Mixing
r lm parameter estimates
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-1
error variable length differs
I am confused with this error and I don't know what to do.
n1<-20
m1<-0
sd1<-1
y<-rnorm(n1,m1, sd1)
x<-rnorm(n1,m1, sd1)
e<-rnorm(n1,m1, sd1)
b0<-0
b1<-1
modelfit1<-lm(y~ b0 + b1*x + e)
Error in model.frame.default(formula = y ~ b0 + b1 * x + e:
variable lengths differ (found for 'b0')
edited:
I am working on such case where n=20, the parameters b0=0, and b=1 are true and the independent and the error are normally distributed with mean=0 and sd=1.
Is this possible?
Thanks a lot!
r parameters lm
asked Jan 3 at 15:56
belbel
32
add a comment |
-1
error variable length differs
I am confused with this error and I don't know what to do.
n1<-20
m1<-0
sd1<-1
y<-rnorm(n1,m1, sd1)
x<-rnorm(n1,m1, sd1)
e<-rnorm(n1,m1, sd1)
b0<-0
b1<-1
modelfit1<-lm(y~ b0 + b1*x + e)
Error in model.frame.default(formula = y ~ b0 + b1 * x + e:
variable lengths differ (found for 'b0')
edited:
I am working on such case where n=20, the parameters b0=0, and b=1 are true and the independent and the error are normally distributed with mean=0 and sd=1.
Is this possible?
Thanks a lot!
r parameters lm
asked Jan 3 at 15:56
belbel
32
2
What do you want to do?
– coffeinjunky
Jan 3 at 16:00
1
Hello, welcome to StackOverflow. I think your simulated example does not really match what you want to do. Maybe this post would help? stats.stackexchange.com/questions/99924/…
– Vincent Guillemot
Jan 3 at 16:01
1
A comment on where the error comes from: R thinks that the model you want to fit contains a response variable y (so far so good), and a complicated combination of explanatory variables (one called b0, one called b1, one called x, one interaction term between the variables x and b1 and a final term that I don't even know how R will interpret it). If your intention was to generate a simple dataset to fit a simple linear model on (and I think it was your intention), then you need to fix your simulated data.
– Vincent Guillemot
Jan 3 at 16:10
@coffeinjunky hello, i want to see the estimate standard error of the parameter b0 and b1.
– bel
Jan 3 at 16:28
add a comment |
-1
-1
-1
error variable length differs
I am confused with this error and I don't know what to do.
n1<-20
m1<-0
sd1<-1
y<-rnorm(n1,m1, sd1)
x<-rnorm(n1,m1, sd1)
e<-rnorm(n1,m1, sd1)
b0<-0
b1<-1
modelfit1<-lm(y~ b0 + b1*x + e)
Error in model.frame.default(formula = y ~ b0 + b1 * x + e:
variable lengths differ (found for 'b0')
edited:
I am working on such case where n=20, the parameters b0=0, and b=1 are true and the independent and the error are normally distributed with mean=0 and sd=1.
Is this possible?
Thanks a lot!
r parameters lm
asked Jan 3 at 15:56
belbel
32
error variable length differs
I am confused with this error and I don't know what to do.
n1<-20
m1<-0
sd1<-1
y<-rnorm(n1,m1, sd1)
x<-rnorm(n1,m1, sd1)
e<-rnorm(n1,m1, sd1)
b0<-0
b1<-1
modelfit1<-lm(y~ b0 + b1*x + e)
Error in model.frame.default(formula = y ~ b0 + b1 * x + e:
variable lengths differ (found for 'b0')
edited:
I am working on such case where n=20, the parameters b0=0, and b=1 are true and the independent and the error are normally distributed with mean=0 and sd=1.
Is this possible?
Thanks a lot!
r parameters lm
r parameters lm
asked Jan 3 at 15:56
belbel
32
asked Jan 3 at 15:56
belbel
32
edited Jan 3 at 16:35
bel
asked Jan 3 at 15:56
belbel
32
asked Jan 3 at 15:56
belbel
32
asked Jan 3 at 15:56
belbel
32
32
2
What do you want to do?
