《极大似然估计》由会员分享，可在线阅读，更多相关《极大似然估计（6页珍藏版）》请在金锄头文库上搜索。
1、经济数学期末作业题目经数学期末大作业姓名学号专业班级指导教师学 院经济与贸易学院完成日期2013年6月17日1利用数据“probit作业xls”用stata软件中的线性回归模 型估计参数d =卩 + 卩 pe + 卩 roe + 卩 qual + 卩 debt + 卩 sd012345+ 卩 cq + 卩 incr + 卩 ln z + 卩 ma + u6789利用stata软件做线性回归:SourceSSdfMSNumber of obs=992F( 9,982)=97.43Model112.233061912.4703402Prob F=0.0000Residual125.68629398
2、2. 127990115R-squared=0.4717Adj R-squared=0.4669Total237.919355991. 240080076Root MSE=.35776d10Coef.Std. Err.tP| t|95% Conf.Intervalpel-.0027299.0015998-1.710.088-.0058692.0004095roe2.0113235.00124939.060.000.0088719.0137752qual3.0004199.0001442.920.004.0001374.0007024debt4.0029588.00141762.090.037.
3、0001769.0057407sd5.0487167.01841252.650.008.0125843.0848492cq6-.0481151.015907-3.020.003-.0793307-.0168994incr7-.0001327.0000536-2.480.013-.0002379-.0000275lnz8.1913293.010714317.860.000.1703037.2123549ma9.00003.00001931.560.119-7.78e-06.0000678cons-3013602.0757553-3.980.000-4500211-1526994d = -0.30
4、13602-0.0027299pe+0.0113235roe+0.0004199qual+0.0029588debt+0.0487167sd-0.0481151cq -0.0001327incr+ 0.19132931n z+0.000CB* ma+ u2利用数据“probit作业xls”用stata软件中的probit模型 估计参数，对比结果，指出哪些系数估计量的数值和显著性 明显不一样。Probit regressionNumber of obs=992LR chi2(9)=740.29Prob chi2=0.0000Log likelihood=-297.15455Pseudo R2=0
5、.5547d10Coef.Std. Err.zP|z|95% Conf.Intervalpe1.0107877.00795431.360.175-.0048025.0263779roe2.1025684.01052689.740.000.0819364.1232005qual3.0039084.00103863.760.000.0018728.005944debt4 0177304.00780832.270.023.0024264.0330345sd5.1864054.10712811.740.082-.0235618.3963725cq6-.3977655.0911208-4.370.000
6、-.5763589-2191721incr7-.000576.0002595 2.220.026-.0010846-.0000674lnz81.118548.081638413.700.000.95854021.278557ma9.0001081.0001520.710.477-.0001898.000406cons-5.032097.4703416-10.700.000-5.953949-4.110244d = 1-5.03209+0.0107877 pe+ 0.1025684oe+ 0.0039084qual+ 0.0177304/ebt+ 0.186405细d 一0.3977655cq-
7、0.000576 incr+1.118548 ln z + 0.000108-1ma+ u 0在取a =0.05为显著性水平条件下，stata软件操作结果显示：P pe1 = 0.175 a Pma9 = 0.477a，所以pe1、ma9两项的系数和其显著新不一致。3利用数据“probit作业xls”用gauss软件，编写probit模 型程序计算参数，初值采用题1的结果，对比gauss程序和 stata软件的结果。理论叙述：1, y = 1x-0 + u 0二元选择模型：x y = 010-假设：x和u独立CMLE法，找出条件概率(密度)f (y | x; 0Q)条件分布率：(Pr y =
8、1| x = Prx - 0 + u 01 x = Pru - x-0 I x*= Pru -x - 0 = O(x - 0).Pr y = 01 x = 1 一(x - 0 )条件概率：f ( y | x; 0 )=(x-0 ) y - 1-(x-0 )1-y截面数据：(x , y )已知i i i =1,.,n单一样本点：(X , y )发生的概率为：(x -P ) y, - 1(x -P )1 - y,i iii全部样本点：(x，y.)-i i i 1，.，n发生的概率为:(X -Pbi-1-(X P)1-y,极值问题:max卩Qmax rf (X - P )y - 1 一(X - P
9、)1-yi1GAUSS程序：use gpe2;load data992,10=e:111.txt;call reset;_nlopt=1;_method=4;_iter=100;_b=-0.3013602,-0.0027299,0.0113235,0.0004199,0.0029588,0.0487167,-0.0481151,-0.0001327, 0.1913293,0.00003;_print=-1;call estimate (&g,data);proc g(data,b);local m;m=sumc(data.,10.*ln(cdfn(b1+b2*data.,1+b3*data.,
10、2+b4*data.,3+b *data.,4+b6 *data.,5+b7*data.,6+b8*data.,7+b9*data.,8+b10*data.,9)+(1-data.,10).*ln(1-cdf n(b1+b2*data.,1+b3*data.,2+b4*data.,3+b 5*data.,4+b *data.,5+b7*data.,6+ b8*data.,7+b9*data.,8+b10*data.,9);retp(m);endp;b=b;print b;end;-5.03210.0107880.102570.00390840.0177310.18641-0.39777-0.0
11、00575951.11850.00010813结论与stata做对比，所有的参数完全一致。4.选择一个实际例子构建似然函数估计模型，收集数据并 编程算出估计值。用东方财富通选取所有股票的净资产收益率roe、资产负债率zbfu、流动比 率Id、每股收益eps作为数据。净资产收益率是许多股民们衡量股票投资价值的 重要指标，而流动比率、资产负债率、每股收益能大致得了解这个公司的财务状 况。本题将roe处于15%40%的数据当做1，而小于15%或者大于40%的数据 则当做0。运用STATA软件做Probit模型估计函数：Probit regressionNumber of obs = LR chi2(
12、3) = Prob chi2 =Pseudo R2 =2422219.47 0.0000 0.0992Log likelihooc=-996.41532roeCoef.Std. Err.zP|z|95% Conf.Intervaleps6.498263.57738311.250.0005.3666137.629912zbfu-.0456343.0558837-0.820.414-.1551643.0638958ld.046375.01053324.400.000.0257303.0670196_cons.4616938.06225917.420.000.3396681.5837195Note: 0 failures and 7 successes completely determined.若以a =0.05为显著性水平，Pzbfu=0.414a =0.05,显著性检验不通过。Roe=0.4617-0.04641d+6.4983eps注:资产负债率=总负债/总资产流动比率二流动资产/流动负债每股收益=净利润/股份总数净资产收益率=净利润/平均所有者权益
