Loopholes
 f  f A  xi ; c  
(1)
 c    i

i
i 1 

Frequently we do not know I or in may cases do not want it to be constant, but would
rather have it given by
2
N
2
 i  aerr  berr fi  cerr fi 2
(2)
In the case that a=0, b=1, and c=0, this is the Poisson distribution. It cannot handle cases
where fi = 0 owing to their infinite contribution to the sum in (1). This is usually handled
by making a equal to a minimum value of 1 in the case that b is equal to 1. It might be
better to change  to
  xi ; c   aerr  berr f A  x; c   cerr f A2  x, c 
(3)
So that
N

 2  c     fi  f A  xi ; c   
2

1
 (4)
f A  x; c   cerr f A2  x, c  
 aerr  berr

Note that for cerr > 0, that a large value of fA makes (4)  a sum of 1’s. This
means that 2  N/cerr . This is usually not significant because a good fit is expected to
have the numerator in (1) equal to the denominator on the average leading to N, but if
the data is barely out of the noise, one might have cerr=1 Then it is almost a test of
..\..\optimization\Robmin\Robmin.doc. If it does not find this loophole it is not working
correctly.
The loophole is diminished by setting cerr to zero but as shown in
..\Poisson\Fitting with Poisson random variables.doc, it still produces a different error
than expected. In summary, the value of fA becomes larger than expected owing to the
fact that a larger value of fA makes 2 smaller.
This is related to the fact that better approximations to fA may raise 2. The
reason that (1) produces a fit that we like is the relative zeroing of the first term in (4), not
the increasing in size of the second.
The recommendation for the best compromise between (2) and (3) is to iterate.
1. Fit the data using a reasonable a, b, c and f1 – option C in nlfit. 2 = Ans1
2. Begin with the last set of coefficients and use option c in nlfit. This will cause fA
to be calculated using the last set of coefficients. 2 = Ans2. Frequently Ans2 >
Ans1. This is due to the use of a different set of {i}.
3. Iterate to completion.
i 1