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 $g^{″}(x)=g^{\prime }(x)(1-2g(x))=g(x)(1-g(x))(1-2g(x))=g(x)g(-x)(1-2g(x))=g(x)g(-x)(g(-x)-g(x))$
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:

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

Second derivative

Using the expression $g(x)(1-g(x))$ for $g^{\prime }$

From the above, we have:

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

Differentiating both sides, we obtain:

$g^{″}(x)=g^{\prime }(x)-2g(x)g^{\prime }(x)$

This simplifies to:

$g^{″}(x)=g^{\prime }(x)(1-2g(x))$

We can now re-use the expression for $g^{\prime }$ and obtain:

$g^{″}(x)=g(x)(1-g(x))(1-2g(x))$

Using the expression $g(x)g(-x)$ for $g^{\prime }$

We have:

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

Using the product rule for differentiation and the chain rule for differentiation, we get:

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

Note from the expression that $g^{\prime }$ shows that $g^{\prime }$ is even, so we can rewrite $g^{\prime }(-x)$ as $g^{\prime }(x)$, and we get:

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

We can re-use the expression $g^{\prime }(x)=g(x)g(-x)$ and obtain:

$g^{″}(x)=g(x)g(-x)(g(-x)-g(x))$

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.

Points and intervals of interest

Critical points

The function has no critical points. To see this, note that the derivative is:

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

Note that the numerator is never zero, nor is the denominator. Therefore, the function is always defined and nonzero.

Intervals of increase and decrease

The derivative:

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

is always positive. So the function is increasing on all of $\mathbb{R}$.

The asymptotic values are:

$\underset{x\to \mathrm{\infty }}{lim}\frac{1}{1+e^{-x}}=\frac{1}{1}=1$

and:

$\underset{x\to -\mathrm{\infty }}{lim}\frac{1}{1+e^{-x}}=\frac{1}{\to \mathrm{\infty }}=0$

In other words, the range of the function is the open interval $(0,1)$, and it increases throughout its domain.

Points of inflection

The second derivative is:

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

We already noted that $g^{\prime }(x)$ is always defined and nonzero, so the only way for $g^{″}(x)$ to be zero is if $1-2g(x)=0$< or $g(x)=1/2$. This solves to:

$\frac{1}{1+e^{-x}}=\frac{1}{2}$

This solves to $e^{-x}=1$, or $x=0$.

Thus, the second derivative is 0 at the point $(0,1/2)$, i.e., with $x=0$ and $g(x)=1/2$.

Intervals of concave up and down

As above, we have:

$g^{″}(x)=g^{\prime }(x)(1-2g(x))$

We also noted that $g^{\prime }(x)>0$ for all $x$. Therefore, $g^{″}(x)>0$ for $x<0$ and $g^{″}(x)<0$ for $x>0$. Therefore, $g$ is:

• concave up for $x<0$, i.e., $x\in (-\mathrm{\infty },0)$
• concave down for $x>0$, i.e., $x\in (0,\mathrm{\infty })$

Symmetry

We discussed above a functional equation satisfied by $g$:

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

From this, the following can be deduced:

• The graph of $g$ has half-turn symmetry about the point $(0,1/2)$.
• $g^{\prime }$ is an even function. Note that this can also be seen from the actual expression: $g^{\prime }(x)=g(x)g(-x)$. But we don't need the actual expression to deduce that it is even -- the functional equation above gives evenness.
• $g^{″}$ is an odd function. This can be directly deduced from $g^{\prime }$ being even, but can also be verified from the actual expression: $g^{″}(x)=g^{\prime }(x)(1-2g(x))=g^{\prime }(x)(g(-x)-g(x))$.

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.