I've been trying to manually get the response values given by the predict.glm function from the stats package in R. However, I'm unable to do so. I only know how to manually get the value with a binomial distribution. I would really appreciate some help. I created two small models (one with Gamma family and one with inverse Gaussian family).
library(stats)
library(dplyr)
data("USArrests")
#Gamma distribution model
model_gam <- glm(Rape~Murder + Assault + UrbanPop, data=USArrests, family=Gamma)
print(summary(model_gam))
responses_gam <- model_gam %>% predict(USArrests[1,], type="response")
print(responses_gam)
#Trying to manually get responses for gamma model
paste(coef(model_gam), names(coef(model_gam)), sep="*", collapse="+")
# "0.108221470842499*(Intercept)+-0.00122165587689519*Murder+-9.47425665022909e-05*Assault+-0.000467789606041651*UrbanPop"
print(USArrests[1,])
#Murder: 13.2, Assault: 236, UrbanPop: 58
x = 0.108221470842499 - 0.00122165587689519 * 13.2 - 9.47425665022909e-05 * 236 - 0.000467789606041651 * 58
# This is wrong. Do I have to include the dispersion? (which is 0.10609)
print (exp(x)/(1+exp(x)))
# result should be (from predict function): 26.02872
# exp(x)/(1+exp(x)) gives: 0.510649
# Gaussian distribution model
model_gaus <- glm(Rape~Murder + Assault + UrbanPop, data=USArrests, family=inverse.gaussian(link="log"))
responses_gaus <- model_gaus %>% predict(USArrests[1,], type="response")
print(summary(model_gaus))
print(responses_gaus)
#Trying to manually get responses for gaussian model
paste(coef(model_gaus), names(coef(model_gaus)), sep="*", collapse="+")
# "0.108221470842499*(Intercept)+-0.00122165587689519*Murder+-9.47425665022909e-05*Assault+-0.000467789606041651*UrbanPop"
x = 1.70049202188329-0.0326196928618521* 13.2 -0.00234379099421488*236-0.00991369000675323*58
# Dispersion in this case is 0.004390825
print(exp(x)/(1+exp(x)))
# result should be (from predict function): 26.02872
# exp(x)/(1+exp(x)) it is: 0.5353866
predict()predict(model_gaus)["Alabama"] ## 3.259201
cat(paste(round(coef(model_gaus),5), names(coef(model_gaus)), sep="*", collapse="+"),"\n")
## 1.70049*(Intercept)+0.03262*Murder+0.00234*Assault+0.00991*UrbanPop
USArrests["Albama",]
## Murder Assault UrbanPop Rape
## Alabama 13.2 236 58 21.2
The value of the intercept is always 1, so we have
1.70049*1+0.03262*13.2+0.00234*236+0.00991*58
## [1] 3.258094
(close enough, since I rounded some things)
You don't need to do anything with the dispersion or the inverse-link function, since the Gaussian model uses an identity link.
Mathematically, the regression equation is defined as X %*% beta where beta is the vector of coefficients and X is the model matrix (for your example, it's a column of ones for the intercept plus your predictors; for models with categorical predictors or more complex terms like splines, it's a little more complicated). You can extract the model matrix from the matrix with model.matrix():
Xg <- model.matrix(model_gaus)
drop(Xg["Alabama",] %*% coef(model_gaus))
For the Gamma model, you would use exactly the same procedure, but at the end you would transform the linear expression you computed (the linear predictor) by 1/x (the inverse link function for the Gamma). (Note that you need predict(..., type = "response") to get the inverse-transformed prediction; otherwise [default type = "link"] R will give you just the plain linear expression.] If you used a log link instead you would exponentiate. More generally,
invlinkfun <- family(fitted_model)$linkinv
X <- model.matrix(fitted_model)
beta <- coef(fitted_model)
invlinkfun(X %*% beta)
The inverse Gaussian model uses a 1/mu^2 link by default; inverse.gaussian()$linkinv is function(eta) { 1/sqrt(eta) }