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Quick formulation of the problem:
Given two points: $(x_l, y_l)$ and $(x_u, y_u)$
with: $x_l < x_u$ and $y_l < y_u$,
and given lower asymptote=0 and higher asymptote=1, what's the logistic function that passes through the two points?

Explanatory image:
logistic

Other details:
I'm given two points in the form of Pareto 90/10 (green in the example above) or 80/20 (blue in the example above), and I know that the upper bound is one and the lower bound is zero.

How do I get the formula of a sigmoid function (such as the logistic function) that has a lower asymptote on the left and higher asymptote on the right and passes via the two points?

2 Answers 2

9

I believe you're looking for constants $a$ and $b$ so that $f(x_\ell) = y_\ell$ and $f(x_u) = y_u$ where $f(x) = \exp(a + bx) / (1 + \exp(a + bx))$.

This is equivalent to the linear system $a + b x_\ell = g(y_\ell)$ and $a + b x_u = g(y_u)$ where $g(y) = f^{-1}(y) = \log(y/(1-y))$.

0

To elaborate on the accepted answer, if we have a logistic function using the common notation:

$$f(x) = \frac{1}{1 + e^{-k(x-x_0)}}$$

... and we want to solve for $k$ and $x_0$ given two points, $(x_l, y_l)$ and $(x_u, y_u)$:

First we can group the unknowns in a single term $b \equiv k(x-x_0)$. So:

$$y = \frac{1}{1 + e^{-b}}$$ $$y(1 + e^{-b}) = 1$$ $$e^{-b} = \frac{1-y}{y}$$ $$-b = \log(\frac{1-y}{y})$$ $$ b = \log(\frac{y}{1-y})$$

Now we expand b: $$k(x-x_0) = \log(\frac{y}{1-y})$$

... which gives us a linear system to solve for $k$ and $x_0$ given the values of two $(x, y)$ coordinates.