Signaling games are used to model the following situation: Player 1, the Sender, receives some private information and sends a message m E M to Player 2, the Receiver. The latter, in turn, observes m but not 0, and chooses response r E R. Players'payoffs depend on 0, m and r. What could be simpler? Yet, there is a huge number of economically interesting games that fit nicely within this framework: Spence's job market signaling model is the leading example, but applications abound in IO (limit pricing, disclosure...) finance (security design) and political economics

一、频率与频率稳定性 频率 设A是一个事件在相同的条件下进 行n次试验,在这n次试验中事件A发生 了m次。则称m为事件A在n次试验中 发生的频数或频次,称m与n的比值m/n 为事件A在n次试验中发生的频率,记 为fn(A)

性质6设M及m分别是函数f(x)在区间[a,b]上的最大值及最小值,则证明因为m≤f(x)≤M,所以

一、选择题(共5题25分) 1.5分(4381)4381 Na(Hg)()NaCK(m )AgCl+Ag-Ag+AgCINaCK(m2)Na(Hg)(a), m=.02 mol-kg,,=0.875,m2=0.100 mol-kg,y+2=0.780,在98K时电池的电动势为:() (A)0.0384V (B)0.154V (C)0.0413V (D)0.0768V

1、设f(x)=,在[-1,1上求f(x)的一次最佳一致逼近多项式。 2、设f(x)∈Ca,b]试证明:f(x)的零次最佳一致逼近多项式p(x)=(M+m),其中M,m分别为f(x)在[a,b]上的最大值和最小值

1、设f(x)=,在[-1,1上求f(x)的一次最佳一致逼近多项式。 2、设f(x)∈Ca,b]试证明:f(x)的零次最佳一致逼近多项式p(x)=(M+m),其中M,m 分别为f(x)在[a,b]上的最大值和最小值

5.1 已知银是单价金属，费米面近似为球形，银的密度ρm = 10.5×103 kg·m -3，原 子量A = 107.87，电阻率在 295K时为 1.61×10-3Ω·m，在 20K时为 0.0038×10-3Ω·m

An interesting property of an M/M/ 1 queue, which greatly simplifies combining these queues into a network, is the surprising fact that the output of an M/M/ queue with arrival rate is a Poisson process of rate This is part of Burke's theorem, which follows from reversibility A Markov chain has the property that P[future present, past] P[future present] Conditional on the present state, future states and past states are independent

三维复式格子:晶体由N=N1N2N3个原胞构成,一个原胞中有n个原子。 n个原子的质量分别是:m1,m2,m3…m 第l个原胞的位置:R()=la1+l2a2+l3 原胞中各原子的位置

一维复式格子的情形一维无限长链 —P和Q两种不同原子:m、M(M>m)构成的一维复式格子 相邻同种原子间的距离为2a一复式格子的晶格常数