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Consider two logistic functions that are multiplied $$ f(x_1,x_2) = \frac{e^{x_1}}{1+e^{x_1}} \times\frac{e^{x_2}}{1+e^{x_2}}=\frac{e^{x_1+x_2}}{1+e^{x_1+x_2}+e^{x_1}+e^{x_2}} $$

To what extent (or under which non-trivial conditions) can this be approximated by $$ f(x_1,x_2) \approx \frac{e^{x_1+x_2}}{1+e^{x_1+x_2 }} $$ ?

Glorfindel
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Jorge Guzman
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  • You can calculate the error of the approximation explicitly by calculating $r(x_1,x_2) = f(x_1,x_2) - \exp(x_1+x_2)/(1+\exp(x_1+x_2))$. – user403813 Apr 04 '18 at 12:00
  • This is quite useful. I will do some simulations. Thanks! – Jorge Guzman Apr 04 '18 at 12:14
  • When $x_1$ and $x_2$ are both positive (and are large enough), $e^{x_1 + x_2}$ will be much larger than $e^{x_1}$ or $e^{x_2}$, so we can disregard the latter terms. Similarly, when $x_1$ and $x_2$ are both negative (and large enough in absolute value), $1$ will be much larger than $e^{x_1}$ or $e^{x_2}$, so we can disregard the latter terms. – Martin L Apr 04 '18 at 12:45
  • I understand the point when they are both positive and large, but not why it would be the case if they are negative and large. Could you explain? – Jorge Guzman Apr 04 '18 at 13:31

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If $x_1 = z$ and $x_2 = -z$, then it most definitely is not a logistic-shaped curve (it would be bell-shaped). So in that case, it cannot be approximated by such a function.

Gregg H
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