nl estimate for a logistic function

[River Huang]

Dear All, I found this question here (http://bbs.pinggu.org/forum.php?mod=...=1#pid50990852). The end to fit a logistic function such as

y=k/(1+a*exp(bx))， k>0  a>0  b<0

The dataset and a sample code (not really working) are

* Example generated by -dataex-. To install: ssc install dataex
clear
input float(x y)
   167029 28706.89
   161561 33845.98
   185859 36910.36
   218371  43732.3
   258063 52074.97
291307.38 57348.29
268989.72 62685.77
286250.03 66478.91
   266881 69358.89
281825.53 75624.97
end

nl (y={k}/(1+{a}*exp({b}*x))), initial(k 1 a 1 b -0.1)

The result is

. nl (y={k}/(1+{a}*exp({b}*x))), initial(k 1 a 1 b -0.1)
(obs = 10)

Iteration 0:  residual SS =  2.38e+09
Iteration 1:  residual SS =  2.38e+09


      Source |      SS            df       MS
-------------+----------------------------------    Number of obs =         10
       Model |          0          0           .    R-squared     =     0.0000
    Residual |  2.376e+09          9   263946028    Adj R-squared =     0.0000
-------------+----------------------------------    Root MSE      =   16246.42
       Total |  2.376e+09          9   263946028    Res. dev.     =   221.2377

------------------------------------------------------------------------------
           y |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
          /k |   52676.73   5137.568    10.25   0.000     41054.75    64298.72
          /a |          1          .        .       .            .           .
          /b |        -.1          .        .       .            .           .
------------------------------------------------------------------------------
  Parameter b taken as constant term in model & ANOVA table

Any suggestions to obtain a solution are highly appreciated.

[Nick Cox]

If you plot the data it doesn't look a very convincing logistic. At best you have something from the steep midsection but there is little information about asymptotes and it's not clear that y -> 0 as x decreases.

In general nl often requires really good guesses and is not smart about recovering from lousy guesses. When data are implausible for the function fitted the problem can't be easier than that.

[River Huang]

Hi, Nick, Got it and thanks.