– coffeinjunky
Jan 3 at 16:00
1
Hello, welcome to StackOverflow. I think your simulated example does not really match what you want to do. Maybe this post would help? stats.stackexchange.com/questions/99924/…
– Vincent Guillemot
Jan 3 at 16:01
1
A comment on where the error comes from: R thinks that the model you want to fit contains a response variable y (so far so good), and a complicated combination of explanatory variables (one called b0, one called b1, one called x, one interaction term between the variables x and b1 and a final term that I don't even know how R will interpret it). If your intention was to generate a simple dataset to fit a simple linear model on (and I think it was your intention), then you need to fix your simulated data.
– Vincent Guillemot
Jan 3 at 16:10
@coffeinjunky hello, i want to see the estimate standard error of the parameter b0 and b1.
– bel
Jan 3 at 16:28
add a comment |
2
What do you want to do?
– coffeinjunky
Jan 3 at 16:00
1
Hello, welcome to StackOverflow. I think your simulated example does not really match what you want to do. Maybe this post would help? stats.stackexchange.com/questions/99924/…
– Vincent Guillemot
Jan 3 at 16:01
1
A comment on where the error comes from: R thinks that the model you want to fit contains a response variable y (so far so good), and a complicated combination of explanatory variables (one called b0, one called b1, one called x, one interaction term between the variables x and b1 and a final term that I don't even know how R will interpret it). If your intention was to generate a simple dataset to fit a simple linear model on (and I think it was your intention), then you need to fix your simulated data.
– Vincent Guillemot
Jan 3 at 16:10
@coffeinjunky hello, i want to see the estimate standard error of the parameter b0 and b1.
– bel
Jan 3 at 16:28
2
2
What do you want to do?
– coffeinjunky
Jan 3 at 16:00
What do you want to do?
– coffeinjunky
Jan 3 at 16:00
1
1
Hello, welcome to StackOverflow. I think your simulated example does not really match what you want to do. Maybe this post would help? stats.stackexchange.com/questions/99924/…
– Vincent Guillemot
Jan 3 at 16:01
Hello, welcome to StackOverflow. I think your simulated example does not really match what you want to do. Maybe this post would help? stats.stackexchange.com/questions/99924/…
– Vincent Guillemot
Jan 3 at 16:01
1
1
A comment on where the error comes from: R thinks that the model you want to fit contains a response variable y (so far so good), and a complicated combination of explanatory variables (one called b0, one called b1, one called x, one interaction term between the variables x and b1 and a final term that I don't even know how R will interpret it). If your intention was to generate a simple dataset to fit a simple linear model on (and I think it was your intention), then you need to fix your simulated data.
– Vincent Guillemot
Jan 3 at 16:10
A comment on where the error comes from: R thinks that the model you want to fit contains a response variable y (so far so good), and a complicated combination of explanatory variables (one called b0, one called b1, one called x, one interaction term between the variables x and b1 and a final term that I don't even know how R will interpret it). If your intention was to generate a simple dataset to fit a simple linear model on (and I think it was your intention), then you need to fix your simulated data.
– Vincent Guillemot
Jan 3 at 16:10
@coffeinjunky hello, i want to see the estimate standard error of the parameter b0 and b1.
– bel
Jan 3 at 16:28
@coffeinjunky hello, i want to see the estimate standard error of the parameter b0 and b1.
– bel
Jan 3 at 16:28
add a comment |
2 Answers
2
active
oldest
votes
1
I suggest you put everything in a data.frame and deal with it that way:
set.seed(2)
m1<-0
sd1<-1
y<-rnorm(n1,m1, sd1)
x<-rnorm(n1,m1, sd1)
b0<-0
b1<-1
d <- data.frame(y,b0,b1,x,e=rnorm(20,0,1))
head(d)
# y b0 b1 x e
# 1 -0.89691455 0 1 2.090819205 -0.3835862
# 2 0.18484918 0 1 -1.199925820 -1.9591032
# 3 1.58784533 0 1 1.589638200 -0.8417051
# 4 -1.13037567 0 1 1.954651642 1.9035475
# 5 -0.08025176 0 1 0.004937777 0.6224939
# 6 0.13242028 0 1 -2.451706388 1.9909204
Now things work nicely:
modelfit1 <- lm(y~b0+b1*x+e, data=d)
modelfit1
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Coefficients:
# (Intercept) b0 b1 x e b1:x
# 0.19331 NA NA -0.06752 0.02240 NA
summary(modelfit1)
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Residuals:
# Min 1Q Median 3Q Max
# -2.5006 -0.4786 -0.1425 0.6211 1.8488
# Coefficients: (3 not defined because of singularities)
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.19331 0.25013 0.773 0.450
# b0 NA NA NA NA
# b1 NA NA NA NA
# x -0.06752 0.21720 -0.311 0.760
# e 0.02240 0.20069 0.112 0.912
# b1:x NA NA NA NA
# Residual standard error: 1.115 on 17 degrees of freedom
# Multiple R-squared: 0.006657, Adjusted R-squared: -0.1102
# F-statistic: 0.05697 on 2 and 17 DF, p-value: 0.9448
answered Jan 3 at 16:28
r2evansr2evans
28.5k33259
thank you. though i am really confused how the parameters b0 and b1 have NA.
