Symmtery of a function

Question

Find the symmetry in the mean function $\mu$ as a function of $x$, where $$y=\mu(x) = \frac{\exp(\beta_0+\beta_1x)}{1+\exp(\beta_0+\beta_1x)}$$

My approach is to change $x$ to $-x$ but how do I find symmetry?

Answer 1

The function is not symmetric around $x=0$, so changing $x$ to $-x$ will not tell you anything useful. You want to look at $$z=\beta_0+\beta_1 x$$ Also, when in the limit $z\to\infty$, $\mu(z)$ goes to $1$, and in the limit $z\to-\infty$, $\mu(z)$ goes to $0$. A quick example might be able to shed some light:

$\frac{\exp(x-3)}{1+exp(x-3)}$" />

So what you want to show is that $$y(z)-\frac12=-\left(y(-z)-\frac12\right)$$ For that $$\nu(z)=\mu(z)-\frac12=\frac{\exp(z)}{1+\exp(z)}-\frac12=\frac12\frac{\exp(z)-1}{\exp(z)+1}$$ Now multiply and divide by $\exp(-z/2)$, and you get $$\nu(z)=\frac12 \frac{\exp(z/2)-\exp(-z/2)}{\exp(z/2)+\exp(-z/2)}=\frac12 \tanh(z/2)$$ Notice that $\nu(z)$ is antisymmetric.