Linear Regression with a Known Fixed Intercept in R

Linear Regression with a known fixed intercept in R

You could subtract the explicit intercept from the regressand and then fit the intercept-free model:

> intercept <- 1.0
> fit <- lm(I(x - intercept) ~ 0 + y, lin)
> summary(fit)

The 0 + suppresses the fitting of the intercept by lm.

edit To plot the fit, use

> abline(intercept, coef(fit))

P.S. The variables in your model look the wrong way round: it's usually y ~ x, not x ~ y (i.e. the regressand should go on the left and the regressor(s) on the right).

How to fit a known linear equation to my data in R?

Let's be clear what you are asking here. You have an existing model, which is "the modelled values are the expected value of the measured values", or in other words, measured = modelled + e, where e are the normally distributed residuals.

You say that the "optimal fit" should be a straight line with intercept 0 and slope 1, which is another way of saying the same thing.

The thing is, this "optimal fit" is not the optimal fit for your actual data, as we can easily see by doing:

summary(lm(measured ~ modelled))
#> 
#> Call:
#> lm(formula = measured ~ modelled)
#> 
#> Residuals:
#>      Min       1Q   Median       3Q      Max 
#> -103.328  -39.130   -4.881   40.428  114.829 
#> 
#> Coefficients:
#>             Estimate Std. Error t value Pr(>|t|)    
#> (Intercept) 23.09461   13.11026   1.762    0.083 .  
#> modelled     0.91143    0.07052  12.924   <2e-16 ***
#> ---
#> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#> 
#> Residual standard error: 55.13 on 63 degrees of freedom
#> Multiple R-squared:  0.7261, Adjusted R-squared:  0.7218 
#> F-statistic:   167 on 1 and 63 DF,  p-value: < 2.2e-16

This shows us the line that would produce the optimal fit to your data in terms of reducing the sum of the squared residuals.

But I guess what you are asking is "How well do my data fit the model measured = modelled + e ?"

Trying to coerce lm into giving you a fixed intercept and slope probably isn't the best way to answer this question. Remember, the p value for the slope only tells you whether the actual slope is significantly different from 0. The above model already confirms that. If you want to know the r-squared of measured = modelled + e, you just need to know the proportion of the variance of measured that is explained by modelled. In other words:

1 - var(measured - modelled) / var(measured)
#> [1] 0.7192672

This is pretty close to the r squared from the lm call.

I think you have sufficient evidence to say that your data is consistent with the model measured = modelled, in that the slope in the lm model includes the value 1 within its 95% confidence interval, and the intercept contains the value 0 within its 95% confidence interval.

Impose Constraint on Intercept in Linear Regression Using R

A linear model.

m0 <- lm(wt ~ qsec + hp + disp, data = mtcars)
m0
# 
# Call:
# lm(formula = wt ~ qsec + hp + disp, data = mtcars)
#
# Coefficients:
# (Intercept)         qsec           hp         disp  
#   -2.450047     0.201713     0.003466     0.006755 

Force the intercept to be zero.

m1 <- lm(wt ~ qsec + hp + disp - 1, data = mtcars)
m1
# 
# Call:
# lm(formula = wt ~ qsec + hp + disp - 1, data = mtcars)
#
# Coefficients:
#      qsec         hp       disp  
# 0.0842281  0.0002622  0.0072967 

You can use nls to apply limits to the paramaters (in this case the lower limit).

m1n <- nls(wt ~ a + b1 * qsec + b2 * hp + b3 * disp, 
    data = mtcars, 
    start = list(a = 1, b1 = 1, b2 = 1, b3 = 1), 
    lower = c(0, -Inf, -Inf, -Inf), algorithm = "port")
m1n
# Nonlinear regression model
#   model: wt ~ a + b1 * qsec + b2 * hp + b3 * disp
#    data: mtcars
#         a        b1        b2        b3 
# 0.0000000 0.0842281 0.0002622 0.0072967 
#  residual sum-of-squares: 4.926
#
# Algorithm "port", convergence message: relative convergence (4)

See here for other example solutions.

Fitting a polynomial with a known intercept

 lm(y~-1+x+I(x^2)+offset(k))

should do it.

• -1 suppresses the otherwise automatically added intercept term
• x adds a linear term
• I(x^2) adds a quadratic term; the I() is required so that R interprets ^2 as squaring, rather than taking an interaction between x and itself (which by formula rules would be equivalent to x alone)
• offset(k) adds the known constant intercept

I don't know whether poly(x,2)-1 would work to eliminate the intercept; you can try it and see. Subtracting the offset from your data should work fine, but offset(k) might be slightly more explicit. You might have to make k a vector (i.e. replicate it to the length of the data set, or better include it as a column in the data set and pass the data with data=...

Linear regression with specified slope

I suppose one approach would be to subtract 1.5*x from y and then fit y using only an intercept term:

mod2 <- lm(I(y-1.5*x)~1)
plot(x, y)
abline(mod2$coefficients, 1.5)

This represents the best linear fit with fixed slope 1.5. Of course, this fit is not very visually appealing because the simulated slope is 1 while the fixed slope is 1.5.

Linear Regression with a known fixed intercept in Accord.NET

Set ols.UseIntercept = False, add the desired intercept value to each opt value, and run ols.Learn(). The resulting Slope will be calculated for the intercept you want.

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