Alright so I am trying to sort out exactly what a logit is in terms of a ratio...

I understand that :

$logit(p) = log(\frac{p}{1-p}) = \beta$

and that

$exp(\beta) =$ odds ratio =$ \frac{\frac{p_1}{1-p_1}}{\frac{p_2}{1-p_2}}$

I guess what's not coming across is how $\beta$, not being a ratio of odds, converts to the odds ratio metric, when taken out of logarithmic space.

To provide a bit more, if this is the logistic regression equation for the constant

$log(\frac{p}{1-p}) = \beta + \beta_1*0 + \beta_2*0 + \beta_3*0 + \beta_4*0 + \beta_5*0$.

then $exp(\beta)$ = odds ratio

so likewise for

$log(\frac{p}{1-p}) = \beta + \beta_1*1$

then $exp(\beta+\beta_1*1)$ = odds ratio for a one unit increase in $\beta_1$

BUT, how does

$$exp(\beta+\beta_1*1) = \frac{\frac{p_1}{1-p_1}}{\frac{p_2}{1-p_2}}\tag{1}$$

because doesn't the $+$ in log terms (e.g. $\beta+\beta_1*1$) equal a multiplication and not a division?


1 Answer 1


The misunderstanding is in (1). In fact $exp(\beta+\beta_1*1)\neq \frac{\frac{p_1}{1-p_1}}{\frac{p_2}{1-p_2}}$

You already know $log(\frac{p_1}{1-p_1}) = \beta + \beta_1*1$ then

$exp(\beta+\beta_1*1)=\frac{p_1}{1-p_1}$ it is an odds not an odds ratio.

while the $exp(\beta_1)$ itself indeed is an odds ration, since $OR=\frac{p_1}{1-p_1}/\frac{p_0}{1-p_0}=\frac{exp(\beta+\beta_1*1)}{exp(\beta+\beta_1*0)}=e^{\beta+\beta_1-\beta-0*\beta_1}=exp{(\beta_1)}$ suppose you have a binary(dummy) predictor variable or when it is a continuous variable you are talking about one unit change of the variable..

Also note $exp(\beta) =$ odds ratio =$ \frac{\frac{p_1}{1-p_1}}{\frac{p_2}{1-p_2}}$ is not correct either.

  • $\begingroup$ Thank you for replying. So, exp(β1) is an odds ratio. and β1 is log odds. $OR=\frac{p_1}{1-p_1}/\frac{p_0}{1-p_0}=\frac{exp(\beta+\beta_1*1)}{exp(\beta+\beta_1*0)}=exp{(\beta_1)}$ If you take the log of both sides of this equation, are you left with $\frac{\beta+\beta_1*1}{\beta+\beta_1*0}=\beta_1$ $\endgroup$
    – fox'
    Commented Oct 7, 2015 at 6:19
  • $\begingroup$ Sorry ran out of editing time, so things are a wee bit wonky. Thanks for looking @deepNorth $\endgroup$
    – fox'
    Commented Oct 7, 2015 at 6:27
  • $\begingroup$ $\beta_1$ is not log odds. You just think this way, suppose the mode is $log(\frac{p}{1-p}) = \beta + \beta_1*x$ and $x=1$ or $x=0$ when $x=1$ you can calculate the odds for $x=1$ from the model and when $x=0$ you can calculate odds for $x=0$ then you compare the two odds you get odds ratio. $\endgroup$
    – Deep North
    Commented Oct 7, 2015 at 6:27
  • $\begingroup$ i see. so the logit does not need to be a ratio of odds, but when you take the exp(logit) it does represent the ratio of odds for that particular variable. $\endgroup$
    – fox'
    Commented Oct 7, 2015 at 6:31
  • 1
    $\begingroup$ yes, the log odds ratio is the logarithm of the odds ratio, wich is $\beta_1$. If you take the exponential of the logarithm of the odds ratio, then you end up with the odds ratio. The log odds is not $\beta_1$, but $\beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k = \ln(\frac{p}{1-p})$. The odds is $\exp(\beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k) = \exp(\ln(\frac{p}{1-p})) = \frac{p}{1-p}$ $\endgroup$ Commented Oct 7, 2015 at 7:57

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.