– bel
Jan 3 at 16:39
In this case, they are singular and invariant, so they can have no impact on the model. Realize that the model is already including an intercept ((Intercept)), which your formula is trying to mimick withb0; if you want to exclude the automatic intercept in the formula, you'll need to subtract one in the formula, as iny ~ b0 + b1*x + e - 1, which now gives you a meaningfulb1but stillb0is invariant ... which doesn't help your model. If this is for a class, I suggest you go ask the professor about what's happening here.
– r2evans
Jan 3 at 16:53
BTW: addingb1is also meaningless here, since thexline in thesummaryoutput is giving you what it thinks the coefficient ofxshould be given this model. If you really want to force it, then perhaps you should not be includingxin the model.
– r2evans
Jan 3 at 16:54
In the long run, I'd think a more meaningful formula could bey ~ x + e.
– r2evans
Jan 3 at 16:55
2
Just a note: in addition, one could also useI(b1*x)in the model formula as it would then not be interpreted as an interaction term, but as the actual product of these two columns. Not clear though that this is what is the OPs intent.
– coffeinjunky
Jan 3 at 16:58
|
show 1 more comment
2
I may be wrong, but I believe you want to simulate an outcome and then estimate it's parameters. If that is true, you would rather do the following:
n1 <- 20
m1 <- 0
sd1<- 1
b0 <- 0
b1 <- 1
x <- rnorm(n1,m1, sd1)
e <- rnorm(n1,m1, sd1)
y <- b0 + b1*x + e
summary(lm(y~x))
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-1.66052 -0.40203 0.05659 0.44115 1.38798
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.3078 0.1951 -1.578 0.132
x 1.1774 0.2292 5.137 6.9e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.852 on 18 degrees of freedom
Multiple R-squared: 0.5945, Adjusted R-squared: 0.572
F-statistic: 26.39 on 1 and 18 DF, p-value: 6.903e-05
And in case you want to do this multiple times, consider the following:
repetitions <- 5
betas <- t(sapply(1:repetitions, function(i){
y <- b0 + b1*x + rnorm(n1,m1, sd1)
coefficients(lm(y~x))
}))
betas
(Intercept) x
[1,] 0.21989182 0.8185690
[2,] -0.12820726 0.7289041
[3,] -0.27596844 0.9794432
[4,] 0.06145306 1.0575050
[5,] -0.31429950 0.9984262
Now you can look at the mean of the estimated betas:
colMeans(betas)
(Intercept) x
-0.08742606 0.91656951
and the variance-covariance matrix:
var(betas)
(Intercept) x
(Intercept) 0.051323041 -0.007976803
x -0.007976803 0.018834711
answered Jan 3 at 16:41
coffeinjunkycoffeinjunky
7,4142143
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
I suggest you put everything in a data.frame and deal with it that way:
set.seed(2)
m1<-0
sd1<-1
y<-rnorm(n1,m1, sd1)
x<-rnorm(n1,m1, sd1)
b0<-0
b1<-1
d <- data.frame(y,b0,b1,x,e=rnorm(20,0,1))
head(d)
# y b0 b1 x e
# 1 -0.89691455 0 1 2.090819205 -0.3835862
# 2 0.18484918 0 1 -1.199925820 -1.9591032
# 3 1.58784533 0 1 1.589638200 -0.8417051
# 4 -1.13037567 0 1 1.954651642 1.9035475
# 5 -0.08025176 0 1 0.004937777 0.6224939
# 6 0.13242028 0 1 -2.451706388 1.9909204
Now things work nicely:
modelfit1 <- lm(y~b0+b1*x+e, data=d)
modelfit1
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Coefficients:
# (Intercept) b0 b1 x e b1:x
# 0.19331 NA NA -0.06752 0.02240 NA
summary(modelfit1)
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Residuals:
# Min 1Q Median 3Q Max
# -2.5006 -0.4786 -0.1425 0.6211 1.8488
# Coefficients: (3 not defined because of singularities)
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.19331 0.25013 0.773 0.450
# b0 NA NA NA NA
# b1 NA NA NA NA
# x -0.06752 0.21720 -0.311 0.760
# e 0.02240 0.20069 0.112 0.912
# b1:x NA NA NA NA
# Residual standard error: 1.115 on 17 degrees of freedom
# Multiple R-squared: 0.006657, Adjusted R-squared: -0.1102
# F-statistic: 0.05697 on 2 and 17 DF, p-value: 0.9448
answered Jan 3 at 16:28
r2evansr2evans
28.5k33259
thank you. though i am really confused how the parameters b0 and b1 have NA.
