--- imach096d/doc/imach.htm 2002/03/11 15:24:05 1.9 +++ imach096d/doc/imach.htm 2002/03/11 22:26:00 1.10 @@ -1,4 +1,4 @@ - +
@@ -430,9 +430,40 @@ 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 expressed as a multiple of newstepm, like newstepm=n stepm, then the following approximation holds: -aij(n stepm) = aij(stepm) +ln(n) +aij(stepm) = aij(n . stepm) - ln(n)and -bij(n stepm) = bij(stepm) .+bij(stepm) = bij(n . stepm) .+ +For example if you already ran for a 6 months interval and +got:
+# Parameters +12 -13.390179 0.126133 +13 -7.493460 0.048069 +21 0.575975 -0.041322 +23 -4.748678 0.030626 ++If you now want to get the monthly estimates, you can guess the aij by +substracting ln(6)= 1,7917
and running
+12 -15.18193847 0.126133 +13 -9.285219469 0.048069 +21 -1.215784469 -0.041322 +23 -6.540437469 0.030626 ++and get
+12 -15.029768 0.124347 +13 -8.472981 0.036599 +21 -1.472527 -0.038394 +23 -6.553602 0.029856 + +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. +-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 ++Guess values for computing variances
This is an output if mle=1. But it can be