proportional weights

Hello all, can help clarify something?

According to R's lm() doc:
Non-NULL weights can be used to indicate that different observations
have different variances (with the values in weights being inversely
*proportional* to the variances); or equivalently, when the elements
of weights are positive integers w_i, that each response y_i is the
mean of w_i unit-weight observations (including the case that there
are w_i observations equal to y_i and the data have been summarized).
Since the idea here is *proportion*, not equality, shouldn't the vectors 
of weights x, 2*x give the same result? And yet they don't, standard 
errors differs:
summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(1,6)))$sigma
[1] 0.07108323
summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(2,6)))$sigma
[1] 0.1005269
So what if I know a-priori, observation A has variance 2 times bigger 
than observation B? Both weights=c(1,2) and weights=c(2,4) (and so on) 
represent very well this knowledge, but we get different regression 
(since sigma is different).

Also, if we do the same thing with a glm() model, than we get a lot of 
other differences like in the deviance.
Hello all, can help clarify something?

According to R's lm() doc:

Non-NULL weights can be used to indicate that different observations
have different variances (with the values in weights being inversely
*proportional* to the variances); or equivalently, when the elements
of weights are positive integers w_i, that each response y_i is the
mean of w_i unit-weight observations (including the case that there
are w_i observations equal to y_i and the data have been summarized).
Since the idea here is *proportion*, not equality, shouldn't the vectors
of weights x, 2*x give the same result? And yet they don't, standard
errors differs:

summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(1,6)))$sigma
[1] 0.07108323
summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(2,6)))$sigma
[1] 0.1005269
The weights are in fact case weights, i.e., a weight of 2 is the same as 
including the corresponding item twice. I agree that the documentation 
is no wonder of clarity in this respect.

Btw, note that, in your example, (0.1005269 / 0.07108323)^2 = 2, your 
constant weight.

G?ran Brostr?m

So what if I know a-priori, observation A has variance 2 times bigger
than observation B? Both weights=c(1,2) and weights=c(2,4) (and so on)
represent very well this knowledge, but we get different regression
(since sigma is different).

Also, if we do the same thing with a glm() model, than we get a lot of
other differences like in the deviance.

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Dear Marco and Goran,

Perhaps the documentation could be clearer, but it is after all a brief help page. Using weights of 2 to lm() is *not* equivalent to entering the observation twice. The weights are variance weights, not case weights.

You can see this by looking at the whole summary() output for the models, not just the residual standard errors:

------------- snip ---------
summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(1,6)))
Call:
lm(formula = c(1, 2, 3, 1, 2, 3) ~ c(1, 2.1, 2.9, 1.1, 2, 3), 
    weights = rep(1, 6))

Residuals:
       1        2        3        4        5        6 
 0.06477 -0.08728  0.07487 -0.03996  0.01746 -0.02986 

Coefficients:
                          Estimate Std. Error t value Pr(>|t|)    
(Intercept)               -0.11208    0.08066   -1.39    0.237    
c(1, 2.1, 2.9, 1.1, 2, 3)  1.04731    0.03732   28.07 9.59e-06 ***
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1

Residual standard error: 0.07108 on 4 degrees of freedom
Multiple R-squared:  0.9949,	Adjusted R-squared:  0.9937 
F-statistic: 787.6 on 1 and 4 DF,  p-value: 9.59e-06
summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(2,6)))
Call:
lm(formula = c(1, 2, 3, 1, 2, 3) ~ c(1, 2.1, 2.9, 1.1, 2, 3), 
    weights = rep(2, 6))

Residuals:
       1        2        3        4        5        6 
 0.09160 -0.12343  0.10589 -0.05652  0.02469 -0.04223 

Coefficients:
                          Estimate Std. Error t value Pr(>|t|)    
(Intercept)               -0.11208    0.08066   -1.39    0.237    
c(1, 2.1, 2.9, 1.1, 2, 3)  1.04731    0.03732   28.07 9.59e-06 ***
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1

Residual standard error: 0.1005 on 4 degrees of freedom
Multiple R-squared:  0.9949,	Adjusted R-squared:  0.9937 
F-statistic: 787.6 on 1 and 4 DF,  p-value: 9.59e-06

------------- snip -------------

Notice that while the residual standard errors differ, the coefficients and their standard errors are identical. There are compensating changes in the residual variance and the weighted sum of squares and products matrix for X. 

