BookmarkSubscribeRSS Feed
seyibote
Calcite | Level 5

Trying to estimate the parameters for the following log-likelihood, have the following code but sas keeps giving me the error "ERROR: (execution) Matrix has not been set to a value." here is my code; if anyone can help will appreciate it attached is also the function I am working with

proc iml;

aarset=

{0.1 0.2 1 1 1 1 1 2 3 6 7 11 12 18 18 18 18 18

21 32 36 40 45 46 47 50 55 60 63 63 67 67 67 67

72 75 79 82 82 83 84 84 84 85 85 85 85 85 86 86};

P = ncol(aarset); m = p;

start f_gglfr(x) global(aarset);

/* use x-a, x[2]-b, x[3]-d, x[4]-delta */

p=ncol(aarset); m=p-4;

sum1 = 0.; sum2 = 0.; sum3 = 0.; sum4 = 0.;

      do i = 1 to p;

         temp = (x[1]+(x[2]/2)*aarset##2);

         if i <= m then sum1 = sum1 + (x[1]+x[2]*aarset);

         sum2 = sum2 + (-x[3]*log(1-exp(-temp)));

  sum3 = sum3 + (x[1]*aarset+(x[2]/2)*aarset##2);

  sum4 = sum4 + (1-exp(-temp));

      end;

      f = m*log(x[3]) - m*log(Gamma(x[4])) + log(sum1) +(x[4]-1)*log(sum2) - sum3 + (x[3]-1)*log(sum4);

      return(f);

   finish f_gglfr;

   start g_gglfr(x) global(aarset);

     /* use x-a, x[2]-b, x[3]-d, x[4]-delta */

      p=ncol(aarset); m=p-4;

      g = j(1,4,0.);

      sum1 = 0.; sum2 = 0.; sum3 = 0.; sum4 = 0.; sum5 = 0.; sum6 = 0.; sum7 = 0.; sum8 = 0.; sum9 = 0.; sum10 = 0.; sum11 = 0.;

      do i = 1 to p;

         temp = (x[1]+(x[2]/2)*aarset##2);

         if i <= m then sum1 = sum1 + (x[1]+x[2]*aarset);

         sum2 = sum2 + (aarset*exp(-temp))/((1-exp(-temp)*log(1-exp(-temp)));

  sum3 = sum3 + (aarset*exp(temp))/(1-exp(temp));

  sum4 = sum4 + aarset;

  sum5 = sum5 + aarset/(x[1]+x[2]*aarset);

  sum6 = sum6 + (aarset##2*exp(-temp))/(log(1-exp(-temp)*(1-exp(-temp));

  sum7 = sum7 + aarset##2/2;

  sum8 = sum8 + (aarset##2*exp(temp))/(1-exp(temp));

  sum9 = sum9 + (log(1-exp(-temp)))/(log(1-exp(-temp)));

  sum10 = sum10 + (log(1-exp(-temp)));

  sum11 = sum11 + log(-x[3]*log(1-exp(-temp)));

      end;

  temp = (1/2)*(2*x[1]+x[2]*aarset##2);

      g[1] = 1/sum1+(x[4]-1)*sum2+sum4+(x[3]-1)*sum3;

      g[2] = sum5+(x[4]-1)/2*sum6+sum7+(x[3]-1)/2*sum8;

      g[3] = m/x[3] - (x[4]-1)/x[3]*sum9+sum10;

      g[4] = -m*psi(x[4])/Gamma(x[4])+sum11;

      return(g);

   finish g_gglfr;

   n = 4;

   x0 = j(1,n,2);

   optn = {1 2};

   con = { 0 0 0 0,

           . . . . };

   call nlptr(rc,xres,"f_gglfr",x0,optn,con,,,,"g_gglfr");

   /*--- Save result in xopt, fopt ---*/

   xopt = xres`; fopt = f_gglfr(xopt);

    

quit;

1 REPLY 1
Rick_SAS
SAS Super FREQ

You should always call your function before you try to optimize. The call gives an error:

x = {2 2 2 2};

f = f_gglfr(x);

The reason is that your TEMP variable has some large numbers (greater than 7000), so 1-exp(-temp) is numerically 0.

I think there might be another problem.When aarset is a vector, the function f_ggfr returns a vector.  I assume you intend for f_ggfr to be a scalar function.

sas-innovate-2026-white.png



April 27 – 30 | Gaylord Texan | Grapevine, Texas

Registration is open

Walk in ready to learn. Walk out ready to deliver. This is the data and AI conference you can't afford to miss.
Register now and lock in 2025 pricing—just $495!

Register now

From The DO Loop
Want more? Visit our blog for more articles like these.
Discussion stats
  • 1 reply
  • 1326 views
  • 3 likes
  • 2 in conversation