You need to review the assumptions of linear models:
y is assumed to be the realization of a random variable,
not a constant (or, more precisely: there are assumed to
be deviations that are N(0, sigma^2).
If you 'know' that y is a constant, then you have
two options:
1. don't do the regression because it makes no sense;
2. if you want to test lm()'s handling of the data:
fm <- lm(y ~ x, data = df, offset = rep(1, nrow(df)))
(or use: offset = y)
-Peter Ehlers
Jan-Henrik Pötter wrote:
Hello.
Consider the response-variable of data.frame df is constant, so analytically
perfect fit of a linear model is expected. Fitting a regression line using
lm result in residuals, slope and std.errors not exactly zero, which is
acceptable in some way, but errorneous. But if you use summary.lm it shows
inacceptable error propagation in the calculation of the t value and the
corresponding p-value for the slope, as well R-Square – just consider the
adj R-Square of 0.6788! This result is independent of which mode used for
the input vectors. Is there any way to get the perfect fitted regression
curve using lm and prevent this error propagation? I consider rounding all
values of the lm-object afterwards to somewhat precision as a bad idea.
Unfortunately there is no option in lm for calculation precision.
df<-data.frame(x=1:10,y=1)
myl<-lm(y~x,data=df)
myl
Call:
lm(formula = y ~ x, data = df)
Coefficients:
(Intercept) x
1.000e+00 9.463e-18
summary(myl)
Call:
lm(formula = y ~ x, data = df)
Residuals:
Min 1Q Median 3Q Max
-1.136e-16 -1.341e-17 7.886e-18 2.918e-17 5.047e-17
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.000e+00 3.390e-17 2.950e+16 <2e-16 ***
x 9.463e-18 5.463e-18 1.732e+00 0.122
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 4.962e-17 on 8 degrees of freedom
Multiple R-squared: 0.7145, Adjusted R-squared: 0.6788
F-statistic: 20.02 on 1 and 8 DF, p-value: 0.002071
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Peter Ehlers
University of Calgary
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