Example 14.7 A Two-Equation Maximum Likelihood Problem

The following example and notation are taken from Bard (1974). A two-equation model is used to fit U.S. production data for the years 1909–1949, where is capital input, is labor input, is real output, is time in years (with 1929 as the origin), and is the ratio of price of capital services to wage scale. The data can be entered by using the following statements:

   proc iml;
      z={ 1.33135 0.64629 0.4026 -20 0.24447,
          1.39235 0.66302 0.4084 -19 0.23454,
          1.41640 0.65272 0.4223 -18 0.23206,
          1.48773 0.67318 0.4389 -17 0.22291,
          1.51015 0.67720 0.4605 -16 0.22487,
          1.43385 0.65175 0.4445 -15 0.21879,
          1.48188 0.65570 0.4387 -14 0.23203,
          1.67115 0.71417 0.4999 -13 0.23828,
          1.71327 0.77524 0.5264 -12 0.26571,
          1.76412 0.79465 0.5793 -11 0.23410,
          1.76869 0.71607 0.5492 -10 0.22181,
          1.80776 0.70068 0.5052  -9 0.18157,
          1.54947 0.60764 0.4679  -8 0.22931,
          1.66933 0.67041 0.5283  -7 0.20595,
          1.93377 0.74091 0.5994  -6 0.19472,
          1.95460 0.71336 0.5964  -5 0.17981,
          2.11198 0.75159 0.6554  -4 0.18010,
          2.26266 0.78838 0.6851  -3 0.16933,
          2.33228 0.79600 0.6933  -2 0.16279,
          2.43980 0.80788 0.7061  -1 0.16906,
          2.58714 0.84547 0.7567   0 0.16239,
          2.54865 0.77232 0.6796   1 0.16103,
          2.26042 0.67880 0.6136   2 0.14456,
          1.91974 0.58529 0.5145   3 0.20079,
          1.80000 0.58065 0.5046   4 0.18307,
          1.86020 0.62007 0.5711   5 0.18352,
          1.88201 0.65575 0.6184   6 0.18847,
          1.97018 0.72433 0.7113   7 0.20415,
          2.08232 0.76838 0.7461   8 0.18847,
          1.94062 0.69806 0.6981   9 0.17800,
          1.98646 0.74679 0.7722  10 0.19979,
          2.07987 0.79083 0.8557  11 0.21115,
          2.28232 0.88462 0.9925  12 0.23453,
          2.52779 0.95750 1.0877  13 0.20937,
          2.62747 1.00285 1.1834  14 0.19843,
          2.61235 0.99329 1.2565  15 0.18898,
          2.52320 0.94857 1.2293  16 0.17203,
          2.44632 0.97853 1.1889  17 0.18140,
          2.56478 1.02591 1.2249  18 0.19431,
          2.64588 1.03760 1.2669  19 0.19492,
          2.69105 0.99669 1.2708  20 0.17912 };

The two-equation model in five parameters is

where the variables and are considered dependent (endogenous) and the variables , , and are considered independent (exogenous).

Differentiation of the two equations and with respect to the endogenous variables and yields the Jacobian matrix for and , where corresponds to rows (equations) and corresponds to endogenous variables (refer to Bard (1974)). You must consider parameter sets for which the elements of the Jacobian and the logarithm of the determinant cannot be computed. In such cases, the function module must return a missing value. Here is the code:

   start fiml(pr) global(z);
      c1 = pr[1]; c2 = pr[2]; c3 = pr[3]; c4 = pr[4]; c5 = pr[5];
     /* 1. Compute Jacobian */
      lndet = 0 ;
      do t= 1 to 41;
         j11 = (-c3/c4) * c1 * 10 ##(c2 * z[t,4]) * (-c4) * c5 *
               z[t,1]##(-c4-1) * (c5 * z[t,1]##(-c4) + (1-c5) *
               z[t,2]##(-c4))##(-c3/c4 -1);
         j12 = (-c3/c4) * (-c4) * c1 * 10 ##(c2 * z[t,4]) * (1-c5) *
               z[t,2]##(-c4-1) * (c5 * z[t,1]##(-c4) + (1-c5) *
               z[t,2]##(-c4))##(-c3/c4 -1);
         j21 = (-1-c4)*(c5/(1-c5))*z[t,1]##( -2-c4)/ (z[t,2]##(-1-c4));
         j22 = (1+c4)*(c5/(1-c5))*z[t,1]##( -1-c4)/ (z[t,2]##(-c4));

         j = (j11 || j12 ) // (j21 || j22) ;
         if any(j = .) then detj = 0.;
            else detj = det(j);
         if abs(detj) < 1.e-30 then do;
            print t detj j;
            return(.);
         end;
         lndet = lndet + log(abs(detj));
      end;

Assuming that the residuals of the two equations are normally distributed, the likelihood is then computed as in Bard (1974). The following code computes the logarithm of the likelihood function:

     /* 2. Compute Sigma */
      sb = j(2,2,0.);
      do t= 1 to 41;
         eq_g1 = c1 * 10##(c2 * z[t,4]) * (c5*z[t,1]##(-c4)
                 + (1-c5)*z[t,2]##(-c4))##(-c3/c4) - z[t,3];
         eq_g2 = (c5/(1-c5)) * (z[t,1] / z[t,2])##(-1-c4) - z[t,5];
         resid = eq_g1 // eq_g2;
         sb = sb + resid * resid`;
      end;
      sb = sb / 41;
     /* 3. Compute log L */
      const = 41. * (log(2 * 3.1415) + 1.);
      lnds = 0.5 * 41 * log(det(sb));
      logl = const - lndet + lnds;
      return(logl);
   finish fiml;

There are potential problems in computing the power and log functions for an unrestricted parameter set. As a result, optimization algorithms that use line search fail more often than algorithms that restrict the search area. For that reason, the NLPDD subroutine is used in the following code to maximize the log-likelihood function:

   pr = j(1,5,0.001);
   optn = {0 2};
   tc = {. . . 0};
   call nlpdd(rc, xr,"fiml", pr, optn,,tc);
   print "Start" pr, "RC=" rc, "Opt Par" xr;

Part of the iteration history is shown in Output 14.7.1.

The results are very close to those reported by Bard (1974). Bard also reports different approaches to the same problem that can lead to very different MLEs.