Properties of an arbitrary (i.e., not fitted) statistical model
Clash Royale CLAN TAG#URR8PPP
Suppose that I have a data and a statistical model that is not the result of fitting, but has a different origin. For example, I have a list of measurements and a theoretical model whose predicting accuracy I want to assess.
Is there a simple way to obtain all the properties of a FittedModel object, but from my theoretical model?
For example, suppose that
mt=Table[i,1+i+RandomReal,i,5]
yields mt=1,2.56508,2,3.58291,3,4.8005,4,5.24265,5,6.38087
If I call
LinearModelFit[mt,x,x]
I get
![FittedModel[1.72701 +0.929131 x]](https://i.stack.imgur.com/8gruv.png)
And I can then get R$^2$ for the adjusted model:
%["RSquared"]
which gives $0.9851$.
But I know the theoretical model is $1.5+x$. Can I have the power of a FittedModel (residuals, ANOVA, RSquared, etc.) for my non-fitted model?
probability-or-statistics fitting
edited Dec 22 '18 at 17:00
Michael E2
145k11195464
asked Dec 21 '18 at 18:17
Rafael
18329
add a comment |
Suppose that I have a data and a statistical model that is not the result of fitting, but has a different origin. For example, I have a list of measurements and a theoretical model whose predicting accuracy I want to assess.
Is there a simple way to obtain all the properties of a FittedModel object, but from my theoretical model?
For example, suppose that
mt=Table[i,1+i+RandomReal,i,5]
yields mt=1,2.56508,2,3.58291,3,4.8005,4,5.24265,5,6.38087
If I call
LinearModelFit[mt,x,x]
I get
And I can then get R$^2$ for the adjusted model:
%["RSquared"]
which gives $0.9851$.
But I know the theoretical model is $1.5+x$. Can I have the power of a FittedModel (residuals, ANOVA, RSquared, etc.) for my non-fitted model?
probability-or-statistics fitting
edited Dec 22 '18 at 17:00
Michael E2
145k11195464
asked Dec 21 '18 at 18:17
Rafael
18329
If you look atFullForm[LinearModelFit[mt,x,x]]from your example above then the internal structure seems reasonably simple and understandable. What happens if you carefully build aFittedModelusing that structure as an example, not by doing a fit but by substituting your theoretical model in that and then you query that constructedFittedModelfor residuals, ANOVA, etc? Can you do this on test cases where you know what the calculations should show to verify that this is being done correctly?
– Bill
Dec 21 '18 at 22:04
add a comment |
Suppose that I have a data and a statistical model that is not the result of fitting, but has a different origin. For example, I have a list of measurements and a theoretical model whose predicting accuracy I want to assess.
Is there a simple way to obtain all the properties of a FittedModel object, but from my theoretical model?
For example, suppose that
mt=Table[i,1+i+RandomReal,i,5]
yields mt=1,2.56508,2,3.58291,3,4.8005,4,5.24265,5,6.38087
If I call
LinearModelFit[mt,x,x]
I get
And I can then get R$^2$ for the adjusted model:
%["RSquared"]
which gives $0.9851$.
But I know the theoretical model is $1.5+x$. Can I have the power of a FittedModel (residuals, ANOVA, RSquared, etc.) for my non-fitted model?
probability-or-statistics fitting
edited Dec 22 '18 at 17:00
Michael E2
145k11195464
asked Dec 21 '18 at 18:17
Rafael
18329
Suppose that I have a data and a statistical model that is not the result of fitting, but has a different origin. For example, I have a list of measurements and a theoretical model whose predicting accuracy I want to assess.
Is there a simple way to obtain all the properties of a FittedModel object, but from my theoretical model?
For example, suppose that
mt=Table[i,1+i+RandomReal,i,5]
yields mt=1,2.56508,2,3.58291,3,4.8005,4,5.24265,5,6.38087
If I call
LinearModelFit[mt,x,x]
I get
And I can then get R$^2$ for the adjusted model:
%["RSquared"]
which gives $0.9851$.
