Hello again,
You are absolutely right about probabilities.. Thanks for reminding me about
that.
I did exactly how you said but in the end I receive the error : "objective
function in optim evaluates to length 12 not 1".
I checked how llfn give a vector instead of scalar, but couldn't figure it
out.
Can you please tell me how did you obtain those estimates?
Thanks again,
Best,
Marc
Rui Barradas wrote
>
> Hello, again.
>
> Bug report:
> 1. Your densities can return negative values, 1 - exp(...) < 0.
> Shouldn't those be 1 PLUS exp()?
>
> P3 <- function(bx,b3,b,tt) {
> P <- exp(bx*x+b3+b*(tt == 1))/(1+exp(bx*x+b3+b*(tt == 1)))
> return(P)
> }
>
> And the same for P2 and P1?
>
> 2. Include 'a' and 'tt' as llfn parameters and call like the following.
>
> llfn <- function(param, a, tt) {
>
> [... etc ...]
> return(-llfn)
> }
>
> start.par <- rep(0, 5)
> est <- optim(start.par, llfn, gr=NULL, a=a, tt=tt)
> est
> $par
> [1] 4.1776294 -0.9952026 -0.7667640 -0.1933693 0.7325221
>
> $value
> [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 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,b,tt))
>>
>> llfn <- (a==1)*lL1+(a==2)*lL2+(a==3)*lL3
>> return(-llfn)
>> }
>>
>> However, it does not fix the problem, I still receive the same error..
>>
>
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