optim 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 m
ction(param, a, tt) {
>
>
> llfn <- sum((a==1)*lL1+(a==2)*lL2+(a==3)*lL3) # sum of logs, it's a
> log-likelihood.
> return(-llfn)
> }
>
> Rui Barradas
>
> infinitehorizon wrote
>>
>> Hello again,
>>
>> You are absolutely right about pro
e
> [1] 0
>
> $counts
> function gradient
> 44 NA
>
> $convergence
> [1] 0
>
> $message
> NULL
>
>
> Note the optimum value of zero, est$value == 0
>
> Rui Barradas
>
> infinitehorizon wrote
>>
>> By the way, in my l
By the way, in my last post I forgot to return negative of llfn, hence the
llfn will be as follows:
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
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
>
> 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
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 gener
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