How can I motivate the derivation of the $p$ below?
From:
$l = log_b\frac{p}{1-p}=\beta_0 + \beta_1x_1+\beta_2x_2 $
$p = \frac{b^{\beta_0 + \beta_1x_1+\beta_2x_2}}{b^{(\beta_0 + \beta_1x_1+\beta_2x_2)} +1}= \frac{1}{1+b^{-( \beta_0 + \beta_1x_1+\beta_2x_2 )}} $
Given that:
$p $= probability of event $Y$=1
$l$ = logit or log-odd of $p$
$x$ = predictors or independent variables
I don't understand how the p above arrived at $ \frac{b^{\beta_0 + \beta_1x_1+\beta_2x_2}}{b^{(\beta_0 + \beta_1x_1+\beta_2x_2)} +1}= \frac{1}{1+b^{-( \beta_0 + \beta_1x_1+\beta_2x_2 )}} $
$\endgroup$2 Answers
$\begingroup$You start by assuming a linear form for the log odds, i.e.,$$ \ln \left( \frac{p}{1-p} \right) = \beta^Tx $$thus, exponentiating both sides you get$$ \frac{p}{1-p} = e^{\beta^Tx} $$re-arranging the equation$$ e^{\beta^Tx} - pe^{\beta^Tx} = p\to p(1+e^{\beta^Tx})=e^{\beta^Tx} $$finally,$$ p = \frac{e^{\beta^Tx}}{1+e^{\beta^Tx}} $$
$\endgroup$ 1 $\begingroup$You've copied the equations from the wikipedia page incorrectly. It should read\begin{align*} p = \dfrac{b^{\beta_0 + \beta_1x_1 + \beta_2x_2}}{b^{\beta_0 + \beta_1x_1 + \beta_2x_2} + 1} \end{align*}where $b$ is the base of the logarithm (e.g. $e$). $b$ is not a parameter nor is it related to $\beta$.
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