But I know the theoretical model is $1.5+x$. Can I have the power of a FittedModel (residuals, ANOVA, RSquared, etc.) for my non-fitted model?
probability-or-statistics fitting
probability-or-statistics fitting
edited Dec 22 '18 at 17:00
Michael E2
145k11195464
asked Dec 21 '18 at 18:17
Rafael
18329
edited Dec 22 '18 at 17:00
Michael E2
145k11195464
asked Dec 21 '18 at 18:17
Rafael
18329
edited Dec 22 '18 at 17:00
Michael E2
145k11195464
edited Dec 22 '18 at 17:00
Michael E2
145k11195464
edited Dec 22 '18 at 17:00
Michael E2
145k11195464
145k11195464
asked Dec 21 '18 at 18:17
Rafael
18329
asked Dec 21 '18 at 18:17
Rafael
18329
asked Dec 21 '18 at 18:17
Rafael
18329
18329
If you look atFullForm[LinearModelFit[mt,x,x]]from your example above then the internal structure seems reasonably simple and understandable. What happens if you carefully build aFittedModelusing that structure as an example, not by doing a fit but by substituting your theoretical model in that and then you query that constructedFittedModelfor residuals, ANOVA, etc? Can you do this on test cases where you know what the calculations should show to verify that this is being done correctly?
– Bill
Dec 21 '18 at 22:04
add a comment |
If you look atFullForm[LinearModelFit[mt,x,x]]from your example above then the internal structure seems reasonably simple and understandable. What happens if you carefully build aFittedModelusing that structure as an example, not by doing a fit but by substituting your theoretical model in that and then you query that constructedFittedModelfor residuals, ANOVA, etc? Can you do this on test cases where you know what the calculations should show to verify that this is being done correctly?
– Bill
Dec 21 '18 at 22:04
If you look at
FullForm[LinearModelFit[mt,x,x]] from your example above then the internal structure seems reasonably simple and understandable. What happens if you carefully build a FittedModel using that structure as an example, not by doing a fit but by substituting your theoretical model in that and then you query that constructed FittedModel for residuals, ANOVA, etc? Can you do this on test cases where you know what the calculations should show to verify that this is being done correctly?– Bill
Dec 21 '18 at 22:04
If you look at
FullForm[LinearModelFit[mt,x,x]] from your example above then the internal structure seems reasonably simple and understandable. What happens if you carefully build a FittedModel using that structure as an example, not by doing a fit but by substituting your theoretical model in that and then you query that constructed FittedModel for residuals, ANOVA, etc? Can you do this on test cases where you know what the calculations should show to verify that this is being done correctly?– Bill
Dec 21 '18 at 22:04
add a comment |
1 Answer
1
active
oldest
votes
You can use NonlinearModelFit but not all output options will either be appropriate and some will require "adjustment".
All of the fitting procedures want to estimate parameters (that's what they're made for) so NonlinearModelFit can be tricked by included a parameter that isn't in the model. Here I used a parameter named a:
SeedRandom[12345];
mt = Table[i, 1 + i + RandomReal, i, 5]
nlm = NonlinearModelFit[mt, 1.5 + x, a, x]
So the following work fine with no adjustment needed:
nlm["FitResiduals"]
(* -0.378754, -0.170078, 0.282753, -0.0698316, -0.276414 *)
nlm["PredictedResponse"]
(* 2.5, 3.5, 4.5, 5.5, 6.5 *)
The "EstimatedVariance" needs adjustment:
Mean[(mt[[All, 2]] - 1.5 - mt[[All, 1]])^2]
(* 0.0667223 *)
n = Length[mt];
nlm["EstimatedVariance"] (n - 1)/n
(* 0.0667223 *)
answered Dec 21 '18 at 19:28
JimB
16.9k12662
Nice! You get some information on the dummy model/parametera(in the ANOVA, ParameterTable, etc.), is it possible to get such information for my coefficients (i.e., 1.5 and 1)? Or is all that kind of information just meaningless for a non-fitted model?