– bel
Jan 3 at 16:39
In this case, they are singular and invariant, so they can have no impact on the model. Realize that the model is already including an intercept ((Intercept)), which your formula is trying to mimick withb0; if you want to exclude the automatic intercept in the formula, you'll need to subtract one in the formula, as iny ~ b0 + b1*x + e - 1, which now gives you a meaningfulb1but stillb0is invariant ... which doesn't help your model. If this is for a class, I suggest you go ask the professor about what's happening here.
– r2evans
Jan 3 at 16:53
BTW: addingb1is also meaningless here, since thexline in thesummaryoutput is giving you what it thinks the coefficient ofxshould be given this model. If you really want to force it, then perhaps you should not be includingxin the model.
– r2evans
Jan 3 at 16:54
In the long run, I'd think a more meaningful formula could bey ~ x + e.
– r2evans
Jan 3 at 16:55
2
Just a note: in addition, one could also useI(b1*x)in the model formula as it would then not be interpreted as an interaction term, but as the actual product of these two columns. Not clear though that this is what is the OPs intent.
– coffeinjunky
Jan 3 at 16:58
|
show 1 more comment
1
I suggest you put everything in a data.frame and deal with it that way:
set.seed(2)
m1<-0
sd1<-1
y<-rnorm(n1,m1, sd1)
x<-rnorm(n1,m1, sd1)
b0<-0
b1<-1
d <- data.frame(y,b0,b1,x,e=rnorm(20,0,1))
head(d)
# y b0 b1 x e
# 1 -0.89691455 0 1 2.090819205 -0.3835862
# 2 0.18484918 0 1 -1.199925820 -1.9591032
# 3 1.58784533 0 1 1.589638200 -0.8417051
# 4 -1.13037567 0 1 1.954651642 1.9035475
# 5 -0.08025176 0 1 0.004937777 0.6224939
# 6 0.13242028 0 1 -2.451706388 1.9909204
Now things work nicely:
modelfit1 <- lm(y~b0+b1*x+e, data=d)
modelfit1
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Coefficients:
# (Intercept) b0 b1 x e b1:x
# 0.19331 NA NA -0.06752 0.02240 NA
summary(modelfit1)
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Residuals:
# Min 1Q Median 3Q Max
# -2.5006 -0.4786 -0.1425 0.6211 1.8488
# Coefficients: (3 not defined because of singularities)
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.19331 0.25013 0.773 0.450
# b0 NA NA NA NA
# b1 NA NA NA NA
# x -0.06752 0.21720 -0.311 0.760
# e 0.02240 0.20069 0.112 0.912
# b1:x NA NA NA NA
# Residual standard error: 1.115 on 17 degrees of freedom
# Multiple R-squared: 0.006657, Adjusted R-squared: -0.1102
# F-statistic: 0.05697 on 2 and 17 DF, p-value: 0.9448
answered Jan 3 at 16:28
r2evansr2evans
28.5k33259
thank you. though i am really confused how the parameters b0 and b1 have NA.
