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-2024.png

Don't miss out on SAS Innovate - Register now for the FREE Livestream!

Can't make it to Vegas? No problem! Watch our general sessions LIVE or on-demand starting April 17th. Hear from SAS execs, best-selling author Adam Grant, Hot Ones host Sean Evans, top tech journalist Kara Swisher, AI expert Cassie Kozyrkov, and the mind-blowing dance crew iLuminate! Plus, get access to over 20 breakout sessions.

 

Register now!

Multiple Linear Regression in SAS

Learn how to run multiple linear regression models with and without interactions, presented by SAS user Alex Chaplin.

Find more tutorials on the SAS Users YouTube channel.

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