1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

$$y \mapsto y^a$$

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – jbx May 1 '14 at 15:47
  • $\begingroup$ You mean to keep $f(0) = 0.5$? You can adjust $c$ to achieve that. $\endgroup$ – PA6OTA May 1 '14 at 16:00
  • $\begingroup$ 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. $\endgroup$ – jbx May 2 '14 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.