– bel
Jan 3 at 16:39
In this case, they are singular and invariant, so they can have no impact on the model. Realize that the model is already including an intercept ((Intercept)), which your formula is trying to mimick withb0; if you want to exclude the automatic intercept in the formula, you'll need to subtract one in the formula, as iny ~ b0 + b1*x + e - 1, which now gives you a meaningfulb1but stillb0is invariant ... which doesn't help your model. If this is for a class, I suggest you go ask the professor about what's happening here.
– r2evans
Jan 3 at 16:53
BTW: addingb1is also meaningless here, since thexline in thesummaryoutput is giving you what it thinks the coefficient ofxshould be given this model. If you really want to force it, then perhaps you should not be includingxin the model.
– r2evans
Jan 3 at 16:54
In the long run, I'd think a more meaningful formula could bey ~ x + e.
– r2evans
Jan 3 at 16:55
2
Just a note: in addition, one could also useI(b1*x)in the model formula as it would then not be interpreted as an interaction term, but as the actual product of these two columns. Not clear though that this is what is the OPs intent.
– coffeinjunky
Jan 3 at 16:58
|
show 1 more comment
1
1
1
I suggest you put everything in a data.frame and deal with it that way:
set.seed(2)
m1<-0
sd1<-1
y<-rnorm(n1,m1, sd1)
x<-rnorm(n1,m1, sd1)
b0<-0
b1<-1
d <- data.frame(y,b0,b1,x,e=rnorm(20,0,1))
head(d)
# y b0 b1 x e
# 1 -0.89691455 0 1 2.090819205 -0.3835862
# 2 0.18484918 0 1 -1.199925820 -1.9591032
# 3 1.58784533 0 1 1.589638200 -0.8417051
# 4 -1.13037567 0 1 1.954651642 1.9035475
# 5 -0.08025176 0 1 0.004937777 0.6224939
# 6 0.13242028 0 1 -2.451706388 1.9909204
Now things work nicely:
modelfit1 <- lm(y~b0+b1*x+e, data=d)
modelfit1
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Coefficients:
# (Intercept) b0 b1 x e b1:x
# 0.19331 NA NA -0.06752 0.02240 NA
summary(modelfit1)
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Residuals:
# Min 1Q Median 3Q Max
# -2.5006 -0.4786 -0.1425 0.6211 1.8488
# Coefficients: (3 not defined because of singularities)
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.19331 0.25013 0.773 0.450
# b0 NA NA NA NA
# b1 NA NA NA NA
# x -0.06752 0.21720 -0.311 0.760
# e 0.02240 0.20069 0.112 0.912
# b1:x NA NA NA NA
# Residual standard error: 1.115 on 17 degrees of freedom
# Multiple R-squared: 0.006657, Adjusted R-squared: -0.1102
# F-statistic: 0.05697 on 2 and 17 DF, p-value: 0.9448
answered Jan 3 at 16:28
r2evansr2evans
28.5k33259
I suggest you put everything in a data.frame and deal with it that way:
set.seed(2)
m1<-0
sd1<-1
y<-rnorm(n1,m1, sd1)
x<-rnorm(n1,m1, sd1)
b0<-0
b1<-1
d <- data.frame(y,b0,b1,x,e=rnorm(20,0,1))
head(d)
# y b0 b1 x e
# 1 -0.89691455 0 1 2.090819205 -0.3835862
# 2 0.18484918 0 1 -1.199925820 -1.9591032
# 3 1.58784533 0 1 1.589638200 -0.8417051
# 4 -1.13037567 0 1 1.954651642 1.9035475
# 5 -0.08025176 0 1 0.004937777 0.6224939
# 6 0.13242028 0 1 -2.451706388 1.9909204
Now things work nicely:
modelfit1 <- lm(y~b0+b1*x+e, data=d)
modelfit1
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Coefficients:
# (Intercept) b0 b1 x e b1:x
# 0.19331 NA NA -0.06752 0.02240 NA
summary(modelfit1)
# Call:
# lm(formula = y ~ b0 + b1 * x + e, data = d)
# Residuals:
# Min 1Q Median 3Q Max
# -2.5006 -0.4786 -0.1425 0.6211 1.8488
# Coefficients: (3 not defined because of singularities)
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.19331 0.25013 0.773 0.450
# b0 NA NA NA NA
# b1 NA NA NA NA
# x -0.06752 0.21720 -0.311 0.760
# e 0.02240 0.20069 0.112 0.912
# b1:x NA NA NA NA
# Residual standard error: 1.115 on 17 degrees of freedom
# Multiple R-squared: 0.006657, Adjusted R-squared: -0.1102
# F-statistic: 0.05697 on 2 and 17 DF, p-value: 0.9448
answered Jan 3 at 16:28
r2evansr2evans
28.5k33259
answered Jan 3 at 16:28
r2evansr2evans
28.5k33259
answered Jan 3 at 16:28
r2evansr2evans
28.5k33259
answered Jan 3 at 16:28
r2evansr2evans
28.5k33259
28.5k33259
thank you. though i am really confused how the parameters b0 and b1 have NA.