In contrast, literally entering each observation twice reduces the coefficient standard errors by a factor of sqrt((6 - 2)/(12 - 2)), i.e., the square root of the relative residual df of the models:

------------- snip --------
summary(lm(rep(c(1,2,3,1,2,3),2)~rep(c(1,2.1,2.9,1.1,2,3),2)))
Call:
lm(formula = rep(c(1, 2, 3, 1, 2, 3), 2) ~ rep(c(1, 2.1, 2.9, 
    1.1, 2, 3), 2))

Residuals:
      Min        1Q    Median        3Q       Max 
-0.087276 -0.039963 -0.006201  0.064768  0.074874 

Coefficients:
                                  Estimate Std. Error t value Pr(>|t|)
(Intercept)                       -0.11208    0.05101  -2.197   0.0527
rep(c(1, 2.1, 2.9, 1.1, 2, 3), 2)  1.04731    0.02360  44.374 8.12e-13

(Intercept)                       .  
rep(c(1, 2.1, 2.9, 1.1, 2, 3), 2) ***
---
Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1

Residual standard error: 0.06358 on 10 degrees of freedom
Multiple R-squared:  0.9949,	Adjusted R-squared:  0.9944 
F-statistic:  1969 on 1 and 10 DF,  p-value: 8.122e-13

---------- snip -------------

I hope this helps,

John

------------------------------------------------
John Fox, Professor
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/

On Thu, 6 Feb 2014 09:27:22 +0100
On 05/02/14 22:40, Marco Inacio wrote:
Hello all, can help clarify something?

According to R's lm() doc:

Non-NULL weights can be used to indicate that different observations
have different variances (with the values in weights being inversely
*proportional* to the variances); or equivalently, when the elements
of weights are positive integers w_i, that each response y_i is the
mean of w_i unit-weight observations (including the case that there
are w_i observations equal to y_i and the data have been summarized).
Since the idea here is *proportion*, not equality, shouldn't the vectors
of weights x, 2*x give the same result? And yet they don't, standard
errors differs:

summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(1,6)))$sigma
[1] 0.07108323
summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(2,6)))$sigma
[1] 0.1005269
The weights are in fact case weights, i.e., a weight of 2 is the same as including the corresponding item twice. I agree that the documentation is no wonder of clarity in this respect.

Btw, note that, in your example, (0.1005269 / 0.07108323)^2 = 2, your constant weight.

G?ran Brostr?m

So what if I know a-priori, observation A has variance 2 times bigger
than observation B? Both weights=c(1,2) and weights=c(2,4) (and so on)
represent very well this knowledge, but we get different regression
(since sigma is different).

Also, if we do the same thing with a glm() model, than we get a lot of
other differences like in the deviance.

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Dear John,

thanks for the clarification! The lesson to be learned is that one 
should be aware of the fact that weights may mean different things in 
different functions, and sometimes different things in the same function 
(glm)!

G?ran
Dear Marco and Goran,

Perhaps the documentation could be clearer, but it is after all a
brief help page. Using weights of 2 to lm() is *not* equivalent to
entering the observation twice. The weights are variance weights, not
case weights.

You can see this by looking at the whole summary() output for the
models, not just the residual standard errors:

------------- snip ---------

summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(1,6)))
Call: lm(formula = c(1, 2, 3, 1, 2, 3) ~ c(1, 2.1, 2.9, 1.1, 2, 3),
weights = rep(1, 6))

Residuals: 1        2        3        4        5        6 0.06477
-0.08728  0.07487 -0.03996  0.01746 -0.02986

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept)
-0.11208    0.08066   -1.39    0.237 c(1, 2.1, 2.9, 1.1, 2, 3)
1.04731    0.03732   28.07 9.59e-06 *** --- Signif. codes:  0 ?***?
0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1

Residual standard error: 0.07108 on 4 degrees of freedom Multiple
R-squared:  0.9949,	Adjusted R-squared:  0.9937 F-statistic: 787.6 on
1 and 4 DF,  p-value: 9.59e-06

summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(2,6)))
Call: lm(formula = c(1, 2, 3, 1, 2, 3) ~ c(1, 2.1, 2.9, 1.1, 2, 3),
weights = rep(2, 6))

