 间
T ＝T＋ T ，在终结时间 T 附近有 p T ＝y T y T T ，代回（2）式得到横
截条件的一般形式： Fy
12
第五讲
古典变分法
樊潇彦 复旦大学经济学院经济系
13
第五讲
古典变分法
樊潇彦 复旦大学经济学院经济系
(完整版)数理经济学课件
结合律：（A U B）U C＝A U（B U C）, （A I B）I C＝A I（B I C）;
分配律： (A U B）I C＝（A I C）U（B I C）, （A I B）UC＝（A UC）I（B UC）;
吸收律； 若A B，则A UB＝B；A I B＝A, A I , A \ B＝，A U＝A;
三、关系与函数
1. 二元关系
定义： 任何有序对(s,t)把一个元素s S,与另一个元 素t T联系起来，则这些有序对的集合R被 认为构成S和T之间的一个二元关系。 若(s,t) R,则写成sRt
显然R S T
定义： 当一个二元关系是一集合S与自身的乘积的子集， 称这是集合S上的一个关系。
定义A1.2 如果对于S中的所有元素x与y, 有xRy或yRx, 那么称S上 的关系R是具有完备性。 定义A1.3 如果对于S中的任何三个元素x、y、z, 有xRy和yRz, 则蕴含着xRz，那么称S上的关系R是具有传递性。
举例2：最优控制问题
min : t1 f (t, x(t),u(t))dt t0
s.t : x&(t) (t, x(t),u(t)) x(t0 ) x0 u(t) U
f : R Rn Rm R, : R Rn Rm R,U Rm
四、授课主要内容
相关数学背景知识
（集合与映射、微积分、微分方程）
R2中的凸集
非凸集： 例2：
R2中的非凸集
因此当且仅当把集合内的任意两点用一条 直线联接，该直线完全处于集合内，那么此集 合为凸集。
凸集本质上没有洞，无断点，在边界没有麻 烦的凸凹。
2. 凸集的性质
定理1：凸集的交集是凸集
数理金融学基本知识ppt课件
风险：如何利用金融工具获利和规避风 险
退休后需要回南京安度晚年，能否提出 个投资方案？或创造一个工具供他们投 资？
整理版课件
9
案例1.1：一个投资家的婚礼
——金融工具无所不在
假设你要结婚了，但是你首先要说服太太：婚纱 照我们不拍了。—别急，你仍然爱她，但是有更 有利可图的办法来表达你的浪漫。
整理版课件
13
证明1：公司A的这种投资是有意义的
由于利息支出计入税前成本，因此公司A实际 税后成本为10%×（1-40%）=6%。
公司A投资于优先股的股息税后收入为： 8% －8%×（1－80%）×40%=7.36% 。
证明2：这笔交易对公司B来说亦是盈利的。
公司B通过借出资金获得利息收入，税后收益 为10％×（1－12％）＝8.8%，而公司B向公 司A支付的优先股股息为8%。
套利的三个基本特征：零投资、无风险、 正收益。
但世上没有免费的午餐！无套利原则说明证券 之间的价格可以从技术角度予以确定。
金融工程的基本思路！
整理版课件
22
1.3.3 风险厌恶（Risk aversion）原则
假设两项投资成本相同，预期回报相同， 投资者将选择风险小的投资项目。
例如：投资项目A会得到确定的回报10％，而 项目B的投资项目获得10％的回报是不确定的， 则投资者必然选择项目A。
金融资产（Financial assets）：实际资产的要求 权（ Claims on real assets ），定义实际资产在 投资者之间的配置。
金融资产的价值与其物质形态没有任何关系：债券可 能并不比印制债券的纸张更值钱。
整个社会财富的总量与金融资产数量无关，金融资产 不是社会财富的代表。
数理经济学课件
Matrix Operation Scalar Multiplication
• The product of a matrix ������ and a number ������, denoted by ������������, is a matrix created by multiplying each entry of ������ by ������. ������11 ⋯ ������1������ ������������11 ⋯ ������������1������ ������������������ ⋮ ⋮ ������������������������ ⋮ ������ ⋮ = ������������1 ⋯ ������������������ ������������������1 ⋯ ������������������������ • In summary, within the class of ������ × ������ matrices, addition, subtraction and scalar multiplication are all defined in terms of the corresponding entry of the matrices.
• Note that is this case, the product taken in reverse order, ������ ������ ������ ������ ������ ������ ������ ������ ������ ������ • is not defined.
