Shuailin Li doc. Student No. 0922994
更新时间:2023-04-12 10:28:01 阅读量: 实用文档 文档下载
- 世界杯推荐度:
- 相关推荐
Quantitative Method Project BST164
NAME: SHUAILIN LI
STUDENT NO.:0922994
DATE: 11TH FEB 2013
1
Question 1
As can be seen from the following
table, the dependent variable is well
explained by the independent
variables since the R-Squared is
relative high and it is very close to
one.Additionally, the p-value of both
independent variables is equal to zero
which
means they are both significant in this equation. (I)
DW test
H 0: ρ=0
H 1: ρ>0 or ρ<0.
Referring to the DW tables with k=2 and n=55 for the 5% significance level, we see that d L =1.49. Since d (II) Breusch-Godfrey LM test H 0: e t ~ N (0, σ2 I) H 1: e t is generated by AR (p) e t is generated by MA (p) process. The null hypothesis of no serial correlation is strongly rejected. (III) Box-Pierce test H 0: ρ1=ρ2=…=0 H 1: ρ1≠ρ2≠ 0 Reject the null hypothesis of autocorrelation coefficients are zero. (IV) Breush-Pagan test H 0:there is NO heteroscedasticity H 1: there is heteroscedasticity. We can reject the null hypothesis of no heteroscedasticity. 2 (V) RESET testis H 0: the linear functional form is correct H 1: the functional form is of higher order. Reject the null hypothesis of the linear functional form is correct. (IV) Chow test H 0: there is NO structural break H 1: there is structural break Reject the null hypothesis. (1.2) Wald test H 0: β=1 H 1: β≠1. The null hypothesis should clearly be rejected Question 2 We can see from the table that the p- value of constant and the inflation are both larger than 0.05, which means these two variables are not significant in this equation. On the other hand, LNK and FD have less standard deviation and zero probability. These two estimations illustrate that they are significant while R-squared is also close to 1 meaning the dependent variables can be explained well by the independent variables. (I) DW-statistics is 0.251243 which is smaller than 2, so there is positive first order serial correlation of the error term.It is also smaller than the critical value, thus reject the null of non-autocorrelated 3 (II) TheBreusch-Godfrey LM test H 0:e t ~ N (0, σ2 I) H 1: e t is generated by AR (p) process e t is generated by MA (p) process TR 2=42.27822>χ2 (0.95, 2) =5.991 so the null hypothesis o f no serial correlation is strongly rejected. (III) Box-Pierce statistics H 0: ρ1 = ρ2 = …= 0 H 1: ρ1 ≠ρ2 ≠…≠ 0 Since Q*=99.530>χ0.952(4)=9.488, Reject the null hypothesis of autocorrelation coefficients are zero. (IV) Breusch-Pagan test H 0: There is NO heteroscedasticity H 1: There is heteroscedasticity. Reject the null hypothesis. (V) RESET test H 0: the linear functional form is correct H 1: the functional form is of higher order. Reject the null hypothesis. (VI) Chow test H 0: there is NO structural break H 1: there is structural break Reject the null hypothesis. Wald test H 0: θ2=θ3=0 H 1: θ2≠θ3≠0 Null hypothesis is rejected 4 (2.1) The table on the left shows the weighted least square using inflation as the weight. As we can see there are some differences between the weighted statistics and the unweighted statistics. As a result of R 2 is bigger in weighted model than in unweighted model, the dependent variable in weighted model is explained better than in OLS. Moreover, the mean of dependent variable in weighted statistics is less than the unweighted one but the S.D. in WLS is much higher than that in OLS. However, the inflation is still not very relevant and significant in both WLS and OLS since the p-value is bigger than 0.05 (2.2) Hausman ’s test of endogeneity H 0: lnk is exogenous H 1: lnk is endogenous Reject the null hypothesis. So lnk is endogenous. (2.3) In TSLS, the coefficients of independent variable have changed a lot but the significance of them has not changed. (2.4) Sargan ’s instrument validity test H 0: instruments (r, lnpop, infd and infl) are valid H 1: instruments (r, lnpop, infd and infl) are not valid. The test statistics is SARG=(54- 5 4)*0.039563=1.97815 comparing with chi-square critical value χ0.952(4)=9.488, so we cannot reject H 0 and the conclusion is the instruments are valid. Question 3 (I) From the table we can see that the dummy variable is highly significant and has the least S.D. By comparing the results with the regression