Adding a custom constraint to weighted least squares regression model












0












$begingroup$


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
    5 hours ago












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    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
    4 hours ago


















0












$begingroup$


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?










share|improve this question











$endgroup$












  • $begingroup$
    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
    5 hours ago












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    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
    4 hours ago
















0












0








0


2



$begingroup$


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?










share|improve this question











$endgroup$




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?







python regression linear-regression






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share|improve this question













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edited 9 mins ago







ste_kwr

















asked 5 hours ago









ste_kwrste_kwr

1063




1063












  • $begingroup$
    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
    5 hours ago












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    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
    4 hours ago




















  • $begingroup$
    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
    5 hours ago












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    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
    4 hours ago


















$begingroup$
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
5 hours ago






$begingroup$
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
5 hours ago














$begingroup$
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
5 hours ago




$begingroup$
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
5 hours ago












$begingroup$
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
4 hours ago






$begingroup$
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
4 hours ago












1 Answer
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0












$begingroup$

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(E)


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






share|improve this answer








New contributor




Juan Esteban de la Calle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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    1 Answer
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    1 Answer
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    0












    $begingroup$

    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(E)


    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






    share|improve this answer








    New contributor




    Juan Esteban de la Calle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$


















      0












      $begingroup$

      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(E)


      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






      share|improve this answer








      New contributor




      Juan Esteban de la Calle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$
















        0












        0








        0





        $begingroup$

        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(E)


        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






        share|improve this answer








        New contributor




        Juan Esteban de la Calle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$



        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(E)


        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







        share|improve this answer








        New contributor




        Juan Esteban de la Calle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|improve this answer



        share|improve this answer






        New contributor




        Juan Esteban de la Calle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered 3 hours ago









        Juan Esteban de la CalleJuan Esteban de la Calle

        36011




        36011




        New contributor




        Juan Esteban de la Calle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        Juan Esteban de la Calle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        Juan Esteban de la Calle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






























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