Matrix Operation Multiplication
Matrix Operation Subtraction
《数理经济学讲义》课件
经济增长模型
总结词
经济增长模型是用于描述一个国家或地区经 济增长规律的数理经济学模型。
详细描述
经济增长模型通常以国民收入或国内生产总 值为研究对象，通过分析影响经济增长的要 素，如劳动力、资本、技术等，来解释经济 增长的原因和规律。该模型对于制定经济发
展战略和政策具有重要的指导意义。
贸易模型
总结词
贸易模型是用于描述国际贸易关系的数理经济学模型 。
详细描述
贸易模型通常以国家间的贸易往来为研究对象，通过 分析比较优势、机会成本等因素，来解释国际贸易的 动因和结果。该模型对于理解国际贸易格局、分析贸 易政策的影响等方面具有重要价值。
2023
REPORTING
THANKS
感谢观看
模型复杂性与解释性
模型复杂性
数理经济学模型往往涉及多个变量和复杂的 相互作用，这使得模型的建立和验证变得困 难。如何简化模型并保持其解释力是数理经 济学家面临的一个重要挑战。
模型解释性
数理经济学模型通常侧重于数学推导和统计 分析，而忽视了模型的经济解释。提高模型 的解释性，使其更好地反映经济现实，是数 理经济学未来的一个重要发展方向。
未来展望
随着数学和计算机科学的进步，数理经济学将继续发展壮大，研究范围将更加广泛，对 经济学的贡献也将更加显著。
2023
PART 02
数理经济学的基本概念
REPORTING述经济现象和过程的抽象数学结构。
02
数学模型能够通过数学方程和不等式来表达经济关系和规律，帮助我 们理解和预测经济行为和结果。
2023
PART 05
数理经济学的挑战与未来 发展
REPORTING
数据质量与可获得性
数理经济学课件
2020/5/24
IV.9/10.10
GuoSipei@CCNUMATH
▪ 严格凹函数:如果在曲线上选择任意两个点M和N并 以一条直线将他们连接起来,线段MN完全位于曲线 下方
▪ 严格凸函数:如果在曲线上选择任意两个点M和N并 以一条直线将他们连接起来,线段MN完全位于曲线 上方
2020/5/24
2020/5/24
IV.9/10.6
GuoSipei@CCNUMATH
• 相对极值的一阶导数检验
▪ 在一阶导数为0的基础上增加一些附加条件,可得到 相对极值检验的重要方法
▪ 若函数f(x)在x=x0处的一阶导数为0,即f’(x0)=0 则函数在x0的值f(x0)将是:
2020/5/24
IV.9/10.7
第四篇 最优化问题
• 第9章 最优化:一类特殊的均衡分析
▪ 最优值与极值 ▪ 相对极大值和极小值：一阶导数检验 ▪ 二阶及高阶导数 ▪ 二阶导数检验 ▪ 麦克劳林级数与泰勒级数 ▪ 一元函数相对极值的n阶导数检验
2020/5/24
IV.9/10.1
GuoSipei@CCNUMATH
第9章 最优化:一类特殊的均衡分析
• 目标均衡将是我们研究的主要内容
▪ 所谓目标均衡是指给定经济单位,如居民户,厂商或 整个经济等的最优状态,而且这些经济单位主动谋 求均衡的实现
• 最优化的古典方法:微积分法 • 现代的方法:数学规划
2020/5/24
IV.9/10.2
GuoSipei@CCNUMATH
最优值与极值
• 最优化问题的实质:从众多方式中选择最适宜的 方式
GuoSipei@CCNUMATH
二阶及高阶导数
• 导数的导数
数理经济学课件第三章
2 2
s.t x1 x2 1 0, a 0, b 0。 解：L( x1 , x2 , ) ax bx ( x1 x2 1)
2 2
L x 2ax1 0 1 L 2bx2 0 x 2 L x1 x2 1 0
1 1 1 1 得驻点： (0,0),(0, 1),(1,0),( , )ห้องสมุดไป่ตู้( , )。 2 2 2 2
3 0 1 1 可知：在驻点( , )，海赛矩阵H 2 3 2 2 0 2 1 负定，函数达到极大值 ； 8 3 0 1 1 在驻点( , )，海赛矩阵H 2 3 2 2 0 2 1 负定，函数达到极小值－ ； 8 在其它驻点，达不到极值。
2 2 2
8 3 H ( x) 2 3 3 8 * 所以x ( , )是局部极大值。 7 7
课堂练习：
2 2
讨论函数f ( x, y) xy( x y 1)的极值。
解： f ( x, y ) 2 2 y (3 x y 1) 0 x f ( x, y ) 2 2 x( x 3 y 1) 0 y
经济学简单应用 : 1、个人的最优消费问题 考虑本期和来期的两期消费选择问题。 设本期消费为C1 , 来期消费为C2，消费效用为 利率为5％，设在两期中可以自由借贷，问该 如何选择两期的消费？
U C1C2。如果本期收入为4000，来期为5880，
解： C1 S 4000, S (1 5%) 5880 C2 max : U C1 (4000*1.05 C1 *1.05 5880) dU 0 C1 4800 C2 5040 dC1
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。