[R] loop in optim

[Joshua Wiley]

Hi Edward,

At least for me, your llik() function returns Inf for the starting
values specified, so optim() never gets to estimate anything.  You
need to alter llik() or find starting parameters that work before
worrying about getting the for loop working.

Cheers,

Josh

On Mon, Jul 4, 2011 at 2:34 AM, EdBo <n.bowora at gmail.com> wrote:
> Hi
>
> May you help me correct my loop function.
>
> I want optim to estimates al_j; au_j; sigma_j;  b_j by looking at 0 to 20,
> 21 to 40, 41 to 60 data points.
>
> The final result should have 4 columns of each of the estimates AND 4 rows
> of each of 0 to 20, 21 to 40, 41 to 60.
>
> ###MY code is
>
> n=20
> runs=4
> out=matrix(0,nrow=runs)
>
> llik = function(x)
>   {
>    al_j=x[1]; au_j=x[2]; sigma_j=x[3];  b_j=x[4]
>    sum(na.rm=T,
>        ifelse(a$R_j< 0, -log(1/(2*pi*(sigma_j^2)))-
>                           (1/(2*(sigma_j^2))*(a$R_j+al_j-b_j*a$R_m))^2,
>         ifelse(a$R_j>0 , -log(1/(2*pi*(sigma_j^2)))-
>                           (1/(2*(sigma_j^2))*(a$R_j+au_j-b_j*a$R_m))^2,
>
> -log(pnorm(au_j,mean=b_j*a$R_m,sd=sqrt(sigma_j^2))-
>                           pnorm(au_j,mean=b_j*a$R_m,sd=sqrt(sigma_j^2)))))
>
>       )
>
>   }
>
> start.par = c(0, 0, 0.01, 1)
> out1 = optim(llik, par=start.par, method="Nelder-Mead")
>
>
> for (i in 1: runs)
> {
>  index_start=20*(i-1)+1
>  index_end= 20*i
>  out[i]=out1[index_start:index_end]
> }
> out
>
>
> Thank you in advance
>
> Edward
> UCT
> ####My data
>
> R_j             R_m
> -0.0625         0.002320654
> 0               -0.004642807
> 0.033333333     0.005936332
> 0.032258065     0.001060848
> 0               0.007114057
> 0.015625        0.005581558
> 0               0.002974794
> 0.015384615     0.004215271
> 0.060606061     0.005073116
> 0.028571429     -0.006001279
> 0               -0.002789594
> 0.013888889     0.00770633
> 0               0.000371663
> 0.02739726      -0.004224228
> -0.04           0.008362539
> 0               -0.010951605
> 0               0.004682924
> 0.013888889     0.011839993
> -0.01369863     0.004210383
> -0.027777778    -0.04658949
> 0               0.00987272
> -0.057142857    -0.062203157
> -0.03030303     -0.119177639
> 0.09375         0.077054642
> 0               -0.022763619
> -0.057142857    0.050408775
> 0               0.024706076
> -0.03030303     0.004043701
> 0.0625          0.004951088
> 0               -0.005968731
> 0               -0.038292548
> 0               0.013381097
> 0.014705882     0.006424728
> -0.014492754    -0.020115626
> 0               -0.004837891
> -0.029411765    -0.022054654
> 0.03030303      0.008936428
> 0.044117647     8.16925E-05
> 0               -0.004827246
> -0.042253521    0.004653096
> -0.014705882    -0.004222151
> 0.029850746     0.000107267
> -0.028985507    -0.001783206
> 0.029850746     -0.006372981
> 0.014492754     0.005492374
> -0.028571429    -0.009005846
> 0               0.001031683
> 0.044117647     0.002800551
>
>
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>
> --
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> Sent from the R help mailing list archive at Nabble.com.
>
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>



-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/