– Rafael
Dec 21 '18 at 19:55
2
Yes, I'd vote for meaningless. I'd also avoid any option that starts with"Parameter..."as the residual variance is the only parameter estimated.
– JimB
Dec 21 '18 at 20:14
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use NonlinearModelFit but not all output options will either be appropriate and some will require "adjustment".
All of the fitting procedures want to estimate parameters (that's what they're made for) so NonlinearModelFit can be tricked by included a parameter that isn't in the model. Here I used a parameter named a:
SeedRandom[12345];
mt = Table[i, 1 + i + RandomReal, i, 5]
nlm = NonlinearModelFit[mt, 1.5 + x, a, x]
So the following work fine with no adjustment needed:
nlm["FitResiduals"]
(* -0.378754, -0.170078, 0.282753, -0.0698316, -0.276414 *)
nlm["PredictedResponse"]
(* 2.5, 3.5, 4.5, 5.5, 6.5 *)
The "EstimatedVariance" needs adjustment:
Mean[(mt[[All, 2]] - 1.5 - mt[[All, 1]])^2]
(* 0.0667223 *)
n = Length[mt];
nlm["EstimatedVariance"] (n - 1)/n
(* 0.0667223 *)
answered Dec 21 '18 at 19:28
JimB
16.9k12662
Nice! You get some information on the dummy model/parametera(in the ANOVA, ParameterTable, etc.), is it possible to get such information for my coefficients (i.e., 1.5 and 1)? Or is all that kind of information just meaningless for a non-fitted model?
– Rafael
Dec 21 '18 at 19:55
2
Yes, I'd vote for meaningless. I'd also avoid any option that starts with"Parameter..."as the residual variance is the only parameter estimated.
– JimB
Dec 21 '18 at 20:14
add a comment |
You can use NonlinearModelFit but not all output options will either be appropriate and some will require "adjustment".
All of the fitting procedures want to estimate parameters (that's what they're made for) so NonlinearModelFit can be tricked by included a parameter that isn't in the model. Here I used a parameter named a:
SeedRandom[12345];
mt = Table[i, 1 + i + RandomReal, i, 5]
nlm = NonlinearModelFit[mt, 1.5 + x, a, x]
So the following work fine with no adjustment needed:
nlm["FitResiduals"]
(* -0.378754, -0.170078, 0.282753, -0.0698316, -0.276414 *)
nlm["PredictedResponse"]
(* 2.5, 3.5, 4.5, 5.5, 6.5 *)
The "EstimatedVariance" needs adjustment:
Mean[(mt[[All, 2]] - 1.5 - mt[[All, 1]])^2]
(* 0.0667223 *)
n = Length[mt];
nlm["EstimatedVariance"] (n - 1)/n
(* 0.0667223 *)
answered Dec 21 '18 at 19:28
JimB
16.9k12662
Nice! You get some information on the dummy model/parametera(in the ANOVA, ParameterTable, etc.), is it possible to get such information for my coefficients (i.e., 1.5 and 1)? Or is all that kind of information just meaningless for a non-fitted model?
– Rafael
Dec 21 '18 at 19:55
2
Yes, I'd vote for meaningless. I'd also avoid any option that starts with"Parameter..."as the residual variance is the only parameter estimated.
– JimB
Dec 21 '18 at 20:14
add a comment |
You can use NonlinearModelFit but not all output options will either be appropriate and some will require "adjustment".
All of the fitting procedures want to estimate parameters (that's what they're made for) so NonlinearModelFit can be tricked by included a parameter that isn't in the model. Here I used a parameter named a:
SeedRandom[12345];
mt = Table[i, 1 + i + RandomReal, i, 5]
nlm = NonlinearModelFit[mt, 1.5 + x, a, x]
So the following work fine with no adjustment needed:
nlm["FitResiduals"]
(* -0.378754, -0.170078, 0.282753, -0.0698316, -0.276414 *)
nlm["PredictedResponse"]
(* 2.5, 3.5, 4.5, 5.5, 6.5 *)
The "EstimatedVariance" needs adjustment:
Mean[(mt[[All, 2]] - 1.5 - mt[[All, 1]])^2]
(* 0.0667223 *)
n = Length[mt];
nlm["EstimatedVariance"] (n - 1)/n
(* 0.0667223 *)
answered Dec 21 '18 at 19:28
JimB
16.9k12662
You can use NonlinearModelFit but not all output options will either be appropriate and some will require "adjustment".