– bel
Jan 3 at 16:39
In this case, they are singular and invariant, so they can have no impact on the model. Realize that the model is already including an intercept ((Intercept)), which your formula is trying to mimick withb0; if you want to exclude the automatic intercept in the formula, you'll need to subtract one in the formula, as iny ~ b0 + b1*x + e - 1, which now gives you a meaningfulb1but stillb0is invariant ... which doesn't help your model. If this is for a class, I suggest you go ask the professor about what's happening here.
– r2evans
Jan 3 at 16:53
BTW: addingb1is also meaningless here, since thexline in thesummaryoutput is giving you what it thinks the coefficient ofxshould be given this model. If you really want to force it, then perhaps you should not be includingxin the model.
– r2evans
Jan 3 at 16:54
In the long run, I'd think a more meaningful formula could bey ~ x + e.
– r2evans
Jan 3 at 16:55
2
Just a note: in addition, one could also useI(b1*x)in the model formula as it would then not be interpreted as an interaction term, but as the actual product of these two columns. Not clear though that this is what is the OPs intent.
– coffeinjunky
Jan 3 at 16:58
|
show 1 more comment
thank you. though i am really confused how the parameters b0 and b1 have NA.
– bel
Jan 3 at 16:39
In this case, they are singular and invariant, so they can have no impact on the model. Realize that the model is already including an intercept ((Intercept)), which your formula is trying to mimick withb0; if you want to exclude the automatic intercept in the formula, you'll need to subtract one in the formula, as iny ~ b0 + b1*x + e - 1, which now gives you a meaningfulb1but stillb0is invariant ... which doesn't help your model. If this is for a class, I suggest you go ask the professor about what's happening here.
– r2evans
Jan 3 at 16:53
BTW: addingb1is also meaningless here, since thexline in thesummaryoutput is giving you what it thinks the coefficient ofxshould be given this model. If you really want to force it, then perhaps you should not be includingxin the model.
– r2evans
Jan 3 at 16:54
In the long run, I'd think a more meaningful formula could bey ~ x + e.
– r2evans
Jan 3 at 16:55
2
Just a note: in addition, one could also useI(b1*x)in the model formula as it would then not be interpreted as an interaction term, but as the actual product of these two columns. Not clear though that this is what is the OPs intent.
– coffeinjunky
Jan 3 at 16:58
thank you. though i am really confused how the parameters b0 and b1 have NA.
– bel
Jan 3 at 16:39
thank you. though i am really confused how the parameters b0 and b1 have NA.
