Adding a custom constraint to weighted least squares regression model
$begingroup$
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?
python regression linear-regression
asked 5 hours ago
ste_kwrste_kwr
1063
$endgroup$
add a comment |
$begingroup$
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?
python regression linear-regression
asked 5 hours ago
ste_kwrste_kwr
1063
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
python regression linear-regression
asked 5 hours ago
ste_kwrste_kwr
1063
$endgroup$
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
python regression linear-regression
asked 5 hours ago
ste_kwrste_kwr
1063
asked 5 hours ago
ste_kwrste_kwr
1063
edited 1 min ago
ste_kwr
asked 5 hours ago
ste_kwrste_kwr
1063
asked 5 hours ago
ste_kwrste_kwr
1063
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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votes
$begingroup$
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
answered 2 hours ago
Juan Esteban de la CalleJuan Esteban de la Calle
35811
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.
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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
answered 2 hours ago
Juan Esteban de la CalleJuan Esteban de la Calle
35811
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.
$endgroup$
add a comment |
$begingroup$
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
answered 2 hours ago
Juan Esteban de la CalleJuan Esteban de la Calle
35811
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.
$endgroup$
add a comment |
$begingroup$
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
answered 2 hours ago
Juan Esteban de la CalleJuan Esteban de la Calle
35811
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
answered 2 hours ago
Juan Esteban de la CalleJuan Esteban de la Calle
35811
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 2 hours ago
Juan Esteban de la CalleJuan Esteban de la Calle
35811
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 2 hours ago
Juan Esteban de la CalleJuan Esteban de la Calle
35811
answered 2 hours ago
Juan Esteban de la CalleJuan Esteban de la Calle
35811
35811
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.
add a comment |
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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