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How do I estimate a nonlinear model using ml?

Title   Use of ml for nonlinear model
Authors Weihua Guan, Gustavo Sanchez, StataCorp
Date February 2002; updated July 2005; minor revisions June 2013

Consider the model

y = f(x) + e

where y is the outcome, f(x) is a nonlinear form of covariate x, and e is the random error. The command nl will estimate the parameters of f(x) by using least squares. Here is an example:

. sysuse auto
1978 Automobile Data)

. nl (rep78 = {b0}*(1-exp(-{b1}*headroom))), initial(b0 1 b1 0.1) nolog
(obs = 69)

Source SS df MS
Number of obs = 69
Model 800.673786 2 400.336893 R-squared = 0.9235
Residual 66.3262137 67 .989943487 Adj R-squared = 0.9212
Root MSE = .994959
Total 867 69 12.5652174 Res. dev. = 193.0865
rep78 Coef. Std. Err. t P>|t| [95% Conf. Interval]
/b0 3.435331 .1604479 21.41 0.000 3.115075 3.755586
/b1 1.947203 1.76742 1.10 0.275 -1.580583 5.474988

We can write an nl program to fit the same model. We may want to do this if we need to be able to fit the model with different variables.

program nlnexpgr, rclass

	version 13
	syntax varlist(min=2 max=2) if
	local lhs : word 1 of `varlist'
	local rhs : word 2 of `varlist'

	return local eq "`lhs'={b0=1}*(1-exp(-{b1=0.1}*`rhs'))"

end

This program will give the same results as we obtained above:

. sysuse auto, clear
(1978 Automobile Data)

. nl nexpgr : rep78 headroom, nolog
(obs = 69)

Source SS df MS
Number of obs = 69
Model 800.673786 2 400.336893 R-squared = 0.9235
Residual 66.3262137 67 .989943487 Adj R-squared = 0.9212
Root MSE = .994959
Total 867 69 12.5652174 Res. dev. = 193.0865
rep78 Coef. Std. Err. t P>|t| [95% Conf. Interval]
/b0 3.435331 .1604479 21.41 0.000 3.115075 3.755586
/b1 1.947203 1.76742 1.10 0.275 -1.580583 5.474988

Now let’s consider how we would fit the same model with ml, which uses maximum likelihood. We can change the equation into

e = y − f(x)

If we assume that the errors are distributed as N(0,sigma), the density function is as follows:

f(e) = 1/(sqrt(2π) σ) exp (− e2/(2σ2))

Then, the log-likelihood function will be

lnL = -0.5*ln(2*_pi) - ln(sigma) - 0.5*e^2/sigma^2
    = -0.5*ln(2*_pi) - ln(sigma) - 0.5*(y-f(x))^2/sigma^2

In our example, we have

f(x) = b0*(1-exp(-b1*x))

Considering the formula, we may treat b0 as a parameter and b1*x as the linear combination xb, which has no constant term. We also need to estimate another parameter, “sigma”, the standard deviation of the normal distribution.

program mlnexpgr
	version 13
	args lnf b1x b0 sigma
	tempvar res
	quietly gen double `res' = $ML_y1 - `b0'*(1-exp(-`b1x'))
	quietly replace `lnf' = -0.5*ln(2*_pi)-ln(`sigma')-0.5*`res'^2/`sigma'^2
end

. ml model lf mlnexpgr (b1: rep78 = headroom, nocons) (b0:) (sigma:)
  
. ml max
(output omitted)

                                                  Number of obs   =         69
                                                  Wald chi2(1)    =       2.85
Log likelihood = -96.543274                       Prob > chi2     =     0.0912

rep78 Coef. Std. Err. z P>|z| [95% Conf. Interval]
b1
headroom 1.947199 1.152899 1.69 0.091 -.3124418 4.206841
b0
_cons 3.435331 .1375274 24.98 0.000 3.165782 3.70488
sigma
_cons .9804333 .08346 11.75 0.000 .8168547 1.144012

The estimates we obtained from ml are close to those from the nl program. The estimated standard errors are a little different, but we know that they are from different algorithms.

We could extend the ml program to estimate robust variances with the vce(robust) option or with the vce(cluster clustvar) option.

. ml model lf mlnexpgr (b1:rep78=headroom,nocons) (b0:) (sigma:), vce(robust)

. ml max
 (output omitted)

                                                  Number of obs   =         69
                                                  Wald chi2(1)    =       3.98
Log pseudolikelihood = -96.543274                 Prob > chi2     =     0.0461

Robust
rep78 Coef. Std. Err. z P>|z| [95% Conf. Interval]
b1
headroom 1.947199 .9764862 1.99 0.046 .0333217 3.861077
b0
_cons 3.435331 .1145489 29.99 0.000 3.210819 3.659843
sigma
_cons .9804333 .0743649 13.18 0.000 .8346808 1.126186

The robust estimation gives the same estimated coefficients, but it gives adjusted standard errors for the three parameters in this model. We may also apply this program for clustered data:

 . ml model lf mlnexpgr (b1:rep78=headroom,nocons) (b0:) (sigma:), vce(cluster turn)
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