Can a logistic function(x+y) be approximately factored into $f_1(x)$ and $f_2(x)$?

Question

I need to somehow factor the logistic function

$$\frac{1}{1+e^{-(\theta-\beta)}}$$

into $f_1(\theta)$$f_2(\beta)$ approximately...

namely $\frac{1}{1+e^{-(\theta-\beta)}} \approx f_1(\theta)$$f_2(\beta)$

Do you have any suggestion?

Answer 1

Taking $\theta=\beta$ it should be $$ f_1(\theta)\,f_2(\theta)\approx\frac12. $$ From here we get $$ \frac{1}{1+e^{-(\theta-\beta)}}\approx\frac12\,\frac{1+e^{-\beta}}{1+e^{-\theta}}. $$ It is not a good approximation.

You can see that there is not a solution to the problem writing $$ \frac{1}{f_2(\beta)}\approx\bigl(1+e^{-(\theta-\beta)}\bigr)\,f_1(\theta). $$ The left hand side is independent of $\theta$, while the right hand side is not, no matter the choice of $f_1$.

Answer 2

@Julian Aguirre, Thank you for the fast answer. I never expected it to be so fast! I am writing this way because I want to post a picture,

For $\beta < -2 $, the difference is huge!

I am going to use the function for a restricted rage $ -3 < \theta < 3, -3 < \beta < 3$. Given this condition, will there be any better function?

Your approx. is the exact function when $\beta=\theta$. So if we give up that property so the approximate function can be slightly different from the actual function when, let's say, $\theta=2, \beta=2$, we might get better approximations... I think, I hope... :)