Magean - 1 year ago 90

R Question

In order to correct heteroskedasticity in error terms, I am running the following weighted least squares regression in R :

`#Call:`

#lm(formula = a ~ q + q2 + b + c, data = mydata, weights = weighting)

#Weighted Residuals:

# Min 1Q Median 3Q Max

#-1.83779 -0.33226 0.02011 0.25135 1.48516

#Coefficients:

# Estimate Std. Error t value Pr(>|t|)

#(Intercept) -3.939440 0.609991 -6.458 1.62e-09 ***

#q 0.175019 0.070101 2.497 0.013696 *

#q2 0.048790 0.005613 8.693 8.49e-15 ***

#b 0.473891 0.134918 3.512 0.000598 ***

#c 0.119551 0.125430 0.953 0.342167

#---

#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

#Residual standard error: 0.5096 on 140 degrees of freedom

#Multiple R-squared: 0.9639, Adjusted R-squared: 0.9628

#F-statistic: 933.6 on 4 and 140 DF, p-value: < 2.2e-16

Where "weighting" is a variable (function of the variable

`q`

`q2`

`q^2`

Now, to double-check my results, I manually weight my variables by creating new weighted variables :

`mydata$a.wls <- mydata$a * mydata$weighting`

mydata$q.wls <- mydata$q * mydata$weighting

mydata$q2.wls <- mydata$q2 * mydata$weighting

mydata$b.wls <- mydata$b * mydata$weighting

mydata$c.wls <- mydata$c * mydata$weighting

And run the following regression, without the weights option, and without a constant - since the constant is weighted, the column of 1 in the original predictor matrix should now equal the variable weighting:

`Call:`

lm(formula = a.wls ~ 0 + weighting + q.wls + q2.wls + b.wls + c.wls,

data = mydata)

#Residuals:

# Min 1Q Median 3Q Max

#-2.38404 -0.55784 0.01922 0.49838 2.62911

#Coefficients:

# Estimate Std. Error t value Pr(>|t|)

#weighting -4.125559 0.579093 -7.124 5.05e-11 ***

#q.wls 0.217722 0.081851 2.660 0.008726 **

#q2.wls 0.045664 0.006229 7.330 1.67e-11 ***

#b.wls 0.466207 0.121429 3.839 0.000186 ***

#c.wls 0.133522 0.112641 1.185 0.237876

#---

#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

#Residual standard error: 0.915 on 140 degrees of freedom

#Multiple R-squared: 0.9823, Adjusted R-squared: 0.9817

#F-statistic: 1556 on 5 and 140 DF, p-value: < 2.2e-16

As you can see, the results are similar but not identical. Am I doing something wrong while manually weighting the variables, or does the option "weights" do something more than simply multiplying the variables by the weighting vector?

Thanks in advance!

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Answer Source

Provided you do manual weighting correctly, you won't see discrepancy.

So the correct way to go is:

```
X <- model.matrix(~ q + q2 + b + c, mydata) ## non-weighted model matrix (with intercept)
w <- mydata$weighting ## weights
rw <- sqrt(w) ## root weights
y <- mydata$a ## non-weighted response
X_tilde <- rw * X ## weighted model matrix (with intercept)
y_tilde <- rw * y ## weighted response
## remember to drop intercept when using formula
fit_by_wls <- lm(y ~ X - 1, weights = w)
fit_by_ols <- lm(y_tilde ~ X_tilde - 1)
```

Although it is generally recommended to use `lm.fit`

and `lm.wfit`

when passing in matrix directly:

```
matfit_by_wls <- lm.wfit(X, y, w)
matfit_by_ols <- lm.fit(X_tilde, y_tilde)
```

But when using these internal subroutines `lm.fit`

and `lm.fit`

, it is required that all input are complete cases without `NA`

, otherwise the underlying FORTRAN routine `dqrdc.f`

will complain.

If you still want to use the formula interface rather than matrix, you can do the following:

```
## weight by square root of weights, not weights
mydata$root.weighting <- sqrt(mydata$weighting)
mydata$a.wls <- mydata$a * mydata$root.weighting
mydata$q.wls <- mydata$q * mydata$root.weighting
mydata$q2.wls <- mydata$q2 * mydata$root.weighting
mydata$b.wls <- mydata$b * mydata$root.weighting
mydata$c.wls <- mydata$c * mydata$root.weighting
fit_by_wls <- lm(formula = a ~ q + q2 + b + c, data = mydata, weights = weighting)
fit_by_ols <- lm(formula = a.wls ~ 0 + root.weighting + q.wls + q2.wls + b.wls + c.wls,
data = mydata)
```

**Reproducible Example**

Let's use R's built-in data set `trees`

. Use `head(trees)`

to inspect this dataset. There is no `NA`

in this dataset. We aim to fit a model:

```
Height ~ Girth + Volume
```

with some random weights between 1 and 2:

```
set.seed(0); w <- runif(nrow(trees), 1, 2)
```

We fit this model via weighted regression, either by passing weights to `lm`

, or manually transforming data and calling `lm`

with no weigths:

```
X <- model.matrix(~ Girth + Volume, trees) ## non-weighted model matrix (with intercept)
rw <- sqrt(w) ## root weights
y <- trees$Height ## non-weighted response
X_tilde <- rw * X ## weighted model matrix (with intercept)
y_tilde <- rw * y ## weighted response
fit_by_wls <- lm(y ~ X - 1, weights = w)
#Call:
#lm(formula = y ~ X - 1, weights = w)
#Coefficients:
#X(Intercept) XGirth XVolume
# 83.2127 -1.8639 0.5843
fit_by_ols <- lm(y_tilde ~ X_tilde - 1)
#Call:
#lm(formula = y_tilde ~ X_tilde - 1)
#Coefficients:
#X_tilde(Intercept) X_tildeGirth X_tildeVolume
# 83.2127 -1.8639 0.5843
```

So indeed, we see identical results.

Alternatively, we can use `lm.fit`

and `lm.wfit`

:

```
matfit_by_wls <- lm.wfit(X, y, w)
matfit_by_ols <- lm.fit(X_tilde, y_tilde)
```

We can check coefficients by:

```
matfit_by_wls$coefficients
#(Intercept) Girth Volume
# 83.2127455 -1.8639351 0.5843191
matfit_by_ols$coefficients
#(Intercept) Girth Volume
# 83.2127455 -1.8639351 0.5843191
```

Again, results are the same.

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