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I am trying to run a weighted least squares model that looks something like this (but could be different):

$y = \beta_0 + \beta_1 x + \beta_2 log(x) + \epsilon$

with weights $w_1, w_2, ..$

However, I know, from external knowledge, that whatever the model the outcome must asymptotically converge to a constant for large values of $x$. How can I get an OLS estimate with this constraint.

As an example, let's say if I knew the asymptote $c$, then I can add two fake data points to my model, with very high values of $x$ and very high weights $w$ and $y=c$, and run the normal WLS model and it would give me what I need - except I don't know the value of $c$. Is there a way to impose this constraint - maybe through adding a custom error term to the model?

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  • $\begingroup$ Perhaps solving the equation $y=max(\beta_0+\beta_1x+\beta_2*log(x),c)+\epsilon$ instead of the original? you will have to add artificial points to the data with $(x_{large},c)$ $\endgroup$ – Juan Esteban de la Calle Apr 16 at 21:19
  • $\begingroup$ I don't know the value of $c$, so somehow I imagine the loss function would need to take care of this. If I knew $c$, I have described in the question how I would go about doing this. $\endgroup$ – ste_kwr Apr 16 at 21:23
  • $\begingroup$ Maybe you can try to fit something like a modified logit model. I have never tried something liike this and I don't know anything about a possible implementation, but a logit regression has a natural limit of $1$, you may work with a unknown limit. The equation would be like this: $Y=\frac{c}{(1+e^{-(\beta_0+\beta_1x+\beta_2log(x))})}$ $\endgroup$ – Juan Esteban de la Calle Apr 16 at 22:23
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The model you are looking for is this:

$Y=\frac{A}{1+e^{-(\beta_0+\beta_1x+\beta_2log(x))}}$, this could not be obtained but a very similar was obtained.

This code in R might work:

R=data.frame(X=c(1,2,3,4,5,6,7,8,9),Y=c(1,2,3,3,3,3,3,3,3)) # Data in which X is a line, and Y has an still unknown limit.
model=nls(formula = Y~A/(1+exp(-(b0+b1*X))),data=R)
summary(model)

In the result you can see how $A$ says that the limit of 3 (previously unknown) is calculated.

There is a limitation to take into account, is explained in this link, is summarized in the impossibility for all possible models to exist, the "most inside" model should be linear.

The model $\beta_0+\beta_1x+\beta_2log(x)$ could not be used, the model $\beta_0+\beta_1x$ could be used, take this into account.

First steps with Non-Linear Regression in R

Singular Gradient Error in nls with correct starting values

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