Logistic function

Definition

The logistic function is a function with domain $\mathbb{R}$ and range the open interval $(0,1)$, defined as:

$x\mapsto \frac{1}{1+e^{-x}}$

Equivalently, it can be written as:

$x\mapsto \frac{e^x}{e^x+1}$

Yet another form that is sometimes used, because it makes some aspects of the symmetry more evident, is:

$x\mapsto \frac{e^{x/2}}{e^{x/2}+e^{-x/2}}$

For this page, we will denote the function by the letter $g$.

We may extend the logistic function to a function $[-\mathrm{\infty },\mathrm{\infty }]\to [0,1]$, where $g(-\mathrm{\infty })=0$ and $g(\mathrm{\infty })=1$.

Probabilistic interpretation

The logistic function transforms the logarithm of the odds to the actual probability. Explicitly, given a probability $p$ (strictly between 0 and 1)of an event occurring, the odds in favor of $p$ are given as:

$\frac{p}{1-p}$

This could take any value in $(0,\mathrm{\infty })$

The logarithm of odds is the expression:

$\ln (\frac{p}{1-p})$

If $x$ equals the above expression, then the function describing $p$ in terms of $x$ is the logistic function.

Key data

Item	Value
default domain	all of $\mathbb{R}$, i.e., all reals
range	the open interval $(0,1)$, i.e., the set $\{x\mid 0\le x\le 1\}$
derivative	the derivative is $\frac{-e^{-x}}{(1+e^{-x}{)}^2}$. If we denote the logistic function by the letter $g$, then we can also write the derivative as $g^{\prime }(x)=g(x)g(-x)=g(x)(1-g(x))$
second derivative	If we denote the logistic function by the letter $g$, then we can also write the derivative as $g^{\prime }(x)=g(x)g(-x)=g(x)(1-g(x))(1-2g(x))$
logarithmic derivative	the logarithmic derivative is $\frac{e^{-x}}{1+e^{-x}}$ If we denote the logistic function by $g$, the logarithmic derivative is $g(-x)$
antiderivative	the function $x\mapsto \ln (e^x+1)+C=-\ln (g(-x))+C$
critical points	none
critical points for the derivative (correspond to points of inflection for the function)	$x=0$; the corresponding point on the graph of the function is $(0,1/2)$.
local maximal values and points of attainment	none
local minimum values and points of attainment	none
intervals of interest	increasing and concave up on $(-\mathrm{\infty },0)$ increasing and concave down on $(0,\mathrm{\infty })$
horizontal asymptotes	asymptote at $y=0$ corresponding to the limit for $x\to -\mathrm{\infty }$ asymptote at $y=1$ corresponding to the limit for $x\to \mathrm{\infty }$
inverse function	inverse logistic function or log-odds function given by $x\mapsto \ln (\frac{x}{1-x})$

Differentiation

First derivative

Consider the expression for $g(x)$:

$g(x)=\frac{1}{1+e^{-x}}=(1+e^{-x}{)}^{-1}$

We can differentiate this using the chain rule for differentiation (the inner function being $x\mapsto 1+e^{-x}$ and the outer function being the reciprocal function $t\mapsto 1/t$. We get:

$g^{\prime }(x)=-(1+e^{-x}{)}^{-2}(-e^{-x})$

Simplifying, we get:

$g^{\prime }(x)=\frac{e^{-x}}{(1+e^{-x}{)}^2}$

We can write this in an alternate way that is sometimes more useful. We split the expression as a product:

$g^{\prime }(x)=(\frac{1}{1+e^{-x}})(\frac{e^{-x}}{1+e^{-x}})$

The first factor on the right is $g(x)$, and the second factor is $1-g(x)$, so this simplifies to:

$g^{\prime }(x)=g(x)(1-g(x))$

Functional equations

Symmetry equation

The logistic function $g$ has the property that its graph $y=g(x)$ has symmetry about the point $(0,1/2)$. Explicitly, it satisfies the functional equation:

$g(x)+g(-x)=1$

We can see this algebraically:

$g(-x)=\frac{1}{1+e^{-(-x)}}=\frac{1}{1+e^x}$

Multiply numerator and denominator by $e^{-x}$, and get:

$g(-x)=\frac{e^{-x}}{e^{-x}+1}=1-\frac{1}{1+e^{-x}}=1-g(x)$

Differential equation

As discussed in the #First derivative section, the logistic function satisfies the condition:

$g^{\prime }(x)=g(x)(1-g(x))$

Therefore, $y=g(x)$ is a solution to the autonomous differential equation:

$\frac{dy}{dx}=y(1-y)$

The general solution to that equation is the function $y=g(x+C)$ where $C\in \mathbb{R}$. The initial condition $y=1/2$ at $x=0$ pinpoints the logistic function uniquely.

Integration

First antiderivative

Direct computation

We have:

$\int g(x)dx=\int \frac{1dx}{1+e^{-x}}=\int \frac{e^{x}dx}{e^x+1}=\ln (e^x+1)+C$

=Computation in terms of functional equations for the logistic function

We have:

$g^{\prime }(x)=g(x)(1-g(x))$

We also have that $1-g(x)=g(-x)$, so we get:

$g^{\prime }(x)=g(x)g(-x)$

This can be rewritten as:

$\frac{d}{dx}(\ln (g(x))=g(-x)$

By the chain rule for differentiation, we get:

$\frac{d}{dx}(\ln (g(-x))=-g(x)$

Thus:

$\int g(x)dx=-\ln (g(-x))+C$

This can be simplified and verified to be the same as the answer obtained by direct computation.