– bel
Jan 3 at 16:39
In this case, they are singular and invariant, so they can have no impact on the model. Realize that the model is already including an intercept (
(Intercept)), which your formula is trying to mimick with b0; if you want to exclude the automatic intercept in the formula, you'll need to subtract one in the formula, as in y ~ b0 + b1*x + e - 1, which now gives you a meaningful b1 but still b0 is invariant ... which doesn't help your model. If this is for a class, I suggest you go ask the professor about what's happening here.– r2evans
Jan 3 at 16:53
In this case, they are singular and invariant, so they can have no impact on the model. Realize that the model is already including an intercept (
(Intercept)), which your formula is trying to mimick with b0; if you want to exclude the automatic intercept in the formula, you'll need to subtract one in the formula, as in y ~ b0 + b1*x + e - 1, which now gives you a meaningful b1 but still b0 is invariant ... which doesn't help your model. If this is for a class, I suggest you go ask the professor about what's happening here.– r2evans
Jan 3 at 16:53
BTW: adding
b1 is also meaningless here, since the x line in the summary output is giving you what it thinks the coefficient of x should be given this model. If you really want to force it, then perhaps you should not be including x in the model.– r2evans
Jan 3 at 16:54
BTW: adding
b1 is also meaningless here, since the x line in the summary output is giving you what it thinks the coefficient of x should be given this model. If you really want to force it, then perhaps you should not be including x in the model.– r2evans
Jan 3 at 16:54
In the long run, I'd think a more meaningful formula could be
y ~ x + e.– r2evans
Jan 3 at 16:55
In the long run, I'd think a more meaningful formula could be
y ~ x + e.– r2evans
Jan 3 at 16:55
2
2
Just a note: in addition, one could also use
I(b1*x) in the model formula as it would then not be interpreted as an interaction term, but as the actual product of these two columns. Not clear though that this is what is the OPs intent.– coffeinjunky
Jan 3 at 16:58
Just a note: in addition, one could also use
I(b1*x) in the model formula as it would then not be interpreted as an interaction term, but as the actual product of these two columns. Not clear though that this is what is the OPs intent.– coffeinjunky
Jan 3 at 16:58
|
show 1 more comment
2
I may be wrong, but I believe you want to simulate an outcome and then estimate it's parameters. If that is true, you would rather do the following:
n1 <- 20
m1 <- 0
sd1<- 1
b0 <- 0
b1 <- 1
x <- rnorm(n1,m1, sd1)
e <- rnorm(n1,m1, sd1)
y <- b0 + b1*x + e
summary(lm(y~x))
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-1.66052 -0.40203 0.05659 0.44115 1.38798
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.3078 0.1951 -1.578 0.132
x 1.1774 0.2292 5.137 6.9e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.852 on 18 degrees of freedom
Multiple R-squared: 0.5945, Adjusted R-squared: 0.572
F-statistic: 26.39 on 1 and 18 DF, p-value: 6.903e-05
And in case you want to do this multiple times, consider the following:
repetitions <- 5
betas <- t(sapply(1:repetitions, function(i){
y <- b0 + b1*x + rnorm(n1,m1, sd1)
coefficients(lm(y~x))
}))
betas
(Intercept) x
[1,] 0.21989182 0.8185690
[2,] -0.12820726 0.7289041
[3,] -0.27596844 0.9794432
[4,] 0.06145306 1.0575050
[5,] -0.31429950 0.9984262
Now you can look at the mean of the estimated betas:
colMeans(betas)
(Intercept) x
-0.08742606 0.91656951
and the variance-covariance matrix:
var(betas)
(Intercept) x
(Intercept) 0.051323041 -0.007976803
x -0.007976803 0.018834711
answered Jan 3 at 16:41
coffeinjunkycoffeinjunky
7,4142143
add a comment |
2
I may be wrong, but I believe you want to simulate an outcome and then estimate it's parameters. If that is true, you would rather do the following:
n1 <- 20
m1 <- 0
sd1<- 1
b0 <- 0
b1 <- 1
x <- rnorm(n1,m1, sd1)
e <- rnorm(n1,m1, sd1)
y <- b0 + b1*x + e
summary(lm(y~x))
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-1.66052 -0.40203 0.05659 0.44115 1.38798
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.3078 0.1951 -1.578 0.132
x 1.1774 0.2292 5.137 6.9e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.852 on 18 degrees of freedom
Multiple R-squared: 0.5945, Adjusted R-squared: 0.572
F-statistic: 26.39 on 1 and 18 DF, p-value: 6.903e-05
And in case you want to do this multiple times, consider the following:
