Type of equation that has the property that $g(z) = 1 - g(-z)$

Question

I am currently working my way through logistic regression, which gives $g(z) = \frac{1}{1+e^{-z}}$ and it also says that $g(-z) = 1 - g(z)$. I understand how to manipulate the equations for this to be true. However, I was wondering if these types of functions have a name and whether or not it is simple to spot functions where this will be true.

Answer 1

Note that if $g(z)=1-g(-z)$ then $$g(z)-\frac{1}{2}=-\left(g(-z)-\frac{1}{2}\right)$$

If $h(z)$ is any odd function, then:

$$g(z)=h(z)+\frac{1}{2}$$

Has the property that $$g(z)=h(z)+\frac{1}{2}=-h(-z)+\frac{1}{2}=1-\left(h(-z)+\frac{1}{2}\right)=1-g(-z)$$

For nice functions (like analytic functions) this means that:

$$g(z)=\frac{1}{2}+zf(z^2)$$ for some nice $f$.

For your particular $g(z)=\frac{1}{1+e^{-z}}$, you get:

$$g(z)-\frac{1}{2}=\tanh \frac{z}{2}$$

Answer 2

Any odd function, that is, one that satisfies $f(-x) = -f(x)$, can be shifted to have the property you describe by defining $g(x) = f(x) + \frac{1}{2}$. So there is not much need to have a new term for functions with this property, since they are in a one-to-one correspondence with odd functions.