Impact of change in independent variable on probability in logistic regression Take the following logistic regression:
$$\log(o) = \beta_0 + \beta_1 x + \epsilon$$
We know that increasing $x$ by 1 increases the log-odds of the event by $\beta_1$, and multiply the odds of the event by $e^{\beta_1}$.
But what is the impact of an increase of 1 in $x$ on the probability of the event?
 A: As you pointed out, if the model computes the log-odds from $x$ by
$$z(x) = \beta_1x + \beta_0,$$
then, if you increase $x$ by one, the log-odds $z(x)$ are increased by $\beta_1$.
Next, the inverse of the logit function is the logistic sigmoid $\sigma(z)$, so the change $\Delta_1 p$ in probability, when increasing $x$ by one, is
$$
\Delta_1 p = \sigma(z(x) + \beta_1) - \sigma(z(x)).
$$
This difference depends not only on $\beta_1$ but also on $x$, and cannot be simplified.
But what you probably want is the effect of $x$ on $p$, i.e. a factor that you can multiply small values of $\Delta x$ with to obtain $\Delta p$. But since $\sigma(z)$ is nonlinear, you can only compute linear approximations. To do this, first note that the derivative of $\sigma(z)$ is given by:
$$
\sigma^\prime(z) = \sigma(z)(1-\sigma(z)).
$$
Then, the following approximation makes sense if $\beta_1$ is small or if $|z| \gg 1$:
$$
\begin{align}
\frac{\Delta p(x)}{\Delta x} &\approx \sigma^\prime(z(x))\;\beta_1\\
              & = \sigma(z(x))\;(1-\sigma(z(x)))\;\beta_1\\
              &= \sigma(\beta_1x + \beta_0)\;(1-\sigma(\beta_1x + \beta_0))\;\beta_1.
\end{align}
$$
Thus, in summary: For small $\Delta x$, small $\beta_1$, or $|z| \gg 1$, an approximation of the effect $\Delta p/\Delta x$ is $\sigma(\beta_1x + \beta_0)\;(1-\sigma(\beta_1x + \beta_0))\;\beta_1$, but this varies with $x$.