Residuals: 1        2        3        4        5        6 0.09160
-0.12343  0.10589 -0.05652  0.02469 -0.04223

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept)
-0.11208    0.08066   -1.39    0.237 c(1, 2.1, 2.9, 1.1, 2, 3)
1.04731    0.03732   28.07 9.59e-06 *** --- Signif. codes:  0 ?***?
0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1

Residual standard error: 0.1005 on 4 degrees of freedom Multiple
R-squared:  0.9949,	Adjusted R-squared:  0.9937 F-statistic: 787.6 on
1 and 4 DF,  p-value: 9.59e-06

------------- snip -------------

Notice that while the residual standard errors differ, the
coefficients and their standard errors are identical. There are
compensating changes in the residual variance and the weighted sum of
squares and products matrix for X.

In contrast, literally entering each observation twice reduces the
coefficient standard errors by a factor of sqrt((6 - 2)/(12 - 2)),
i.e., the square root of the relative residual df of the models:

------------- snip --------

summary(lm(rep(c(1,2,3,1,2,3),2)~rep(c(1,2.1,2.9,1.1,2,3),2)))
Call: lm(formula = rep(c(1, 2, 3, 1, 2, 3), 2) ~ rep(c(1, 2.1, 2.9,
1.1, 2, 3), 2))

Residuals: Min        1Q    Median        3Q       Max -0.087276
-0.039963 -0.006201  0.064768  0.074874

Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept)
-0.11208    0.05101  -2.197   0.0527 rep(c(1, 2.1, 2.9, 1.1, 2, 3),
2)  1.04731    0.02360  44.374 8.12e-13

(Intercept)                       . rep(c(1, 2.1, 2.9, 1.1, 2, 3), 2)
*** --- Signif. codes:  0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ?
1

Residual standard error: 0.06358 on 10 degrees of freedom Multiple
R-squared:  0.9949,	Adjusted R-squared:  0.9944 F-statistic:  1969 on
1 and 10 DF,  p-value: 8.122e-13

---------- snip -------------

I hope this helps,

John

------------------------------------------------ John Fox, Professor
McMaster University Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/   On Thu, 6 Feb 2014 09:27:22 +0100
G?ran Brostr?m <goran.brostrom at umu.se> wrote:
On 05/02/14 22:40, Marco Inacio wrote:
Hello all, can help clarify something?

According to R's lm() doc:

Non-NULL weights can be used to indicate that different
observations have different variances (with the values in
weights being inversely *proportional* to the variances); or
equivalently, when the elements of weights are positive
integers w_i, that each response y_i is the mean of w_i
unit-weight observations (including the case that there are w_i
observations equal to y_i and the data have been summarized).
Since the idea here is *proportion*, not equality, shouldn't the
vectors of weights x, 2*x give the same result? And yet they
don't, standard errors differs:

summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(1,6)))$sigma

[1] 0.07108323
summary(lm(c(1,2,3,1,2,3)~c(1,2.1,2.9,1.1,2,3),weight=rep(2,6)))$sigma

[1] 0.1005269
The weights are in fact case weights, i.e., a weight of 2 is the
same as including the corresponding item twice. I agree that the
documentation is no wonder of clarity in this respect.

Btw, note that, in your example, (0.1005269 / 0.07108323)^2 = 2,
your constant weight.

G?ran Brostr?m

So what if I know a-priori, observation A has variance 2 times
bigger than observation B? Both weights=c(1,2) and weights=c(2,4)
(and so on) represent very well this knowledge, but we get
different regression (since sigma is different).

Also, if we do the same thing with a glm() model, than we get a
lot of other differences like in the deviance.

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the
posting guide http://www.R-project.org/posting-guide.html and
provide commented, minimal, self-contained, reproducible code.

______________________________________________ R-help at r-project.org
mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do
read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Thanks for the answers.
Dear Marco and Goran,

Perhaps the documentation could be clearer, but it is after all a brief help page. Using weights of 2 to lm() is *not* equivalent to entering the observation twice. The weights are variance weights, not case weights.

According to your post here:
   http://tolstoy.newcastle.edu.au/R/e2/help/07/05/16311.html
   there are 3 possible kinds of weights.

