I need both quadratic and linear coefficients in a GLM with binary response. What's the best option?

Question

I have three predictors and one response. What can I do if my response variable is binary?

Answer 1

You can add a quadratic term with logistic regression just as you can with regular old linear regression. That is a simple way to include a 'curve' in your model. Be sure you understand what that means. I suspect you want an R tutorial, which is off-topic on CV. The basic approach to adding a quadratic in R is to include I(x^2) in the formula. Here is a simple example:

lo.to.p = function(lo){                 # we need this function to generate the data
  odds = exp(lo)
  prob = odds/(1+odds)
  return(prob)
}
set.seed(4649)                          # this makes the example exactly reproducible
x1 = runif(100, min=0, max=10)          # you have 3, largely uncorrelated predictors
x2 = runif(100, min=0, max=10)
x3 = runif(100, min=0, max=10)
lo = -78 + 35*x1 - 3.5*(x1^2) + .1*x2   # there is a quadratic relationship w/ x1, a
p  = lo.to.p(lo)                        #  linear relationship w/ x2 & no relationship
y  = rbinom(100, size=1, prob=p)        #  w/ x3

model = glm(y~x1+I(x1^2)+x2+x3, family=binomial)
summary(model)
# Call:
# glm(formula = y ~ x1 + I(x1^2) + x2 + x3, family = binomial)
# 
# Deviance Residuals: 
#      Min        1Q    Median        3Q       Max  
# -1.74280  -0.00387   0.00000   0.04145   1.74573  
# 
# Coefficients:
#              Estimate Std. Error z value Pr(>|z|)   
# (Intercept) -53.65462   19.65288  -2.730  0.00633 **
# x1           24.78164    8.92910   2.775  0.00551 **
# I(x1^2)      -2.49888    0.89344  -2.797  0.00516 **
# x2            0.03318    0.20198   0.164  0.86952   
# x3           -0.09277    0.18650  -0.497  0.61890   
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
# 
# (Dispersion parameter for binomial family taken to be 1)
# 
#     Null deviance: 128.207  on 99  degrees of freedom
# Residual deviance:  18.647  on 95  degrees of freedom
# AIC: 28.647
# 
# Number of Fisher Scoring iterations: 10