I'm looking for a way to specify the value of a predictor variable. When I run a glm with my current data, the coefficient for one of my variables is close to one. I'd like to set it at .8.
I know this will give me a lower R^2 value, but I know a priori that the predictive power of the model will be greater.
The weights component of glm looks promising, but I haven't figured it out yet.
Any help would be greatly appreciated.
I believe you are looking for the offset argument in glm. So for example, you might do something like this:
glm(y ~ x1, offset = x2,...)
where in this case the coefficient of x2 would be set at 1. In your case, you may perhaps want to multiply that column by 0.8?
To expand, here is what ?glm says about the offset argument:
this can be used to specify an a priori known component to be included
in the linear predictor during fitting. This should be NULL or a
numeric vector of length equal to the number of cases. One or more
offset terms can be included in the formula instead or as well, and if
more than one is specified their sum is used. See model.offset.
So you can add offsets in the model formula itself using the offset() function, as well. Here is a simple example illustrating its use:
set.seed(123)
d <- data.frame(y = factor(sample(0:1,size = 100,replace = TRUE)),x1 = runif(100),x2 = runif(100))
glm1 <- glm(y~x1+x2,data = d,family = binomial)
coef(glm1)
(Intercept) x1 x2
0.4307718 -0.4128541 -0.6994810
glm2 <- glm(y~x1,data = d,offset = x2,family = binomial)
coef(glm2)
(Intercept) x1
-0.4963699 -0.2185571
glm3 <- glm(y~x1+offset(x2),data = d,family = binomial)
coef(glm3)
(Intercept) x1
-0.4963699 -0.2185571
Note that the last two have the same coefficients.
Related
I am trying to get a confidence interval for my response in a poisson regression model. Here is my data:
X <- c(1,0,2,0,3,1,0,1,2,0)
Y <- c(16,9,17,12,22,13,8,15,19,11)
What I've done so far:
(i) read my data
(ii) fit a Y by poisson regression with X as a covariate
model <- glm(Y ~ X, family = "poisson", data = mydata)
(iii) use predict()
predict(model,newdata=data.frame(X=4),se.fit=TRUE,interval="confidence",level=0.95, type = "response")
I was expecting to get "fit, lwr, upr" for my response but I got the following instead:
$fit
1
30.21439
$se.fit
1
6.984273
$residual.scale
[1] 1
Could anyone offer some suggestions? I am new to R and struggling with this problem for a long time.
Thank you very much.
First, the function predict() that you are using is the method predict.glm(). If you look at its help file, it does not even have arguments 'interval' or 'level'. It doesn't flag them as erroneous because predict.glm() has the (in)famous ... argument, that absorbs all 'extra' arguments. You can write confidence=34.2 and interval="woohoo" and it still gives the same answer. It only produces the estimate and the standard error.
Second, one COULD then take the fit +/- 2*se to get an approximate 95 percent confidence interval. However, without getting into the weeds of confidence intervals, pivotal statistics, non-normality in the response scale, etc., this doesn't give very satisfying intervals because, for instance, they often include impossible negative values.
So, I think a better approach is to form an interval in the link scale, then transform it (this is still an approximation, but probably better):
X <- c(1,0,2,0,3,1,0,1,2,0)
Y <- c(16,9,17,12,22,13,8,15,19,11)
model <- glm(Y ~ X, family = "poisson")
tmp <- predict(model, newdata=data.frame(X=4),se.fit=TRUE, type = "link")
exp(tmp$fit - 2*tmp$se.fit)
1
19.02976
exp(tmp$fit + 2*tmp$se.fit)
1
47.97273
I was wondering if it might be possible (and perhaps recommended) to obtain standardized coefficients from stan_glm() in the rstanarm package? (did not find anything specific in the documentation)
Can I just standardize all variables as in normal regression? (see below)
Example:
library("rstanarm")
fit <- stan_glm(wt ~ vs*gear, data = mtcars)
Standardization:
design <- wt ~ vs*gear
vars <- all.vars(design)
stand.vars <- lapply(mtcars[, vars], scale)
fit <- stan_glm(stand.vars, data = mtcars)
I would not say that it is affirmatively recommended, but I would recommend that you not subtract the sample mean and divide by the sample standard deviation of the outcome because the estimation uncertainty in those two statistics will not be propagated to the posterior distribution.
Standardizing the predictors is more debatable. You can do it, but it makes doing posterior prediction with new data harder because you have to remember to subtract the old means from the new data and divide by the old standard deviations.
The most computationally efficient approach is to leave the variables as they are but specify the non-default argument QR = TRUE, especially if you are not going to modify the default (normal) priors on the coefficients anyway.
