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version 1.9, 2002/03/11 15:24:05 version 1.10, 2002/03/11 22:26:00
Line 430  a large stepm i.e stepm=12 or 24 and the Line 430  a large stepm i.e stepm=12 or 24 and the
 stepm=1 month. If newstepm is the new shorter stepm and stepm can be  stepm=1 month. If newstepm is the new shorter stepm and stepm can be
 expressed as a multiple of newstepm, like newstepm=n stepm, then the  expressed as a multiple of newstepm, like newstepm=n stepm, then the
 following approximation holds:  following approximation holds:
 <pre>aij(n stepm) = aij(stepm) +ln(n)  <pre>aij(stepm) = aij(n . stepm) - ln(n)
 </pre> and  </pre> and
 <pre>bij(n stepm) = bij(stepm) .</pre>  <pre>bij(stepm) = bij(n . stepm) .</pre>
   
   <p> For example if you already ran for a 6 months interval and
   got:<br>
    <pre># Parameters
   12 -13.390179  0.126133
   13  -7.493460  0.048069
   21   0.575975 -0.041322
   23  -4.748678  0.030626
   </pre>
   If you now want to get the monthly estimates, you can guess the aij by
   substracting ln(6)= 1,7917<br> and running<br>
   <pre>12 -15.18193847  0.126133
   13 -9.285219469  0.048069
   21 -1.215784469 -0.041322
   23 -6.540437469  0.030626
   </pre>
   and get<br>
   <pre>12 -15.029768 0.124347
   13 -8.472981 0.036599
   21 -1.472527 -0.038394
   23 -6.553602 0.029856
   </br>
   which is closer to the results. The approximation is probably useful
   only for very small intervals and we don't have enough experience to
   know if you will speed up the convergence or not.
   <pre>         -ln(12)= -2.484
    -ln(6/1)=-ln(6)= -1.791
    -ln(3/1)=-ln(3)= -1.0986
   -ln(12/6)=-ln(2)= -0.693
   </pre>
   
 <h4><font color="#FF0000">Guess values for computing variances</font></h4>  <h4><font color="#FF0000">Guess values for computing variances</font></h4>
   
 <p>This is an output if <a href="#mle">mle</a>=1. But it can be  <p>This is an output if <a href="#mle">mle</a>=1. But it can be
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