that excluded the dummy variables, it can be seen that the coefficient estimates on the remaining variables change quite a bit. Then the intercept has changed during the period of 2000- 2011. (II) When a slope dummy variable is included in the estimate equation, it can be seen that the dummy variable is not significant because the p-value is bigger than the critical value of t-statistic.However, the R-square does not change very much. Slope coefficient has not changed during the period of 2000-2011 (III) The intercept and slope dummy variables in this equation are also not significant because of the p-value is larger than the significance level. However, the coefficients of other variables and the R-square have not changed very much. So both intercept and slope coefficient have not changed during the period of 1995-2000 6 Question 4 (I) According to these tables and graphs to compare with chi-square critical value with df=4 is 9.488 and the p-value, we can conclude that for FTSE return does not have ARCH effect (2.74<9.488) while for the RPI return do have ARCH effect (24.57>9.488). (II) As the table shows that there is GARCH effect of FTSE return but no GARCH effect of RPI.We can make this summary by comparing the p-value to the significance level and t-statistic to critical value. RET_FTSE RET_RPI 7 (III) As the table shows there is evidence of asymmetric effects of ‘good news ’and‘bad news ’ on conditional volatility of stock returns in both FTSE return and RPI. (IV) Comparing this result to the question (II), it can be seen that in terms of FTSE return the lag and GARCH variables have changed very significantly in GARCH-M than GARCH. However, these variables do not change very much between GARCH-M and GARCH in RPI. (V) There is IGARCH effect in both FTSE and RPI return because of the p-value is smaller than the significance level. 8 Question 5 (I) Unit root test of LNY H 0: lny has a unit root H 1: trend stationary process We cannot reject the null. H 0: lny has a unit root H 1: level stationary process We cannot reject the null H 0: lny has a unit root H 1: stationary mean zero process We cannot reject the null. H 0: lny has s unit root H 1: level stationary process We reject the null and 1st difference of lny is level stationary. Lny is I(1) process Unit root test of LNK H 0: lnk has a unit root H 1: trend stationary process We cannot reject the null H 0: lnk has a unit root H 1: level stationary process We cannot reject the null. H 0: lnk has a unit root H 1: stationary mean zero process We cannot reject the null. 9 H 0: lnk has a unit root H 1: level stationary process We cannot reject the null. H 0: lnk has a unit root H 1: level stationary process We can reject the null and 2rd difference of lnk is level stationary, thus lnk is I(2) process Unit root test of INFL H 0: infl has a unit root H 1: trend stationary process We cannot reject the null H 0: infl has a unit root H 1: level stationary process We cannot reject the null. H 0: infl has a unit root H 1: stationary mean zero process We cannot reject the null. H 0: infl has a unit root H 1: level stationary process Wecan reject the null so 1st difference of infl is level stationary. Then infl is I(1). (II) Since lny(1), lnk(2), lnfl(1), they have different orders of integrations.So we cannot use E-G to test cointegration.lny=F(c, lnk, infl, d2lnk(-1 to -2), d2lnk(1 to 2), dinfl(-1 to -2), dinfl(1 to 2)). There is an autocorrelation so we 10 have to use DGLS method. So from the table on the right, when we add up AR(1), there is no autocorrelation. Then we can reject the null so lnylnk and infl is cointegrated (III) As a result of the analysis above, we can see that inflation and per capita physical capital stock affect the per capita real income while per capita real income affects itself in the following period. (IV) We can see from the statistics that the null hypothesis is rejected so theinfl and lnkp is conitegrated Then we can use the granger causality H 0: lnk does not Granger cause infl H 1: lnk Granger cause infl We can reject the null H 0: infl does not Granger cause lnk H 1: infl Granger cause lnk We can reject the null (V) Because of the inflation does not Granger causes real income of the economy so there is no need to focus on inflation controlling policies.
