How to Find the Inverse of a Sigmoid Equation

The word “sigmoid” can describe several S-shaped functions, but it most often means the standard logistic function

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

To find its inverse, solve for the input $x$ in terms of the output $y$. The result is the logit function. The steps below also show how to invert a shifted or rescaled logistic sigmoid.

Derive the inverse of the standard sigmoid

Set the sigmoid equal to $y$ and rearrange:

$$\begin{aligned} y &= \frac{1}{1+e^{-x}} \\ 1+e^{-x} &= \frac{1}{y} \\ e^{-x} &= \frac{1-y}{y}. \end{aligned}$$

Take the natural logarithm of both sides. Since $\ln(e^{-x})=-x$, solving for $x$ gives

$$\boxed{x=\ln\left(\frac{y}{1-y}\right)}.$$

Thus the inverse function is

$$\sigma^{-1}(y)=\operatorname{logit}(y)=\ln\left(\frac{y}{1-y}\right).$$

The expression $y/(1-y)$ is the odds associated with probability $y$; the logit is the logarithm of those odds.

Check the domain

The standard sigmoid takes every real input to a value strictly between 0 and 1. Therefore its inverse accepts only $0<y<1$ and returns a real number. The endpoints are approached but never reached for finite $x$:

$$\lim_{x\to-\infty}\sigma(x)=0, \qquad \lim_{x\to\infty}\sigma(x)=1.$$

So $y=0$ and $y=1$ do not have finite inverse values. In the extended limiting sense, their corresponding inputs tend to $-\infty$ and $+\infty$ respectively. Values outside $[0,1]$ cannot be outputs of this sigmoid.

Worked example

Find the input whose sigmoid output is $0.8$. Substitute $y=0.8$ into the inverse:

$$\begin{aligned} x &= \ln\left(\frac{0.8}{1-0.8}\right) \\ &= \ln\left(\frac{0.8}{0.2}\right) \\ &= \ln(4) \\ &\approx 1.3863. \end{aligned}$$

Check by putting this value back into the original sigmoid:

$$\sigma(1.3863)\approx\frac{1}{1+e^{-1.3863}}\approx 0.8.$$

The positive input makes sense: the standard sigmoid is above $0.5$ when $x>0$, and $0.8$ is above $0.5$.

Invert a shifted or scaled sigmoid

A common generalized form is

$$y=\frac{L}{1+e^{-k(x-x_0)}},$$

where $L$ sets the upper output limit, $x_0$ is the midpoint, and $k$ controls the steepness and direction. For $L>0$ and $k\ne 0$, its output must satisfy $0<y<L$. Rearranging gives

$$\begin{aligned} \frac{L}{y}-1 &= e^{-k(x-x_0)}, \\ \frac{L-y}{y} &= e^{-k(x-x_0)}, \\ x &= x_0+\frac{1}{k}\ln\left(\frac{y}{L-y}\right). \end{aligned}$$

For example, suppose $y=6/(1+e^{-2(x-3)})$ and you want the input that gives $y=4$. Here $L=6$, $k=2$, and $x_0=3$:

$$x=3+\frac{1}{2}\ln\left(\frac{4}{6-4}\right)=3+\frac{1}{2}\ln(2)\approx 3.347.$$

If the sigmoid is applied to a linear expression, such as $y=\sigma(ax+b)$, first use the standard inverse to obtain $ax+b=\ln(y/(1-y))$. If $a\ne0$, then

$$x=\frac{\ln(y/(1-y))-b}{a}.$$

Common mistakes

• Reversing the odds: The numerator in the logit is $y$, and the denominator is $1-y$. Reversing them changes the sign of the answer.
• Using the formula at an endpoint: The logarithm is not finite at $y=0$ or $y=1$. Do not substitute either endpoint into the formula as though it returned a real number.
• Ignoring the sigmoid’s parameters: For a scaled sigmoid, use $y/L$ or the generalized formula; the unscaled logit alone will give the wrong input.
• Assuming every sigmoid has a logit inverse: Some functions called sigmoid functions, such as the normal cumulative distribution function, have different inverse functions. The formula here applies to the logistic sigmoid and its stated variants.

For values extremely close to 0 or 1, the logit can have a very large magnitude. In numerical work, check that the output is valid and account for rounding before evaluating the logarithm.