CVM数据分析中的半参数模型及实证研究.pdf

?34?9?Vol.34,No.92014?9?Systems Engineering Theory&PracticeSept.,2014?:1000-6788(2014)09-2332-07?:Q141;O212.4?:ACVM?,?,?,?(?,?150030)?(CVM)?.?CVM?,?(WTP)?,?WTP,?.?,?CVM?,?,?CVM?.?,?90.53?/?/?,?.?CVM?.?(CVM);?;COX?;?;?The semi-parametric model and empirical research in CVMAO Chang-lin,WANG Jing,GAO Qin,CHEN Hong-guang(Department of Management Science and Engineering,Northeast Agricultural University,Harbin 150030,China)Abstract The contingent valuation method(CVM)is a availably,flexible non-market valuation methodwhich is widely used to evaluate the environmental goods,especially the landscape evaluation.The doublebounded dichotomous choice is the most effective format of contingent valuation models except that itcan not provide a direct estimate of the willingness to pay(WTP)and its parameter model can notfully fit the WTP as well as exploit the sample data information.This research provides a new pathby addressing an analysis method of the CVM based on survival analysis and builds a semi-parametricproportional hazards model to analyse the impact factors of willingness to pay.Taking Sanjiang plainwetland landscape evaluation as an example,the willingness to pay of Sanjiang wetland landscape is RMB90.53 per year.Individual average annual income,educational background,respondents location and theattitudes to environment protect are the main risk factors which affect the willingness to pay.This researchshows the feasibility and effectiveness of the semi-parametric analysis methods to deal with the data ofdichotomous choice CVM.Keywordscontingent valuation method(CVM);survival analysis;COX proportional hazard model;Sanjiang Plain;landscape value1?(CVM)?,?,?12.CVM?,?Hicks?3.?CVM?(WTP)?(WTA),?.?:?,?.Hoehn?,?4,?:2012-12-27?:?(71171044);?(2013M531012);?(13DJJJ01)?:?(1964),?,?,?,?,?,?:?,E-mail:;?(1988),?,?,?,?,?:?.?9?,?:CVM?2333?,?“?”.?,?“?”?“?”?,?.?NOAA?CVM?CVM?5.?CVM?6?79?.?,?CVM?10,?11.?CVM?,?CVM?,?1213?.?WTP?,?,?.?CVM?,?CVM?1415,?1617?.?CVM?,?1820.?.?CVM?,?CVM?2,?WTP21.?,?CVM?,?,?22.?23?,?.?,?2425,?CVM?WTP?,?CVM?.?WTP?,?WTP?,?WTP?.?,?,?.2?26,?.?,?.?.2.1 WTP?CVM?,?WTP?“?”,?,?,?,?“?”,?“?”.?,?,?,?“?”,?“?”.?“?”?“?”?.?,?,?“?”?“?”.?:?T?“?”,?TU,?TL.?,?4?:“?-?”?“?-?”?“?-?”?“?-?”,?YY?YN?NY?NN.?WTP?,?,?1?.2.2 WTP?2.2.1 COX?h(t)?:h(t)=limt0P(t T t)t=h(t,x)(1)?,T?,x?,?.COX?,?h(t)?:?1?CVM?-?WTPTU?-?TWTPTU?-?TLWTPT?-?WTP ti?.?COX?:L()=n?i=1?h0(t)exp(1xi1+2xi2+mxim)?S?R(ti)h0(t)exp(1xs1+2xs2+mxsm)?i(4)?ti?H0(t)?:ln?h(t,X)h0t?=1x1+2x2+mxm(5)?S0(t)?H0(t)?:S0(t)=expH0(ti)(6)?:S(t;X)=S0(t)exp(1x1+2x2+mxm)=h(t,x)(7)?,S0(t)?t?,x?,?.2.2.2 CVM?CVM?WTP?,?WTP?COX?.?CVM?,?h(A)?:h(A)=limA0P(A T A)A=h(A,x)(8)?:A?,x?,?.WTP?:h(wtp)=h0(wtp)exp(1x1+2x2+mxm)(9)?:h0(wtp)?WTP?,x?WTP?,?,m?.?CVM?:L()=n?i=1?h0(wtp)exp(1xi1+2xi2+mxim)?S?R(Ai)h0(wtp)exp(1xs1+2xs2+mxsm)?i(10)?,?WTP?A,?A+A,?“?”?,?i=0,“?”?,?i=1.?L()?j?.?,xj?.?j 0?,xj?,?xj?:?j 0?,xj?,?xj?.WTP?:S(wtp;X)=S0(wtp)exp(1x1+2x2+mxm)=h(t,x)(11)3?3.1?:?,?.?,?,?.?,?.?27?,0,1,5,10,20,50,100,200?,?,?.?1?.?9?,?:CVM?2335?T?TU?TL?_?_?TL=WTPTT=WTPTU?1?CVM?.?.?.?.?1302?665?.3.2?3.2.1?1197?,?1003?,?194?,?77%,?29.1%.?,?1149?.?,?585?,?564?.?35.6%,?2?.?1?.?749?,?2?.?3?.?,?724?,?3?.?2?0.820.840.860.880.900.920.940.960.981.00050100150200250?3?2?11171140.97420.0171171140.97451161120.96640.0342332260.970101091020.93670.0643423280.9592094800.851120.1284364080.93650106840.792190.1795424920.90810097660.680260.2686395580.873200110720.655360.3277496300.847?3?-?-?-?-?4165103515?0.0570.0900.1420.7112336?34?3.2.2?WTP?(Wald)?,?9?,?edu,?,?4?5?.?4?con?,?5?,?ass?,?5?,?sign?,?5?,?sat?,?5?,?dis?,?,?inc?,?8?,?3000?,?6?5?(?)2?dfSig6000.90037.13760.000WTP?Cox?6?,?:ln?h(t,X)h0t?=0.159con 0.077ass 0.141sign+0.068sat+0.139add 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