# Converting logistic regression coefficient and confidence interval from log-odds scale to probability scale

I am trying to figure out whether exp or expit should be used when converting the regression coefficient from logistic regression.

This paper said that:

the exponential function of the regression coefficient (e^b1) is the odds ratio associated with a one-unit increase in the exposure.

While this webpage said that:

We can also transform the log of the odds back to a probability:

$p = \exp(-1.12546)/(1+\exp(-1.12546)) = .245$, if we like.

The webpage used the inverse of logit function (i.e. expit) but not the exponential function for converting the log-odds.

Which one is correct? Or which one is better?

• The first one looks at converting coefficients to odds ratios, while the latter looks at transforming odds to probabilities. An odds ratio is not the same as an odds. May 6, 2016 at 9:11
• And do not forget that the coefficient for the intercept is log(odds) whereas all the others are log(odds ratio)s. May 6, 2016 at 11:02
• Odds ratio: $\frac{\frac{\pi_0}{1-\pi_0}}{\frac{\pi_3}{1-\pi_3}}$. Odds : $\frac{\pi_0}{1-\pi_0}$. May 6, 2016 at 13:56

1. Logistic regression without explanatory variables:

$\color{red}{\text{log}} \,\left[\color{blue}{\text{ODDS(p(Y=1))}}\right]=\color{red}{\text{log}}\left(\frac{\hat p\,(Y=1)}{1-\hat p\,(Y=1)}\right) = \hat\beta_o$

$\hat\beta_o$ is the estimated log odds.

It is an intercept only construct. Exponentiating we get

$$\color{blue}{\text{ODDS(Y=1)}} = \frac{p\,(Y=1)}{1\,-\,p\,(Y=1)} = e^{\,\hat\beta_0}$$

$\color{blue}{\large e^{\hat\beta_o}}$ are the $\color{blue}{\text{ODDS}}$.

Translating into probabilities:

$\color{green}{\Pr(Y = 1)} = \frac{\color{blue}{\text{odds(Y=1)}}}{1\,+\,\color{blue}{\text{odds(Y=1)}}}=\frac{e^{\,\hat\beta_0 }}{1 \,+\,e^{\,\hat\beta_0}}$

This is the second calculation in the OP (i.e. the one containing $-1.12546$).

2. Logistic regression with explanatory variable:

$\color{red}{\text{log}} \,\left[\color{blue}{\text{ODDS(p(Y=1))}}\right]=\color{red}{\text{log}}\left(\frac{\hat p\,(Y=1)}{1-\hat p\,(Y=1)}\right) = \hat\beta_o+\hat\beta_1x_1$

$\color{blue}{\text{ODDS(Y=1)}} = \frac{p\,(Y=1)}{1\,-\,p\,(Y=1)} = e^{\,\hat\beta_0+\hat\beta_1x_1} \tag{*}$

Introducing the odds ratio:

If instead of $x_1$ in $(*)$ we have $x_1+1$ - a one-unit increase:

$$\color{blue}{\text{ODDS(Y=1)}} = \frac{p\,(Y=1)}{1\,-\,p\,(Y=1)} = e^{\,\hat\beta_0+\hat\beta_1x_1+ \hat \beta_1}$$

and

$$\color{green}{\text{ODDS RATIO}} = \frac{\color{blue}{\text{odds|}x_1+1}} {\color{blue}{\text{odds|}x_1}}= \frac{e^{\,\hat\beta_0+\hat\beta_1x_1+ \hat \beta_1}}{e^{\,\hat\beta_0+\hat\beta_1x_1}}= e^{\hat\beta_1}$$

$\color{green}{\large e^{\hat\beta_1}}$ is the $\color{green}{\text{ODDS RATIO}}$.

This is the first calculation in the OP.

For every unit increase in $x_1$ the odds increased by $e^{\hat\beta_1}$.

Hence,

$\color{red}{\log}\,[\color{green}{\text{ODDS RATIO}}] = \hat\beta_1$