# Biasing sigmoid curve

I wish to use the sigmoid function $1-{1\over1+e^{-x+c}}$ to obtain a value from 0 to 1 (to be used for a probability value), where $c$ is a constant. The higher this constant, the lower the probability.

This is going to be used for a monte-carlo search algorithm, where the deeper the search goes the chances of exploring deeper are lowered. This depth is represented by $x$.

Now I wish to be able to dampen or amplify this function according to some fitness value that I have. I still want it to give me a value from 0 to 1, but I wish to bias the curve a bit upwards with respect to another parameter (this fitness value). If this value is high I want to push the curve up, while if it is low I want to push it down (or vice-versa if its simpler). This way the probability of a 'fit' candidate is a bit higher.

You can assume I can normalise my fitness value to whatever is necessary such that it conforms to the required range of values.

How do I adjust this formula to include this fitness value?

This is my idea.

You'd like to make a 1-1 transform of $x$. Originally, $x \mapsto x$. If you'd like to skew this relation, how about a power of $x$?

For example, $x \mapsto x^a$, for $x$ positive. You'll get a range of behaviors.

Generally, $x \mapsto sgn(x) |x|^a$.

EDIT: I realized this is not quite what you asked for. OK, let's transform the function then. With

$$y = 1-{1\over1+e^{-x+c}}$$

apply the transformation

$$y \mapsto y^a$$

• The power approach seems to squash the sigmoid to the left if $a > 1$, and stretch it to the right if $a < 1$. I was thinking more on the lines of changing the S shape such that its skewed upwards, however your idea might have a good effect too. +1 – jbx May 1 '14 at 15:47
• You mean to keep $f(0) = 0.5$? You can adjust $c$ to achieve that. – PA6OTA May 1 '14 at 16:00
• Lets say 'c = 10', which is where I want $y$ to be almost $0$. I was imagining the S shape to skew a little upwards such that $f(5) > 0.5$ (for fit values). But I guess your approach is good too. It still promotes fit values a bit more than the average, and unfit less. – jbx May 2 '14 at 15:22