Discrete choice model maximum likelihood estimation

Your function will not evaluate as coded. i.e., llfn(start.par) doesn't
"work", as there
are unequal numbers of arguments. Also, while R allows you to use
variables that are not
explicitly defined for a function, I've almost always got into trouble if
I don't pass
them VERY carefully.

Finally, as author of the code used to put CG in optim, I'll advise
against its use. One
of my least successful pieces of code. Rcgmin is better, but you really do
need analytic
derivatives to make it sing. For 5 parameters, use NM, or better the nmk
from dfoptim package.

Best, JN


On 05/15/2012 06:00 AM, r-help-request@ wrote:

Message: 13
Date: Mon, 14 May 2012 04:21:57 -0700 (PDT)
From: infinitehorizon <barisvardar@>
To: r-help@
Subject: Re: [R] Discrete choice model maximum likelihood estimation
Message-ID: <1336994517063-4629910.post at .nabble>
Content-Type: text/plain; charset=us-ascii

Hello again,

I changed the name to tt. 
and for a and tt actually I was getting them from data, I didn't put them
here in the question. Now I restructured my code and below I copy the
full
code, I tried many things but still getting the same error, I don't
understand where is the mistake.

I also defined one more variable to increase comprehension of the
problem.
Instead of data, I define three representative vectors in the beginning.
If
you run this code, you will see the error message.

# Variables, a: discrete choice variable-dependent, x and tt are
independent
variables, tt is binary
a  <- c(1,1,2,3,2,3,1,2,2,3,1,1)
x  <- c(23,26,12,27,10,30,13,20,23,44,17,15)
tt <- c(1,0,0,1,1,1,0,0,1,0,1,1)

# First, just to see, the linear model

LM	<-lm(a ~ x+tt)
coefLM	<- coefficients(LM)
summary(LM)

# Probabilities for discrete choices for a=3, a=2 and a=1 respectively 
P3 <- function(bx,b3,b,tt) { 
P <- exp(bx*x+b3+b*(tt==1))/(1-exp(bx*x+b3+b*(tt==1))) 
return(P) 
} 
P2 <- function(bx,b2,b,tt) { 
P <- exp(bx*x+b2+b*(tt==1))/(1-exp(bx*x+b2+b*(tt==1))) 
return(P) 
} 
P1 <- function(bx,b1,b,tt) { 
P <- exp(bx*x+b1+b*(tt==1))/(1-exp(bx*x+b1+b*(tt==1))) 
return(P) 
}

# Likelihood functions for discrete choices for a=3, a=2 and a=1
respectively

L3 <- function(bx,b1,b2,b3,b,tt) { 
P11 <- P1(bx,b1,b,tt) 
P22 <- P2(bx,b2,b,tt) 
P33 <- P3(bx,b3,b,tt) 

L3l <- P11*P22*P33 
return(L3l) 
} 

L2 <- function(bx,b1,b2,b,tt) { 
P11 <- P1(bx,b1,b,tt) 
P22 <- P2(bx,b2,b,tt) 

L2l <- P11*P22 
return(L2l) 
} 

L1 <- function(bx,b1,b,tt) { 
P11 <- P1(bx,b1,b,tt) 

L1l <- P11 
return(L1l) 
}

# Log-likelihood function 

llfn <- function(param) { 

bx <- param[1] 
b1 <- param[2] 
b2 <- param[3] 
b3 <- param[4] 
b <- param[5] 

lL1 <- log(L1(bx,b1,b2,b,tt)) 
lL2 <- log(L2(bx,b1,b2,b3,b,tt)) 
lL3 <- log(L3(bx,b1,b2,b3,b,tt)) 

llfn <- (a==1)*lL1+(a==2)*lL2+(a==3)*lL3 
} 
start.par <- c(0,0,0,0,0) 
est <- optim(param=start.par,llfn,x=x, a=a, tt=tt, method =
c("CG"),control=list(trace=2,maxit=2000), hessian=TRUE)