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I know I'm missing something in my understanding of logistic regression, and would really appreciate any help.

As far as I understand it, the logistic regression assumes that the probability of a '1' outcome given the inputs, is a linear combination of the inputs, passed through an inverse-logistic function. This is exemplified in the following R code:

#create data:
x1 = rnorm(1000)           # some continuous variables 
x2 = rnorm(1000)
z = 1 + 2*x1 + 3*x2        # linear combination with a bias
pr = 1/(1+exp(-z))         # pass through an inv-logit function
y = pr > 0.5               # take as '1' if probability > 0.5

#now feed it to glm:
df = data.frame(y=y,x1=x1,x2=x2)
glm =glm( y~x1+x2,data=df,family="binomial")

and I get the following error message:

Warning messages: 1: algorithm did not converge 2: fitted probabilities numerically 0 or 1 occurred

I've worked with R for some time now; enough to know that probably I'm the one to blame.. what is happening here?

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The way you simulate your data looks weird to me. If you want, for an alternative more standard way, you can have a look here:… – ocram Dec 25 '12 at 15:31
@ocram: you are right; this is a duplicate question! – user603 Dec 25 '12 at 15:34
I did run an erroneous simulation, as @Stéphane Laurent explained. However, the problem was perfect separation in logistic regression, a problem I was not familiar with, and that I was rather surprised to learn about. – zorbar Dec 26 '12 at 7:42
@zorbar: it was in my response to your question (now deleted). – user603 Dec 26 '12 at 9:23
@user603: I probably missed your response; Thanks anyway – zorbar Dec 26 '12 at 9:53

1 Answer 1

up vote 21 down vote accepted

No. The response variable $y_i$ is a Bernoulli random variable taking value $1$ with probability $pr(i)$.

> set.seed(666)
> x1 = rnorm(1000)           # some continuous variables 
> x2 = rnorm(1000)
> z = 1 + 2*x1 + 3*x2        # linear combination with a bias
> pr = 1/(1+exp(-z))         # pass through an inv-logit function
> y = rbinom(1000,1,pr)      # bernoulli response variable
> #now feed it to glm:
> df = data.frame(y=y,x1=x1,x2=x2)
> glm( y~x1+x2,data=df,family="binomial")

Call:  glm(formula = y ~ x1 + x2, family = "binomial", data = df)

(Intercept)           x1           x2  
     0.9915       2.2731       3.1853  

Degrees of Freedom: 999 Total (i.e. Null);  997 Residual
Null Deviance:      1355 
Residual Deviance: 582.9        AIC: 588.9 
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You're right - I've missed this step. thanks a lot for your help! – zorbar Dec 25 '12 at 16:09
I had a question regarding the way you simulate data. When we simulate data for linear regression, we also simulate a noise (\epsilon). I understand that logistic function is a function of the expectation which by itself cancels the noise out. Is that the reason that you don't have any noise in your z ? – Sam Oct 28 '14 at 17:08
@Sepehr Linear regression assumes a Gaussian distribution. The "noise" is just an interpretation of the variability around the mean, but this is not a noise added to the response : the response is written as $\text{mean response} + \text{noise}$, this is just an interpretation. The logistic regression assumes the response has a binomial distribution, and similarly there's no noise added to the response. – Stéphane Laurent Oct 29 '14 at 8:10
@StéphaneLaurent, exactly. I completely get. Thanks very much for your answer. – Sam Nov 1 '14 at 17:53

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