Understanding glm$residuals and resid(glm)

Question

Can you tell me what is returned by glm$residuals and resid(glm) where glm is a quasipoisson object. e.g. How would I create them using glm$y and glm$linear.predictors.

glm$residuals

     n missing  unique    Mean     .05     .10   .25  .50     .75     .90     .95

 37715   10042    2174 -0.2574 -2.7538 -2.2661 -1.4480 -0.4381  0.7542  1.9845  2.7749



lowest : -4.243 -3.552 -3.509 -3.481 -3.464
highest:  8.195  8.319  8.592  9.089  9.416

resid(glm)

        n    missing     unique       Mean        .05        .10        .25
    37715          0       2048 -2.727e-10    -1.0000    -1.0000    -0.6276
      .50        .75        .90        .95
  -0.2080     0.4106     1.1766     1.7333

lowest : -1.0000 -0.8415 -0.8350 -0.8333 -0.8288
highest:  7.2491  7.6110  7.6486  7.9574 10.1932

Answer 1

Calling resid(model) will default to the deviance residuals, whereas model$resid will give you the working residuals. Because of the link function, there is no single definition of what a model residual is. There are the deviance, working, partial, Pearson, and response residuals. Because these only rely on the mean structure (not the variance), the residuals for the quasipoisson and poisson have the same form. You can take a look at the residuals.glm function for details, but here is an example:

counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
glm.D93 <- glm(counts ~ outcome + treatment, family=quasipoisson())
glm.D93$resid


#working
resid(glm.D93,type="working")
(counts - glm.D93$fitted.values)/exp(glm.D93$linear)

#deviance
resid(glm.D93,type="dev")
fit <- exp(glm.D93$linear)
poisson.dev <- function (y, mu) 
    sqrt(2 * (y * log(ifelse(y == 0, 1, y/mu)) - (y - mu)))
poisson.dev(counts,fit) * ifelse(counts > fit,1,-1)

#response
resid(glm.D93,type="resp")
counts - fit

#pearson
resid(glm.D93,type="pear")
(counts - fit)/sqrt(fit)