Equivalent code to regress(), using fitglm()

2 Ansichten (letzte 30 Tage)
the cyclist
the cyclist am 3 Sep. 2014
Kommentiert: the cyclist am 4 Sep. 2014
Given the same input data, I would expect
b = regress(log(Y),[ones(size(X,1),1) X]);
and
mdl = fitglm(X,Y,'Link','log');
B = mdl.Coefficients.Estimate;
to give the same coefficients.
Below is some sample code where they do not. Am I missing something simple?
NTRIALS = 31;
dateString = {'01-04-1993','01-07-1995','01-03-1996','01-03-1997', ...
'01-02-1998','01-07-1998','01-11-1998','01-08-1999', ...
'01-09-1999','01-09-2002','01-01-2006','01-03-2006', ...
'01-02-2008','01-05-2008','01-10-2008','01-01-2010', ...
'01-02-2010','01-03-2010','01-10-2011','01-11-2011', ...
'01-01-2012','01-04-2012','01-03-2012','01-04-2012', ...
'01-09-2012','01-09-2012','01-01-2012','01-01-2012', ...
'01-04-2013','01-06-2013','01-02-2014'...
}';
ARR = [1.27 0.84 0.90 1.26 ...
1.61 0.50 1.28 0.98 ...
1.08 1.24 0.61 0.67 ...
0.70 0.44 0.41 0.33 ...
0.40 0.90 0.54 0.80 ...
0.28 1.30 0.39 1.20 ...
0.40 0.36 0.501 0.40 ...
1.04 0.505 0.34 ]';
N = [372 251 301 150 ...
40 127 560 168 ...
293 49 1651 942 ...
104 267 257 1326 ...
1272 29 1088 218 ...
66 165 1106 120 ...
1417 1234 1169 1083 ...
80 1404 1331 ]';
dateNumber = datenum(dateString,'dd-mm-yyyy');
dateVector = datevec(dateNumber);
year = dateVector(:,1);
[b,bint,r,rint,stats] = regress(log(ARR),[ones(NTRIALS,1) year]);
mdl = fitglm(year,ARR,'Link','log');
B = mdl.Coefficients.Estimate;
figure
hold on
h = line(datenum({'1993-01-01','2014-01-01'}),exp(1).^[b(1)+b(2)*1994 b(1)+b(2)*2014]);
hg = line(datenum({'1993-01-01','2014-01-01'}),exp(1).^[B(1)+B(2)*1994 B(1)+B(2)*2014]);
set(h,'Color','r')
set(hg,'Color','g')
semilogy(dateNumber,ARR,'o')
axis tight
set(gca,'YLim',[0.1 2],'YTick',[0.2 0.4 0.6 0.8 1.0 1.5 2])
set(gca,'XTick',1994:2:2014)
datetick('x')

Akzeptierte Antwort

Tom Lane
Tom Lane am 4 Sep. 2014
I believe these are not the same. For the fitglm model we are saying that Y has a normal distribution whose log(mean) has a linear relationship with the predictors. Y could take on negative values in that case. For the regress model we are saying that the log of Y has a normal distribution whose mean has a linear relationship with the predictors. Y would have to be positive in that case.
  1 Kommentar
the cyclist
the cyclist am 4 Sep. 2014
Proving once again that I am not a statistician, but just play one on TV. Thanks for the clear explanation.

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