Example 19.12 Cauchy Distribution Estimation

In this example a nonlinear model is estimated by using the Cauchy distribution. Then a simulation is done for one observation in the data.

The following DATA step creates the data for the model.

/* Generate a Cauchy distributed Y */
data c;
   format date monyy.;
   call streaminit(156789);
   do t=0 to 20 by 0.1;
      e=rand('cauchy') + 10 ;

The model to be estimated is

$\displaystyle  y  $
$\displaystyle = $
$\displaystyle  e^{-a~ x} + \epsilon  $
$\displaystyle \epsilon  $
$\displaystyle \sim  $
$\displaystyle  \textrm{Cauchy} ( nc )  $

That is, the residuals of the model are distributed as a Cauchy distribution with noncentrality parameter $nc$.

The log likelihood for the Cauchy distribution is

\[  \textrm{ll} = -\log \pi (1+(x-nc)^2)  \]

The following SAS statements specify the model and the log-likelihood function.

title1 'Cauchy Distribution';

proc model data=c ;
   dependent y;
   parm a -2 nc 4;

       /* Likelihood function for the residuals */
   obj = log(constant('pi')*(1+(-resid.y-nc)**2));

   errormodel y ~ general(obj) cdf=cauchy(nc);

   fit y / outsn=s1 method=marquardt;
   solve y / sdata=s1 data=c(obs=1) random=1000
             seed=256789 out=out1;

title 'Distribution of Y';
proc sgplot data=out1;
   histogram y;

The FIT statement uses the OUTSN= option to output the $\bSigma $ matrix for residuals from the normal distribution. The $\bSigma $ matrix is $1\times 1$ and has value $1.0$ because it is a correlation matrix. The OUTS= matrix is the scalar $2989.0$. Because the distribution is univariate (no covariances), the OUTS= option would produce the same simulation results. The simulation is performed by using the SOLVE statement.

The distribution of $y$ is shown in the following output.

Output 19.12.1: Distribution of Y

Distribution of Y