Another view of Sigmoid function

Sigmoid function (standard logistic function) is defined as

\(\begin{align} \sigma (x) = \frac{1}{1 + e^{-x}}. \end{align}\)

Now, let’s show \(1- \sigma (x) = \sigma(-x)\).

\(\begin{align} 1 - \frac{1}{1 + e^{-x}} &= \frac{1 + e^{-x} - 1}{1 + e^{-x}}\\[10pt] &= \frac{e^{-x} \cdot e^{x}}{(1 + e^{-x})e^{x}}\\[10pt] &= \frac{1}{1 + e^{x}} \end{align}\)

Code for the figure:

sigmoid <- function(x){
	return (1/(1+exp(-x)))
}
x <- as.matrix(seq(-6, 6, 0.01))
y <- apply(x, 2, sigmoid)
plot(x, y, type='l')
abline(v = 0)
lines(x=c(-2, 0), y=c(sigmoid(-2), sigmoid(-2)), lty=2,col=2,lwd=2)
lines(x=c(-2,-2), y=c(0, sigmoid(-2)), lty=2,col=2,lwd=2)
lines(x=c(2, 0), y=c(sigmoid(2), sigmoid(2)), lty=2,col=4,lwd=2)
lines(x=c(2,2), y=c(0, sigmoid(2)), lty=2,col=4,lwd=2)
text(-0.5, sigmoid(2), expression(sigma(2)))
text(1.25, sigmoid(-2), expression(sigma(-2)==1-sigma(2)))

Note:

\(\begin{align} {\rm logit}(p) &= \log \left( \frac{p}{1-p} \right) = a \\[10pt] {\rm logit}^{-1}(a) &= \frac{\exp (a)}{\exp (a) + 1} = \frac{e^a \cdot e^{-a}}{(e^a + 1) \cdot e^{-a}} = \frac{1}{1 + e^{-a}} = p \end{align}\)