[R] Discrete choice model maximum likelihood estimation

[Rui Barradas]

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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