The person in this one:
   http://tolstoy.newcastle.edu.au/R/e2/help/07/06/18743.html
   includes 2 others making a distinction between weights inverse 
proportional to variance and weight equal to inverse variance.

(looking at other posts in the thread shows that other people also make 
confusions on this matter)

So R's lm(), glm(), etc weights **are** the inverse of the variance of 
the observations, right?
They'are not **proportional** to the inverse of variance because if this 
were true, then weight and 2*weight would archive the same results, right?

I needed a method to use proportional weights on observations as I know 
their proportion of variance among each other.
And it doesn't need to be a R function, just an explanation on how 
construct the likehood would be fine. If anybody know an article on the 
subject, would be of great help to.
Dear Marco,

What I said in the 2007 r-help posting to which you refer is, "The weights
used by lm() are (inverse-)'variance weights,' reflecting the variances of
the errors, with observations that have low-variance errors therefore being
accorded greater weight in the resulting WLS regression." ?lm says,
"Non-NULL weights can be used to indicate that different observations have
different variances (with the values in weights being inversely proportional
to the variances)."

If I understand your situation correctly, you know the error variances up to
a constant of proportionality, in which case you can set the weights
argument to lm() to the inverses of these values. As I showed you in the
example I just posted, weight and 2*weight *do* produce the same coefficient
estimates and standard errors, with the difference between the two absorbed
by the residual standard error, which is the square-root of the estimated
constant of proportionality.

If this is insufficiently clear, I'm afraid that I'll have to defer to
someone with greater powers of explanation.

Best,
 John
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
project.org] On Behalf Of Marco Inacio
Sent: Thursday, February 06, 2014 9:06 AM
To: r-help at r-project.org
Subject: Re: [R] proportional weights

Thanks for the answers.

Dear Marco and Goran,

Perhaps the documentation could be clearer, but it is after all a
brief help page. Using weights of 2 to lm() is *not* equivalent to
entering the observation twice. The weights are variance weights, not
case weights.

According to your post here:
   http://tolstoy.newcastle.edu.au/R/e2/help/07/05/16311.html
   there are 3 possible kinds of weights.

The person in this one:
   http://tolstoy.newcastle.edu.au/R/e2/help/07/06/18743.html
   includes 2 others making a distinction between weights inverse
proportional to variance and weight equal to inverse variance.

(looking at other posts in the thread shows that other people also make
confusions on this matter)

So R's lm(), glm(), etc weights **are** the inverse of the variance of
the observations, right?
They'are not **proportional** to the inverse of variance because if
this
were true, then weight and 2*weight would archive the same results,
right?

I needed a method to use proportional weights on observations as I know
their proportion of variance among each other.
And it doesn't need to be a R function, just an explanation on how
construct the likehood would be fine. If anybody know an article on the
subject, would be of great help to.

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
I think we can blame Tim Hesterberg for the confusion:

He writes

"
I'll add: 
* inverse-variance weights, where var(y for observation) = 1/weight   (as opposed to just being inversely proportional to the weight) *
"

And, although I'm not a native English speaker, I think there's a spurious comma in there. The intention was clearly to have this as a  4th type of weight which is a special case of inverse-variance weights, not as an elaboration on the definition of inv.var. weights.

I.e., it is the difference between

Motorists who are reckless drivers...

and

Motorists, who are reckless drivers...

-pd

Dear Marco,

What I said in the 2007 r-help posting to which you refer is, "The weights
used by lm() are (inverse-)'variance weights,' reflecting the variances of
the errors, with observations that have low-variance errors therefore being
accorded greater weight in the resulting WLS regression." ?lm says,
"Non-NULL weights can be used to indicate that different observations have
different variances (with the values in weights being inversely proportional
to the variances)."

If I understand your situation correctly, you know the error variances up to
a constant of proportionality, in which case you can set the weights
argument to lm() to the inverses of these values. As I showed you in the
example I just posted, weight and 2*weight *do* produce the same coefficient
estimates and standard errors, with the difference between the two absorbed
by the residual standard error, which is the square-root of the estimated
constant of proportionality.

If this is insufficiently clear, I'm afraid that I'll have to defer to
someone with greater powers of explanation.