repetitions <- 5
betas <- t(sapply(1:repetitions, function(i){
y <- b0 + b1*x + rnorm(n1,m1, sd1)
coefficients(lm(y~x))
}))
betas
(Intercept) x
[1,] 0.21989182 0.8185690
[2,] -0.12820726 0.7289041
[3,] -0.27596844 0.9794432
[4,] 0.06145306 1.0575050
[5,] -0.31429950 0.9984262
Now you can look at the mean of the estimated betas:
colMeans(betas)
(Intercept) x
-0.08742606 0.91656951
and the variance-covariance matrix:
var(betas)
(Intercept) x
(Intercept) 0.051323041 -0.007976803
x -0.007976803 0.018834711
answered Jan 3 at 16:41
coffeinjunkycoffeinjunky
7,4142143
add a comment |
2
2
2
I may be wrong, but I believe you want to simulate an outcome and then estimate it's parameters. If that is true, you would rather do the following:
n1 <- 20
m1 <- 0
sd1<- 1
b0 <- 0
b1 <- 1
x <- rnorm(n1,m1, sd1)
e <- rnorm(n1,m1, sd1)
y <- b0 + b1*x + e
summary(lm(y~x))
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-1.66052 -0.40203 0.05659 0.44115 1.38798
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.3078 0.1951 -1.578 0.132
x 1.1774 0.2292 5.137 6.9e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.852 on 18 degrees of freedom
Multiple R-squared: 0.5945, Adjusted R-squared: 0.572
F-statistic: 26.39 on 1 and 18 DF, p-value: 6.903e-05
And in case you want to do this multiple times, consider the following:
repetitions <- 5
betas <- t(sapply(1:repetitions, function(i){
y <- b0 + b1*x + rnorm(n1,m1, sd1)
coefficients(lm(y~x))
}))
betas
(Intercept) x
[1,] 0.21989182 0.8185690
[2,] -0.12820726 0.7289041
[3,] -0.27596844 0.9794432
[4,] 0.06145306 1.0575050
[5,] -0.31429950 0.9984262
Now you can look at the mean of the estimated betas:
colMeans(betas)
(Intercept) x
-0.08742606 0.91656951
and the variance-covariance matrix:
var(betas)
(Intercept) x
(Intercept) 0.051323041 -0.007976803
x -0.007976803 0.018834711
answered Jan 3 at 16:41
coffeinjunkycoffeinjunky
7,4142143
I may be wrong, but I believe you want to simulate an outcome and then estimate it's parameters. If that is true, you would rather do the following:
n1 <- 20
m1 <- 0
sd1<- 1
b0 <- 0
b1 <- 1
x <- rnorm(n1,m1, sd1)
e <- rnorm(n1,m1, sd1)
y <- b0 + b1*x + e
summary(lm(y~x))
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-1.66052 -0.40203 0.05659 0.44115 1.38798
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.3078 0.1951 -1.578 0.132
x 1.1774 0.2292 5.137 6.9e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.852 on 18 degrees of freedom
Multiple R-squared: 0.5945, Adjusted R-squared: 0.572
F-statistic: 26.39 on 1 and 18 DF, p-value: 6.903e-05
And in case you want to do this multiple times, consider the following:
repetitions <- 5
betas <- t(sapply(1:repetitions, function(i){
y <- b0 + b1*x + rnorm(n1,m1, sd1)
coefficients(lm(y~x))
}))
betas
(Intercept) x
[1,] 0.21989182 0.8185690
[2,] -0.12820726 0.7289041
[3,] -0.27596844 0.9794432
[4,] 0.06145306 1.0575050
[5,] -0.31429950 0.9984262
Now you can look at the mean of the estimated betas:
colMeans(betas)
(Intercept) x
-0.08742606 0.91656951
and the variance-covariance matrix:
var(betas)
(Intercept) x
(Intercept) 0.051323041 -0.007976803
x -0.007976803 0.018834711
answered Jan 3 at 16:41
coffeinjunkycoffeinjunky
7,4142143
edited Jan 3 at 17:01
answered Jan 3 at 16:41
coffeinjunkycoffeinjunky
7,4142143
answered Jan 3 at 16:41
coffeinjunkycoffeinjunky
7,4142143
answered Jan 3 at 16:41
coffeinjunkycoffeinjunky
7,4142143
7,4142143
add a comment |
add a comment |
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2
What do you want to do?
– coffeinjunky
Jan 3 at 16:00
1
Hello, welcome to StackOverflow. I think your simulated example does not really match what you want to do. Maybe this post would help? stats.stackexchange.com/questions/99924/…
– Vincent Guillemot
Jan 3 at 16:01
1
A comment on where the error comes from: R thinks that the model you want to fit contains a response variable y (so far so good), and a complicated combination of explanatory variables (one called b0, one called b1, one called x, one interaction term between the variables x and b1 and a final term that I don't even know how R will interpret it). If your intention was to generate a simple dataset to fit a simple linear model on (and I think it was your intention), then you need to fix your simulated data.
– Vincent Guillemot
Jan 3 at 16:10
@coffeinjunky hello, i want to see the estimate standard error of the parameter b0 and b1.
– bel
Jan 3 at 16:28