One possibility would to -expand 2- the dataset and add a small amount
of noise to the dependent variable. For example:
clear
set obs 10
gen id = _n
gen double X = .
gen double be = 10 + 10*uniform()
gen double la = 2.5 + uniform()
gen double wespe = 2.5 + uniform()
gen double wvari = 2.5 + uniform()
expand 2
bysor id: replace be = be + .00000001 if _n == 2
count
forv i = 2(2) `=r(N)' {
qui {
local j = `i' - 1
nl (be = {X=1} - (la*((wespe*exp(0.5*wvari)*norm(ln({X} ) ///
-ln(wespe)-wvari))/wvari^.5) - ({X}*norm(ln({X})) - ///
{X}*norm((ln({X})-ln(wespe))/wvari^0.5)) )) in `j'/`i'
replace X = _b[/X] in `j'/`i'
}
}
duplicates drop id, force
gen double test = X - be - (la*((wespe*exp(0.5*wvari)*norm(ln(X) ///
-ln(wespe)-wvari))/wvari^.5) - (X*norm(ln(X)) - ///
X*norm((ln(X)-ln(wespe))/wvari^0.5)) )
l, noobs
. clear
. set obs 10
obs was 0, now 10
. gen id = _n
. gen double X = .
(10 missing values generated)
. gen double be = 10 + 10*uniform()
. gen double la = 2.5 + uniform()
. gen double wespe = 2.5 + uniform()
. gen double wvari = 2.5 + uniform()
.
.
. expand 2
(10 observations created)
. bysor id: replace be = be + .00000001 if _n == 2
(10 real changes made)
. count
20
.
. forv i = 2(2) `=r(N)' {
2. qui {
3. local j = `i' - 1
4. nl (be = {X=1} - (la*((wespe*exp(0.5*wvari)*norm(ln
({X} ) ///
> -ln(wespe)-wvari))/wvari^.5) - ({X}*norm(ln({X})) - ///
> {X}*norm((ln({X})-ln(wespe))/wvari^0.5)) )) in `j'/`i'
5. replace X = _b[/X] in `j'/`i'
6. }
7. }
.
. duplicates drop id, force
Duplicates in terms of id
(10 observations deleted)
. //Test
. gen double test = X - be - (la*((wespe*exp(0.5*wvari)*norm(ln
(X) ///
> -ln(wespe)-wvari))/wvari^.5) - (X*norm(ln(X)) - ///
> X*norm((ln(X)-ln(wespe))/wvari^0.5)) )
. l, noobs
+--------------------------------------------------------------------
--------+
| id X be la wespe
wvari test |
|--------------------------------------------------------------------
--------|
| 1 13.243148 13.101069 2.7776166 3.123684 2.5789619
5.000e-09 |
| 2 12.91888 14.241573 2.8396291 2.8949698 3.1121321
5.000e-09 |
| 3 12.367111 14.218242 3.3797715 2.7676705 3.4745156
4.989e-09 |
| 4 24.262301 19.075531 3.3698243 3.0092984 2.5247363
5.000e-09 |
| 5 19.050771 18.731844 2.9128101 3.1045931 2.9258794
5.000e-09 |
|--------------------------------------------------------------------
--------|
| 6 12.60056 14.491571 2.7960302 2.8251088 3.3989561
4.993e-09 |
| 7 8.6867876 10.236076 3.4760952 2.5443994 3.2996814
4.996e-09 |
| 8 16.66885 18.424199 2.9968444 3.1121078 3.3819863
5.000e-09 |
| 9 10.029968 11.252171 3.2660736 3.2177827 2.8250082
5.000e-09 |
| 10 14.449689 16.639751 3.3539697 3.3019271 3.4364441
4.962e-09 |
+--------------------------------------------------------------------
--------+
.
Hope this helps,
Scott
----- Original Message -----
From: cortega <[email protected]>
Date: Tuesday, January 31, 2006 9:47 am
Subject: st: Nonlinear equation
> I would like to solve the following nonlinear equation for each
> individual in my sample.
> �
> 0=X-be-(la/r)*(wespe*exp(0.5*wvari)*norm(ln(X)-ln(wespe)-
> wvari)/((wvari)^(0.5)))-(X)*norm((ln(X)-
> ln(wespe))/((wvari)^(0.5))))
> �
> where "X" is the parameter I want to obtain and "be", "la",
> "wespe", "wvari" are variables that vary across individuals.
> I have tried using the nl command. The problem is that I only have
> one observation per individual and it could not solve the
> equation.
> �
> Thank you very much,
> �
> Carolina Ortega�
> �
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