In order to estimate the following logistic equation using nonlinear regression (NLS):
Code:
y(t) = m [ 1/(1+exp(-(c+qt) )) - 1/(1+exp(-(c+q(t-1)))) ] + u (t)
Code:
program nlmymodel version 14 syntax varlist(min=3 max=3) [aw fw iw pw] if, at(name) local y: word 1 of `varlist' local t: word 2 of `varlist' local tlag: word 3 of `varlist' tempname m c q scalar `m' = `at'[1,1] scalar `c' = `at'[1,2] scalar `q' = `at'[1,3] tempvar expterm explterm generate double `expterm' = exp(-1*(`c'+`q'*`t')) `if' generate double `explterm' = exp(-1*(`c'+`q'*`tlag')) `if' replace `y' = `m'*((1/(1+`expterm'))-(1/(1+`explterm'))) `if' end
Code:
gen tlag=t[_n-1]
Code:
nl mymodel @ y t tlag, parameters(m c q) initial(m 100 c 37.21 q 0.245)
Code:
Iteration 0: residual SS = 1.08e+14 Source | SS df MS -------------+---------------------------------- Number of obs = 85 Model | -1.943e+13 -1 1.9434e+13 R-squared = -0.2201 Residual | 1.077e+14 85 1.2676e+12 Adj R-squared = -0.2057 -------------+---------------------------------- Root MSE = 1125877 Total | 8.831e+13 84 1.0513e+12 Res. dev. = 2610.012 ------------------------------------------------------------------------------ y | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- /m | 100 . . . . . /c | 37.21 . . . . . /q | .245 . . . . . ------------------------------------------------------------------------------ Parameter q taken as constant term in model & ANOVA table
0 Response to Zero Iteration when using nl regression !
Post a Comment