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I'm a bit confused about odds ratio in logistic regression (LR). In a logistic regression textbook (pdf), it says that the odds ratio (OR) is $OR = \exp(\beta_0 + \beta_1X)$. But, I don't understand how the $\exp(\beta)$ is linked with the odds ratio.

This is my intuition about LR. The equation of the LR is: \begin{align} \ln\bigg(\frac{p}{1-p}\bigg) &= \beta_0 + \beta_1X \\[5pt] &\quad\text{or} \\[5pt] \frac{p}{1-p} &= \exp(\beta_0 + \beta_1X) \end{align} Suppose that $p$ is the probability of winning a lottery, $\frac{p}{1-p}$ is then the odds of winning the lottery. Therefore, ${\rm Odds}_{{\rm win}} = \exp(\beta_0 + \beta_1X)$. So can anyone explain how it is then linked to the OR?

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  • $\begingroup$ What "logistic regression textbook" says this? Can you quote the passage? $\endgroup$ – gung Mar 17 '17 at 17:58
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    $\begingroup$ Hi, I was looking at a few places, for example, web.pdx.edu/~newsomj/da2/ho_logistic.pdf. It says "The odds ratio is equal to exp(B)," $\endgroup$ – user1480478 Mar 17 '17 at 18:11
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You are right. If the book said that, it is wrong. I do wonder if it is a typo or a poorly phrased passage that lends itself to misunderstanding, though. As you show, $\exp(\beta_0 + \beta_1X)$ is the odds of 'success' predicted by the model. The odds ratio associated with a $1$-unit change in $X$ is $\exp(\beta)$. You can also think of is as the factor by which you would multiply the odds of 'success' associated with $x_i$ to get the odds associated with $(x_i + 1)$ (these are the same, phrased differently).

Since $p$ is the probability of success, $p/(1-p)$ is the odds of success by definition. Imagine $X=0$, then the right hand side simplifies to $\exp(\beta_0)$, but the left hand side is unchanged. Thus we can see that $\exp(\beta_0)$ is the odds of success when $X=0$. Now imagine we move to where $X=1$, then we can rewrite the RHS as $\exp(\beta_0)\times\exp(\beta_1)$ using the rules for exponentiation. Thus we can see that $\exp(\beta_1)$ is the factor by which you would multiply the odds of success to determine the odds of success associated with a $1$-unit increase in $X$.

I think the pdf you link to is sloppily written. At one point it says, "Odds Ratio. The odds ratio is equal to exp(B)", but there are no subscripts, whereas above that the intercept and the coefficient on X are differentiated by subscripts. That sets up the reader for a misunderstanding here.

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  • $\begingroup$ Aha, I think it is my wrong interpretation. I thought $\exp(B)$ is for the whole term i.e. $\beta_0 + \beta_1X$. So based on what you answer, the result of the equation is the odds and the exponential of the coefficient is the OR. Is this correct? $\endgroup$ – user1480478 Mar 17 '17 at 18:15
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    $\begingroup$ @user1480478, that's the gist of it: $\exp(\beta)$ is the odds ratio when holding all else constant. $\endgroup$ – gung Mar 17 '17 at 18:18
  • $\begingroup$ Thank you so much I totally got it now from what you edited. Say odds of success given X=0 is $\exp(\beta)$ and the odds of success given X = 1 is $\exp(\beta) * \exp(\beta_1)$. Then the odds ratio of success to failure given the factor X is $\frac{\exp(\beta) * \exp(\beta_1)}{\exp(\beta)}$ which is equal $ \exp(\beta_1)$. Now that is my linkage from odds to odds ratio! $\endgroup$ – user1480478 Mar 17 '17 at 18:35
  • $\begingroup$ You're welcome, @user1480478. $\endgroup$ – gung Mar 17 '17 at 18:38

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