1

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?

Andrei
  • 37,370
Proababilty
  • 141

1 Answers1

2

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:

Graph for <span class=$\frac{\exp(x-3)}{1+exp(x-3)}$" />

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.

Someone
  • 136
Andrei
  • 37,370
  • Thank you, i understand it but i have a quick one to ask. Why are we using y(z)-1/2. where does that 1/2 come from. – Proababilty Oct 13 '20 at 00:03
  • 1
    Also, when you are calculating v(z), the final expression should have a 2 in the denominator as well. – Proababilty Oct 13 '20 at 00:08
  • The $y(z)-1/2$ is coming from the fact that along the vertical axis the function varies between $0$ and $1$. That would not be a problem if the function is even, but for odd functions you need to get to $-1$ as well. So from the graph of the function, I can shift to the middle ($1/2$). As for the final expression, I just forgot the 1/2. I will fix that. – Andrei Oct 13 '20 at 02:52