3
$\begingroup$

Studying logistic regression I have come across this table / passage:

table text

I have seen a few questions answered on stack exchange addressing the interpretation of a specific intercept from a specific logistic regression, but I have not seen any answers that could clarify the meaning of this text for me.

How does the intercept adjust the average fitted probabilities to the proportion of ones in the data?

$\endgroup$

1 Answer 1

2
$\begingroup$

One way that I like to think about the interpretation of logistic regression is the following.

An assumption of logistic regression is that $ \ln(\frac{p}{1-p}) = \alpha + \beta_1 x_1 + \cdots + \beta_p x_p $ where $ p $ is the parameter your response Bernoulli random variable.

This implies that

$$ \frac{p}{1-p} = \exp(\alpha + \beta_1 x_1 + \cdots + \beta_p x_p) = \exp(a) \exp(\beta_1 x_1 + \cdots + \beta_p x_p) $$

So a nice interpretation (for me at least) is that the odds of $ Y = 1 $ (where $ Y $ is your response variable) has a "baseline" of $ \exp(a) $, and the variables $ x_1, \cdots, x_p $ affect the odds of $Y $ occurring multiplicatively.

$\endgroup$
3
  • $\begingroup$ I think I understand. So the $\exp{\{\alpha\}}$ represents the probabilities that $Y=1$ that are not part of the subset of probabilities conditioned on the predictors $x_i$. Does that sound correct? $\endgroup$
    – Hanzy
    Commented Jul 18, 2018 at 0:42
  • $\begingroup$ Yeah, pretty much. It's like what you should expect the odds to be before including relevant effects. $\endgroup$
    – Kevin Li
    Commented Jul 18, 2018 at 2:01
  • $\begingroup$ To be exact, this `baseline' scenario is essentially the case when $x_1 = x_2 = \cdots = x_p =0.$ $\endgroup$
    – XGS
    Commented Aug 31, 2022 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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