Question

我正在创建一个代码来估计条件logit模型。与我使用“ mlogit”软件包调整的代码相比，我得到的系数有很大不同，请您帮我确定我错了吗？

function(b,x,y,lambda){               

  xb1       <-      x[,1]*b[1]+x[,2]*b[2]+x[,3]*b[3]       
  xb2       <-      x[,4]*b[4]+x[,5]*b[5]+x[,6]*b[6]       
  xb3       <-      x[,7]+b[7]+x[,8]*b[8]
  exb1       <-      exp(xb1)       
  exb2       <-      exp(xb2)      
  exb3       <-      exp(xb3)
  exb_total <-      exb1+exb2+exb3                            
  p1        <-      exb1/exb_total                         
  p2        <-      exb2/exb_total                         
  p3        <-      exb3/exb_total                                
  l1        <-      y[,1]*xb1+y[,2]*xb2+y[,3]*xb3                    
  l2        <-      log(exb_total)                         
  likelihood   <-      -sum(l1-l2)                            
  return(likelihood)                                          
}

 model_8coef<-optim(par=B,fn=LL.MNLR,x=X2,y=Y,lambda=0,control = list(maxit=10000),method = "BFGS")

mlogit代码

mlogit_mode<-mlogit.data(mode,shape = "wide",choice = "mod",varying = c(2:3,6:9))

modelogit<-mlogit(mod~0|1|t+c,mlogit_mode,reflevel = "rail")

summary(modelogit)



 $par
[1]   6.55061382   0.08901337   1.01372853   9.15278946   0.21330866  -0.19448328
[7] -18.70340647  -0.59548734

$value
[1] 1856.872

$counts
function gradient 
  44       18 

$convergence
[1] 0

$message
NULL

与mlogit相比

Call:
mlogit(formula = mod ~ 0 | 1 | t + c, data = mlogit_mode, reflevel = "rail", 
method = "nr", print.level = 0)

Frequencies of alternatives:
   rail     bus     car 
0.28979 0.19240 0.51781 

nr method
6 iterations, 0h:0m:0s 
g'(-H)^-1g = 0.000241 
successive function values within tolerance limits 

Coefficients :
             Estimate Std. Error z-value  Pr(>|z|)    
bus:(intercept) -0.888491   0.977076 -0.9093   0.36317    
car:(intercept)  2.977079   0.996054  2.9889   0.00280 ** 
rail:t          -0.079409   0.010740 -7.3937 1.428e-13 ***
bus:t           -0.100321   0.013061 -7.6808 1.577e-14 ***
car:t           -0.093099   0.011628 -8.0065 1.110e-15 ***
rail:c          -1.033796   0.411701 -2.5110   0.01204 *  
bus:c           -0.594146   0.492629 -1.2061   0.22779    
car:c           -0.865582   0.123303 -7.0199 2.220e-12 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Log-Likelihood: -246.07
McFadden R^2:  0.42518 
Likelihood ratio test : chisq = 364.03 (p.value = < 2.22e-16)

在计算输出中，观察到我获得了非常不同的系数，因此我无法理解我的错误在哪里。数据库的每一行都包含每个可用替代方案的时间和成本变量，在这种情况下，替代方案是：汽车，公共汽车和火车。因此，我必须估算8个系数$ \ beta ^ {auto} {0} $，$ \ beta ^ {auto} {time} $，$ \ beta ^ {auto} { cost} $，$ \ beta ^ {bus} {0} $，$ \ beta ^ {bus} {time} $，$ \ beta ^ {bus} {cost} $ ，$ \ beta ^ {rail} {time} $和$ \ beta ^ {rail} {cost} $ 谢谢！

估计条件对数

0 个答案: