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)
Thank you very much,
Best,
Marc
Rui Barradas wrote
>
> Ok, I forgot to say that 't' is also an R function, the matrix transpose.
> Sorry, but after 'par' I thought (in my mind) I had said it when in fact I
> even talked about 't'!
> Use 'tt'.
> If 'tt' is a vector you must first define it, in your code it doesn't
> exist. That's why R searches for and finds an object that does exist,
> function t().
>
> You must also create 'a' beforehand. Maybe it makes sense to assign the
> choice and then call optim.
>
> In functions L2 and L1 you don't use the values of, resp., P33 and P22.
> The calls to P3 and P2 are a waste of time.
>
> And, as a final note, optim minimizes it's objective function, so the
> standard trick is to minimize the symmetric of the log like.
> In your case, pass -llfn to optim.
>
> Rui Barradas
>
> infinitehorizon wrote
>>
>> Hello Rui,
>>
>> First of all, thanks a lot!
>> 1. I changed par to param,
>> 2. t is a variable too, a binary one, b is the parameter associated to
>> it,
>>
>> 4. Yes, this is where I am stuck actually.
>>
>> I fixed the code for likelihood functions as follows, but still getting
>> the same error:
>>
>> L3 <- function(b1,b2,b3,b,t) {
>> P11 <- P1(b1,b,t)
>> P22 <- P2(b2,b,t)
>> P33 <- P3(b3,b,t)
>>
>> L3l <- P11*P22*P33
>> return(L3l)
>> }
>>
>> L2 <- function(b1,b2,b3,b,t) {
>> P11 <- P1(b1,b,t)
>> P22 <- P2(b2,b,t)
>> P33 <- P3(b3,b,t)
>>
>> L2l <- P11*P22
>> return(L2l)
>> }
>>
>> L1 <- function(b1,b2,b,t) {
>> P11 <- P1(b1,b,t)
>> P22 <- P2(b2,b,t)
>>
>> L1l <- P11
>> return(L1l)
>> }
>>
>> # Log-likelihood function
>>
>> llfn <- function(param,a,t) {
>>
>> b1 <- param[1]
>> b2 <- param[2]
>> b3 <- param[3]
>> b <- param[4]
>>
>> lL1 <- log(L1(b1,b2,b,t))
>> lL2 <- log(L2(b1,b2,b3,b,t))
>> lL3 <- log(L3(b1,b2,b3,b,t))
>>
>> llfn <- (a==1)*lL1+(a==2)*lL2+(a==3)*lL3
>> }
>> start.par <- c(1,1,1,1)
>> est <- optim(param=start.par,llfn, method =
>> c("CG"),control=list(trace=2,maxit=2000), hessian=TRUE)
>>
>> In the end, i receive the same error message,
>> actually, first I tried without start.par and I got the error of "object
>> param not found", then I defined start.par, it lead me to same error
>> "cannot coerce type 'closure' to vector of type 'double'".
>>
>> I am aware that my questions might be too basic, but as I said I am not
>> familiar with syntax :/
>> Thanks for your help!
>>
>> Best,
>>
>> Marc
>>
>>
>>
>> Rui Barradas wrote
>>>
>>> Hello,
>>>
>>> There are several issues with your code.
>>>
>>> 1. The error message. Don't use 'par' as a variable name, it's already
>>> an R function, tyo get or set graphics parameters.
>>> Call it something else, say, 'param'.
>>> This is what causes the error. You must pass initial values to optim,
>>> but the variable you're passing doesn't exist, you haven't created it so
>>> R finds an object with that name, the graphics parameters function.
>>> Avoid the confusion.
>>> And create 'param' with as many values as expected by llfn before the
>>> call.
>>>
>>> 2. 't' is also a parameter. Take it out of llfn formals and put it in
>>> 'param'. Then, inside llfn's body,
>>>
>>> t <- param[5]
>>>
>>> 3. It still won't work. llfn will not be passed a value for 'a', for the
>>> same reason it can't find 't'.
>>>
>>> 4. Then, look at L3 and the others. The line just before return.
>>>
>>> L3l <- (P11=1)*(P22=1)*(P33=1)
>>>
>>> After computing P11, etc, you're discarding those values and assigning 1
>>> to each of them.
>>> Your likelihood functions just became constants...
>>> And if this is a typo, if you meant P11 == 1, etc, it's even worse. You
>>> can't expect that ratios of exponentials to be equal to that one real
>>> value.
>>>
>>> Points 1-3 are workable but this last one means you have to revise your
>>> likelihood.
>>> Good luck.
>>>
>>> Hope this helps,
>>>
>>> Rui Barradas
>>>
>>> infinitehorizon wrote
>>>>
>>>> Hello,
>>>>
>>>> I am new to R and I am trying to estimate a discrete model with three
>>>> choices. I am stuck at a point and cannot find a solution.
>>>>
>>>> I have probability functions for occurrence of these choices, and then
>>>> I build the likelihood functions associated to these choices and
>>>> finally I build the general log-likelihood function.
>>>>
>>>> There are four parameters in the model, three of them are associated to
>>>> three discrete choices I mentioned, and one of them is for a binary
>>>> variable in the data (t). There are also latent variables but I didn't
>>>> put them in this question because if I figure out how to do this, I
>>>> will be able to add them as well.
>>>>
>>>> I am not familiar with the syntax I have to write in the likelihood
>>>> functions, so I really doubt that they are true. Below I simplify the
>>>> problem and provide the code I've written:
>>>>
>>>> # Probabilities for discrete choices for a=3, a=2 and a=1 respectively
>>>> P3 <- function(b3,b,t) {
>>>> P <- exp(b3+b*(t==1))/(1-exp(b3+b*(t==1)))
>>>> return(P)
>>>> }
>>>> P2 <- function(b2,b,t) {
>>>> P <- exp(b2+b*(t==1))/(1-exp(b2+b*(t==1)))
>>>> return(P)
>>>> }
>>>> P1 <- function(b1,b,t) {
>>>> P <- exp(b1+b*(t==1))/(1-exp(b1+b*(t==1)))
>>>> return(P)
>>>> }
>>>>
>>>> # Likelihood functions for discrete choices for a=3, a=2 and a=1
>>>> respectively
>>>>
>>>> L3 <- function(b1,b2,b3,b,t) {
>>>> P11 <- P1(b1,b,t)
>>>> P22 <- P2(b2,b,t)
>>>> P33 <- P3(b3,b,t)
>>>>
>>>> L3l <- (P11=1)*(P22=1)*(P33=1)
>>>> return(L3l)
>>>> }
>>>>
>>>> L2 <- function(b1,b2,b3,b,t) {
>>>> P11 <- P1(b1,b,t)
>>>> P22 <- P2(b2,b,t)
>>>> P33 <- P3(b3,b,t)
>>>>
>>>> L2l <- (P11=1)*(P22=1)*(P33=0)
>>>> return(L2l)
>>>> }
>>>>
>>>> L1 <- function(b1,b2,b,t) {
>>>> P11 <- P1(b1,b,t)
>>>> P22 <- P2(b2,b,t)
>>>>
>>>> L1l <- (P11=1)*(P22=0)
>>>> return(L1l)
>>>> }
>>>>
>>>> # Log-likelihood function
>>>>
>>>> llfn <- function(par,a,t) {
>>>>
>>>> b1 <- par[1]
>>>> b2 <- par[2]
>>>> b3 <- par[3]
>>>> b <- par[4]
>>>>
>>>> lL1 <- log(L1(b1,b2,b,t))
>>>> lL2 <- log(L2(b1,b2,b3,b,t))
>>>> lL3 <- log(L3(b1,b2,b3,b,t))
>>>>
>>>> llfn <- (a==1)*lL1+(a==2)*lL2+(a==3)*lL3
>>>> }
>>>> est <- optim(par,llfn, method =
>>>> c("CG"),control=list(trace=2,maxit=2000), hessian=TRUE)
>>>>
>>>> And when I run this code I get "cannot coerce type 'closure' to vector
>>>> of type 'double'" error.
>>>> I will really appreciate your help. Thanks,
>>>>
>>>
>>
>
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