You can then standardize the posterior coefficients after-the-fact if standardized coefficients are of interest. To do so, you can do
X <- model.matrix(fit)
sd_X <- apply(X, MARGIN = 2, FUN = sd)[-1]
sd_Y <- apply(posterior_predict(fit), MARGIN = 1, FUN = sd)
beta <- as.matrix(fit)[ , 2:ncol(X), drop = FALSE]
b <- sweep(sweep(beta, MARGIN = 2, STATS = sd_X, FUN = `*`),
MARGIN = 1, STATS = sd_Y, FUN = `/`)
summary(b)
However, standardizing regression coefficients just gives the illusion of comparability across variables and says nothing about how germane a one standard deviation difference is, particularly for dummy variables. If your question is really whether manipulating this predictor or that predictor is going to make a bigger difference on the outcome variable, then simply simulate those manipulations like
PPD_0 <- posterior_predict(fit)
nd <- model.frame(fit)
nd[ , 2] <- nd[ , 2] + 1 # for example
PPD_1 <- posterior_predict(fit, newdata = nd)
summary(c(PPD_1 - PPD_0))
and repeat that process for other manipulations of interest.
R's standard way of doing regression on categorical variables is to select one factor level as a reference level and constraining the effect of that level to be zero. Instead of constraining a single level effect to be zero, I'd like to constrain the sum of the coefficients to be zero.
I can hack together coefficient estimates for this manually after fitting the model the standard way:
x <- lm(data = mtcars, mpg ~ factor(cyl))
z <- c(coef(x), "factor(cyl)4" = 0)
y <- mean(z[-1])
z[-1] <- z[-1] - y
z[1] <- z[1] + y
z
## (Intercept) factor(cyl)6 factor(cyl)8 factor(cyl)4
## 20.5021645 -0.7593074 -5.4021645 6.1614719
But that leaves me without standard error estimates for the former reference level that I just added as an explicit effect, and I need to have those as well.
I did some searching and found the constrasts functions, and tried
lm(data = mtcars, mpg ~ C(factor(cyl), contr = contr.sum))
but this still only produces two effect estimates. Is there a way to change which constraint R uses for linear regression on categorical variables properly?
Think I've figured it out. Using contrasts actually is the right way to go about it, you just need to do a little work to get the results into a convenient looking form. Here's the fit:
fit <- lm(data = mtcars, mpg ~ C(factor(cyl), contr = contr.sum))
Then the matrix cs <- contr.sum(factor(cyl)) is used to get the effect estimates and the standard error.
The effect estimates just come from multiplying the contrast matrix by the effect estimates lm spits out, like so:
cs %*% coef(fit)[-1]
The standard error can be calculated using the contrast matrix and the variance-covariance matrix of the coefficients, like so:
diag(cs %*% vcov(fit)[-1,-1] %*% t(cs))
I'd like to use the ols() (ordinary least squares) function from the rms package to do a multivariate linear regression, but I would not like it to calculate the intercept. Using lm() the syntax would be like:
model <- lm(formula = z ~ 0 + x + y, data = myData)
where the 0 stops it from calculating an intercept, and only two coefficients are returned, on for x and the other for y. How do I do this when using ols()?
Trying
model <- ols(formula = z ~ 0 + x + y, data = myData)
did not work, it still returns an intercept and a coefficient each for x and y.
Here is a link to a csv file
It has five columns. For this example, can only use the first three columns:
model <- ols(formula = CorrEn ~ intEn_anti_ncp + intEn_par_ncp, data = ccd)
Thanks!
rms::ols uses rms:::Design instead of model.frame.default. Design is called with the default of intercept = 1, so there is no (obvious) way to specify that there is no intercept. I assume there is a good reason for this, but you can try changing ols using trace.
I have a linear model generated using lm. I use the coeftest function in the package lmtest go test a hypothesis with my desired vcov from the sandwich package. The default null hypothesis is beta = 0. What if I want to test beta = 1, for example. I know I can simply take the estimated coefficient, subtract 1 and divide by the provided standard error to get the t-stat for my hypothesis. However, there must be functionality for this already in R. What is the right way to do this?
MWE:
require(lmtest)
require(sandwich)
set.seed(123)
x = 1:10
y = x + rnorm(10)
mdl = lm(y ~ x)
z = coeftest(mdl, df=Inf, vcov=NeweyWest)
b = z[2,1]
se = z[2,2]
mytstat = (b-1)/se
print(mytstat)
the formally correct way to do this:
require(multcomp)
zed = glht(model=mdl, linfct=matrix(c(0,1), nrow=1, ncol=2), rhs=1, alternative="two.sided", vcov.=NeweyWest)
summary(zed)
Use an offset of -1*x
mdl<-lm(y~x)
mdl2 <- lm(y ~ x-offset(x) )
> mdl
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
0.5255 0.9180
> mdl2
Call:
lm(formula = y ~ x - offset(x))
Coefficients:
(Intercept) x
0.52547 -0.08197
You can look at summary(mdl2) to see the p-value (and it is the same as in mdl.
As far as I know, there is no default function to test the model coefficients against arbitrary value (1 in your case). There is the offset trick presented in the other answer, but it's not that straightforward (and always be careful with such model modifications). So, your expression (b-1)/se is actually a good way to do it.
I have two notes on your code:
You can use summary(mdl) to get the t-test for 0.
You are using lmtest with covariance structure (which will change the t-test values), but your original lm model doesn't have it. Perhaps this could be a problem? Perhaps you should use glm and specify the correlation structure from the start.