All of the fitting procedures want to estimate parameters (that's what they're made for) so NonlinearModelFit can be tricked by included a parameter that isn't in the model. Here I used a parameter named a:
SeedRandom[12345];
mt = Table[i, 1 + i + RandomReal, i, 5]
nlm = NonlinearModelFit[mt, 1.5 + x, a, x]
So the following work fine with no adjustment needed:
nlm["FitResiduals"]
(* -0.378754, -0.170078, 0.282753, -0.0698316, -0.276414 *)
nlm["PredictedResponse"]
(* 2.5, 3.5, 4.5, 5.5, 6.5 *)
The "EstimatedVariance" needs adjustment:
Mean[(mt[[All, 2]] - 1.5 - mt[[All, 1]])^2]
(* 0.0667223 *)
n = Length[mt];
nlm["EstimatedVariance"] (n - 1)/n
(* 0.0667223 *)
answered Dec 21 '18 at 19:28
JimB
16.9k12662
answered Dec 21 '18 at 19:28
JimB
16.9k12662
answered Dec 21 '18 at 19:28
JimB
16.9k12662
answered Dec 21 '18 at 19:28
JimB
16.9k12662
16.9k12662
Nice! You get some information on the dummy model/parametera(in the ANOVA, ParameterTable, etc.), is it possible to get such information for my coefficients (i.e., 1.5 and 1)? Or is all that kind of information just meaningless for a non-fitted model?
– Rafael
Dec 21 '18 at 19:55
2
Yes, I'd vote for meaningless. I'd also avoid any option that starts with"Parameter..."as the residual variance is the only parameter estimated.
– JimB
Dec 21 '18 at 20:14
add a comment |
Nice! You get some information on the dummy model/parametera(in the ANOVA, ParameterTable, etc.), is it possible to get such information for my coefficients (i.e., 1.5 and 1)? Or is all that kind of information just meaningless for a non-fitted model?
– Rafael
Dec 21 '18 at 19:55
2
Yes, I'd vote for meaningless. I'd also avoid any option that starts with"Parameter..."as the residual variance is the only parameter estimated.
– JimB
Dec 21 '18 at 20:14
Nice! You get some information on the dummy model/parameter
a (in the ANOVA, ParameterTable, etc.), is it possible to get such information for my coefficients (i.e., 1.5 and 1)? Or is all that kind of information just meaningless for a non-fitted model?– Rafael
Dec 21 '18 at 19:55
Nice! You get some information on the dummy model/parameter
a (in the ANOVA, ParameterTable, etc.), is it possible to get such information for my coefficients (i.e., 1.5 and 1)? Or is all that kind of information just meaningless for a non-fitted model?– Rafael
Dec 21 '18 at 19:55
2
2
Yes, I'd vote for meaningless. I'd also avoid any option that starts with
"Parameter..." as the residual variance is the only parameter estimated.– JimB
Dec 21 '18 at 20:14
Yes, I'd vote for meaningless. I'd also avoid any option that starts with
"Parameter..." as the residual variance is the only parameter estimated.– JimB
Dec 21 '18 at 20:14
add a comment |
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If you look at
FullForm[LinearModelFit[mt,x,x]]from your example above then the internal structure seems reasonably simple and understandable. What happens if you carefully build aFittedModelusing that structure as an example, not by doing a fit but by substituting your theoretical model in that and then you query that constructedFittedModelfor residuals, ANOVA, etc? Can you do this on test cases where you know what the calculations should show to verify that this is being done correctly?– Bill
Dec 21 '18 at 22:04