正在阅读:
Shuailin Li doc. Student No. 092299404-12
2014许昌县面向社会教师资格证认定技能测试工作日程安排初中、小学及幼儿园12-18
RHEL6 Postfix+Dovecot邮件系统配置04-06
案例分析03-31
数电大作业(智能数字钟) - 图文06-14
外出培训汇报材料03-08
直接和间接控制下合并财务报表实务探讨11-22
苏教版科学五年级下册单元测试卷全册01-28
成功经理人十二项自我管理能力(徐剑)07-25
2014年中国儿童用品行业发展战略08-11
- 1OracleSQL讲义(Student)
- 2OracleSQL讲义(Student)
- 3Research Student Monitoring Group
- 4建立student类,并建立一个student对象数组
- 5LI Salon preparation sheet中级
- 6unit1 - economy - student version
- 72019中考专题复习之专题五 分子、原子、离子、元素精品教育 doc.
- 8Student Standardized Testing Does Not Show
- 9Evaluating holonomic quantum computation beyond adiabatic li
- 10英语作文-how to be a qualified college student
- 教学能力大赛决赛获奖-教学实施报告-(完整图文版)
- 互联网+数据中心行业分析报告
- 2017上海杨浦区高三一模数学试题及答案
- 招商部差旅接待管理制度(4-25)
- 学生游玩安全注意事项
- 学生信息管理系统(文档模板供参考)
- 叉车门架有限元分析及系统设计
- 2014帮助残疾人志愿者服务情况记录
- 叶绿体中色素的提取和分离实验
- 中国食物成分表2020年最新权威完整改进版
- 推动国土资源领域生态文明建设
- 给水管道冲洗和消毒记录
- 计算机软件专业自我评价
- 高中数学必修1-5知识点归纳
- 2018-2022年中国第五代移动通信技术(5G)产业深度分析及发展前景研究报告发展趋势(目录)
- 生产车间巡查制度
- 2018版中国光热发电行业深度研究报告目录
- (通用)2019年中考数学总复习 第一章 第四节 数的开方与二次根式课件
- 2017_2018学年高中语文第二单元第4课说数课件粤教版
- 上市新药Lumateperone(卢美哌隆)合成检索总结报告
- Shuailin
- Student
- 0922994
- doc
- Li
- 2022年春高一地理湘教版必修2课后练习题:1.2 人口合理容量
- 专科疾病常规护理工作指引
- 机场代码表_2014_09_24_09_12_08
- 最新北师大版四年级上册数学《期中考试试题》(带答案)
- 0603固定资产管理办法(QW06-03:2010/1)
- 人教版2022年七年级英语上册Unit5—9单元练习汇编(无答案)
- 精选儿童常用顺口溜大全.doc
- 【最新】高尔夫球如何握杆-推荐word版 (3页)
- 2022年华南师范大学01603分析化学考研复试核心题库
- 国产智能手机市场分析精选文档
- 第三章-----金属及其化合物--导学案2
- 部编版语文六年级上册《第六单元测试题》(附答案)
- 最新仁爱英语八年级(下) Unit 8 Topic 1 Section B 说 课 稿
- 全套技术交底-技术交底
- (完整版)单级直齿圆柱齿轮减速器说明书毕业设计
- 肥城六中9月高三月考卷数学
- 软件产品开发及技术服务合同样本
- 山东省济南市 2022-2022学年人教版八年级上册 期末英语模拟题三(
- 2014福州大学研究生招生简章,招生目录,考试书籍,分专业统计
- 旅游专业实习报告范文