R: Force regression coefficients to add up to 1 - r

Question

I'm trying to run a simple OLS regression with a restriction that the sum of the coefficients of two variables add up to 1.
I want:
Y = α + β1 * x1 + β2 * x2 + β3 * x3,
where β1 + β2 = 1
I have found how to make a relation between coefficients like:
β1 = 2* β2
But I haven't found how to make restrictions like:
β1 = 1 - β2
How would I do it in this simple example?
data <- data.frame(
A = c(1,2,3,4),
B = c(3,2,2,3),
C = c(3,3,2,3),
D = c(5,3,3,4)
)
lm(formula = 'D ~ A + B + C', data = data)
Thanks!

Answer 1

β1 + β2 = 1
To have β1 + β2 = 1 the model you have to fit is
fit <- lm(Y ~ offset(x1) + I(x2 - x1) + x3, data = df)
That is
Y = α + x1 + β2 * (x2 - x1) + β3 * x3
after substituting β1 = 1 - β2; x_new = x2 - x1 and the coefficient for x1 is 1.
β1 + β2 + β3 = 1
fit <- lm(Y ~ offset(x1) + I(x2 - x1) + I(x3 - x1), data = df)
Y = α + x1 + β2 * (x2 - x1) + β3 * (x3 - x1)
after substituting β1 = 1 - β2 - β3
β1 + β2 + β3 + ... = 1
I think the pattern is clear... you just have to subtract one variable, x1, from the remaining variables(x2, x3, ...) and have the coefficient of that variable, x1, to 1.
Example β1 + β2 = 1
# Data
df <- iris[, 1:4]
colnames(df) <- c("Y", paste0("x", 1:3, collaapse=""))
# β1 + β2 = 1
fit <- lm(Y ~ offset(x1) + I(x2 - x1) + x3, data = df)
coef_2 <- coef(fit)
beta_1 <- 1 - coef_2[2]
beta_2 <- coef_2[2]

Answer 2

1) CVXR We can compute the coefficients using CVXR directly by specifying the objective and constraint. We assume that D is the response, the coefficients of A and B must sum to 1, b[1] is the intercept and b[2], b[3] and b[4] are the coefficients of A, B and C respectively.
library(CVXR)
b <- Variable(4)
X <- cbind(1, as.matrix(data[-4]))
obj <- Minimize(sum((data$D - X %*% b)^2))
constraints <- list(b[2] + b[3] == 1)
problem <- Problem(obj, constraints)
soln <- solve(problem)
bval <- soln$getValue(b)
bval
## [,1]
## [1,] 1.6428605
## [2,] -0.3571428
## [3,] 1.3571428
## [4,] -0.1428588
The objective is the residual sum of squares and it equals:
soln$value
## [1] 0.07142857
2) pracma We can also use the pracma package to compute the coefficients. We specify the X matrix, response vector, the constraint matrix (in this case the vector given as the third argument is regarded as a one row matrix) and the right hand side of the constraint.
library(pracma)
lsqlincon(X, data$D, Aeq = c(0, 1, 1, 0), beq = 1) # X is from above
## [1] 1.6428571 -0.3571429 1.3571429 -0.1428571
3) limSolve This package can also solve for the coefficients of regression problems with constraints. The arguments are the same as in (2).
library(limSolve)
lsei(X, data$D, c(0, 1, 1, 0), 1)
giving:
$X
A B C
1.6428571 -0.3571429 1.3571429 -0.1428571
$residualNorm
[1] 0
$solutionNorm
[1] 0.07142857
$IsError
[1] FALSE
$type
[1] "lsei"
4) nls This can be formulated as a problem for nls with the B coefficient equal to one minus the A coefficient.
nls(D ~ b0 + b1 * A + (1-b1) * B + b2 * C, data,
start = list(b0 = 1, b1 = 1, b2 = 1))
## D ~ b0 + b1 * A + (1 - b1) * B + b2 * C
## data: data
## b0 b1 b2
## 1.6429 -0.3571 -0.1429
## residual sum-of-squares: 0.07143
##
## Number of iterations to convergence: 1
## Achieved convergence tolerance: 2.803e-08
Check
We can double check the above by using the lm approach in the other answer:
lm(D ~ I(A-B) + C + offset(B), data)
giving:
Call:
lm(formula = D ~ I(A - B) + C + offset(B), data = data)
Coefficients:
(Intercept) I(A - B) C
1.6429 -0.3571 -0.1429
The I(A-B) coefficient equals the coefficient of A in the original formulation and one minus it is the coefficient of C. We see that all approaches do lead to the same coefficients.