3
$\begingroup$
mydata <- read.csv("https://stats.idre.ucla.edu/stat/data/binary.csv")
#create a binary variable just for the purpose of experimentation
mydata$bin = rbinom(nrow(mydata), p = 0.7, size = 1)
mydata$rank <- factor(mydata$rank)
mydata$bin <- factor(mydata$bin)
#Run the model with just bin
my.mod <- glm(admit ~ bin, data = mydata, family = "binomial")
summary(my.mod)

Coefficients: Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.8434294 0.1966085 -4.28989 0.000017876
bin1 0.1121304 0.2347628 0.47763 0.63291

The intercept is simply logodds of admit == 1 when bin1 = 0

Check:

x = subset(mydata, bin == "0")
mean(x$admit)
log(0.3008130081/(1-0.3008130081))

Result:
-0.843429383 Tallied

Next, do the same with rank variable alone in the model

my.mod <- glm(admit ~ rank, data = mydata, family = "binomial")
summary(my.mod)

Coefficients: Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.1643031 0.2569384 0.63946 0.522521
rank2 -0.7500300 0.3079693 -2.43540 0.014875
rank3 -1.3646980 0.3353867 -4.06903 0.000047210
rank4 -1.6867296 0.4093073 -4.12094 0.000037733

The intercept is simply logodds of admit == 1 when rank2 = 0 & rank3 = 0 & rank4 = 0; in other words rank == 1

Check:

x = subset(mydata, rank == "1" )
mean(x$admit)
log(0.5409836066/(1-0.5409836066))

Result:
0.1643030515 Tallied

Next, I add both the categorical variables in the model

my.mod <- glm(admit ~ rank + bin, data = mydata, family = "binomial")
summary(my.mod)

Coefficients: Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.23245129 0.32715773 0.71052 0.477383
rank2 -0.75673758 0.30869909 -2.45138 0.014231
rank3 -1.38523274 0.34119770 -4.05991 0.000049091
rank4 -1.69956417 0.41127752 -4.13240 0.000035899
bin1 -0.08308446 0.24667503 -0.33682 0.736255

Question: How to make sense of this intercept like above 2 examples? I don't think in this case intercept equals logodds of admit == 1 when (bin = 0 and rank = 1).

Check:

 x = subset(mydata, rank == "1" & bin == "0" )
 mean(x$admit)
 log(0.6363636364/(1-0.6363636364))

Result: 0.5596157881 Not tallied!

$\endgroup$
1
$\begingroup$

So, you have two binary predictors, say $X$ and $Z$, and the logistic model $$\DeclareMathOperator{\P}{\mathbb{P}} \P(Y=1 \mid X=x) = \frac{e^{\beta_0 + \beta_1 x}}{1+e^{\beta_0 + \beta_1 x}} $$ setting $X=0$ you can see that $\beta_0$ is the log odds of probability that $Y=1$ given $X=0$. Now adding the second predictor in the model: $$ \P(Y=1 \mid X=x,Z=z) = \frac{e^{\beta_0 + \beta_1 x+\beta_2 z}}{1+e^{\beta_0 + \beta_1 x + \beta_2 z}} $$ and you can see that in this model $\beta_0$ is the log odds of the probability that $Y=1$ given both $X=0$ and $Z=0$ (in your code example you look at the conditioning $X=0, Z=1$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ dummy variables for the categorical variable with rank = 1 as the baseline are all zero. bin1 is also zero for binary predictor for bin = 0. This will wipe out all the terms in the linear predictor except for the intercept - β0. Ex - > model.matrix(my.mod)[13,] (Intercept) bin1 rank2 rank3 rank4 1 0 0 0 0 ..I still cannot see the answer, some elaboration would surely help. $\endgroup$ – SrikanthRaja Sep 9 '17 at 8:01
  • $\begingroup$ > mydata[13,] ADMIT GRE GPA RANK bin rank 13 1 760 4 1 0 1 $\endgroup$ – SrikanthRaja Sep 9 '17 at 8:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.