Best,
John

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
project.org] On Behalf Of Marco Inacio
Sent: Thursday, February 06, 2014 9:06 AM
To: r-help at r-project.org
Subject: Re: [R] proportional weights

Thanks for the answers.

Dear Marco and Goran,

Perhaps the documentation could be clearer, but it is after all a
brief help page. Using weights of 2 to lm() is *not* equivalent to
entering the observation twice. The weights are variance weights, not
case weights.

According to your post here:
  http://tolstoy.newcastle.edu.au/R/e2/help/07/05/16311.html
  there are 3 possible kinds of weights.

The person in this one:
  http://tolstoy.newcastle.edu.au/R/e2/help/07/06/18743.html
  includes 2 others making a distinction between weights inverse
proportional to variance and weight equal to inverse variance.

(looking at other posts in the thread shows that other people also make
confusions on this matter)

So R's lm(), glm(), etc weights **are** the inverse of the variance of
the observations, right?
They'are not **proportional** to the inverse of variance because if
this
were true, then weight and 2*weight would archive the same results,
right?

I needed a method to use proportional weights on observations as I know
their proportion of variance among each other.
And it doesn't need to be a R function, just an explanation on how
construct the likehood would be fine. If anybody know an article on the
subject, would be of great help to.

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com
I think we can blame Tim Hesterberg for the confusion:

He writes

"
I'll add:
* inverse-variance weights, where var(y for observation) = 1/weight   (as opposed to just being inversely proportional to the weight) *
"

And, although I'm not a native English speaker, I think there's a spurious comma in there. The intention was clearly to have this as a  4th type of weight which is a special case of inverse-variance weights, not as an elaboration on the definition of inv.var. weights.

I.e., it is the difference between

Motorists who are reckless drivers...

and

Motorists, who are reckless drivers...

-pd
In fact, that wasn't what caused the confusion as I have understood what 
he meant despite the problem with the comma.

But I got the idea now, R uses weighted least squares and:

"Var[\epsilon | X] = \Omega" (equal, not proportion)
(https://en.wikipedia.org/wiki/Generalized_least_squares) (since WLS is 
just a special case of GLS)

"The weights should, ideally, be equal to the reciprocal of the variance 
of the measurement."
(https://en.wikipedia.org/wiki/Linear_least_squares_(mathematics)#Weighted_linear_least_squares)

I guess I need to find another strategy to use proportional weights 
(weights know up to a constant, as John says).

So, thank you much to you all, and sorry the inconvenience I caused.
Dear Marco,
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
project.org] On Behalf Of Marco Inacio
Sent: Thursday, February 06, 2014 12:41 PM
To: R help
Subject: Re: [R] proportional weights

I think we can blame Tim Hesterberg for the confusion:

He writes

"
I'll add:
* inverse-variance weights, where var(y for observation) = 1/weight
(as opposed to just being inversely proportional to the weight) *
"

And, although I'm not a native English speaker, I think there's a
spurious comma in there. The intention was clearly to have this as a
4th type of weight which is a special case of inverse-variance weights,
not as an elaboration on the definition of inv.var. weights.
I.e., it is the difference between

Motorists who are reckless drivers...

and

Motorists, who are reckless drivers...

-pd
In fact, that wasn't what caused the confusion as I have understood
what
he meant despite the problem with the comma.

But I got the idea now, R uses weighted least squares and:

"Var[\epsilon | X] = \Omega" (equal, not proportion)
(https://en.wikipedia.org/wiki/Generalized_least_squares) (since WLS is
just a special case of GLS)

"The weights should, ideally, be equal to the reciprocal of the
variance
of the measurement."
(https://en.wikipedia.org/wiki/Linear_least_squares_(mathematics)#Weigh
ted_linear_least_squares)

I guess I need to find another strategy to use proportional weights
(weights know up to a constant, as John says).
No, you are perfectly fine using WLS. The constant of proportionality is the
estimated error variance, i.e., the square of the residual standard error
(as I think I said earlier).

John
So, thank you much to you all, and sorry the inconvenience I caused.

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
No, you are perfectly fine using WLS. The constant of proportionality is the
estimated error variance, i.e., the square of the residual standard error
(as I think I said earlier).

John
You're right. That was a little hard for me to grasp. Thanks for the 
patience.