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$X$ is random variable with the domain $(x_0,x_1)$. Under what conditions, can the function:

$$\log\left(\frac{1+e^{a+bx}}{1+e^{a+c+bx}}\right)$$ be approximately linear in $x$ (i.e., $k_0+k_1 x$)? $a, b, c$ are constant parameters.

Thanks,

InsideOut
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Vincent
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  • Any differentiable function is approximately linear in any neighborhood of any interior point in the domain, with a suitable definition of "approximately linear". – lisyarus Aug 08 '19 at 18:45
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    A professor of mine used to say, "An approximation of a number is any other number." The same is true for functions. The question is, how are you measuring whether a function is a good estimator? – Thomas Andrews Aug 08 '19 at 18:46

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Its a sufficiently smooth function, so you can just calculate its Taylor series: $$ f(x) = f(x_*) + f'(x_*)(x-x_*) + \mathcal{O}(x^2). $$ where $x_* \in (x_0, x_1)$. You should calculate it for some arbitrary $x_*$ in your domain. As an example, Wolfram Alpha returns the following Taylor series for $x = 0$: $$ \log \left(\frac{1+e^{a+b x}}{1+e^{a+c+b x}}\right) \approx \log \left(\frac{e^{a}+1}{e^{a+c}+1}\right)-\left[\frac{e^{a} b\left(e^{c}-1\right)}{\left(e^{a}+1\right)\left(e^{a+c}+1\right)} \right] x$$ So your original function is "well approximated" by a linear function so long as $x \approx 0$.

talrefae
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