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This is a General Linear Models topic but I believe it's just basic failure to remember some more basic math rules that's making it difficult for me.


If the link function is

$$g(\pi) = \log(\frac{\pi}{1-\pi}) = x^T\beta$$

show that this is equivalent to modeling the probability $\pi$ as,

$$\pi = \frac{e^{x^T}\beta}{1+e^{x^T}\beta}$$

Again I think this is a simply inverse. But I'm not quite sure how to do it.

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You're right, it is just the inverse

$$ \ln\frac{\pi}{1 - \pi} = x^T\beta ~~~\Rightarrow~~~ \frac{\pi}{1 - \pi} = e^{x^T\beta} ~~~\Rightarrow~~~ \pi = (1 - \pi)e^{x^T\beta} = e^{x^T\beta} - \pi e^{x^T\beta} $$

Now rearrange a bit things

$$ \pi (1 + e^{x^T\beta}) = e^{x^T\beta} $$

and from here

$$ \pi = \frac{e^{x^T\beta}}{1 + e^{x^T\beta}} $$

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  • $\begingroup$ Thank you kindly! I'm not sure why I kept stopping myself at the part where we multiply the exponential by $(1-\pi)$. I appreciate your time! $\endgroup$ – Nicklovn Feb 4 at 19:24
  • $\begingroup$ @Nicklovn Happy to help $\endgroup$ – caverac Feb 4 at 19:25

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