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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?

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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.

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  • $\begingroup$ 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' Oct 7 '15 at 6:19
  • $\begingroup$ Sorry ran out of editing time, so things are a wee bit wonky. Thanks for looking @deepNorth $\endgroup$ – fox' Oct 7 '15 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 Oct 7 '15 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' Oct 7 '15 at 6:31
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    $\begingroup$ 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 $\exp(\beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k) = \exp(\ln(\frac{p}{1-p})) = \frac{p}{1-p}$ $\endgroup$ – Maarten Buis Oct 7 '15 at 7:57

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