# Monotonic transformation to smooth the probabilities

I am studying some event for a set of objects that can be plotted on a square $$[0, 100] ^ 2$$. I have used logistic regression to calculate probabilities that event occur for different objects and the output can be plotted as below:

We can observe a high density around point $$(100, 50)$$ and that basically the farther the less probable the event is. A distance to this point is one of the predictors, but there are also few others. What I am interested in is to capture a potential movement, i.e. how valuable is to move from point $$A$$ to point $$B$$. But I would like to reward also those movements which are made far from the point $$(100, 50)$$, not only those near to the point. And for the former the absolute increase in probability will be small even if a movement is pretty significant (long distance), while for the latter we can observe a large increase in probabilities even with a tiny move. So I think what I would need is to somehow nullify/smooth effect of distance from $$(100, 50)$$, i.e. make it less significant for calculating our movement gain. And I do not really know how to accomplish that. I cannot calculate the percentage gain, because this results in a really large gains. I think I need some monotonic transformation that would make a large probabilities lower and low probabilities higher. Any ideas what I could apply here?

# Edit

@AlexFrancisco, I decided to edit a post to include an image that should help me in explanations. My question is: is there a way to restrict the parameters $$a, b, c, M$$ to such intervals that changing values within those intervals would make $$1 - \mathcal{F}(f(x))$$ different concave functions ranging from identity (blue curve), by slightly concave function (purple curve) to maximally concave function (red curve)?

In other words, I would like to exclude those values of $$a, b, c, M$$ that makes $$1 - \mathcal{F}(f(x))$$ not concave.

• If $f$ is known beforehand, the range of all parameters can be presumably determined as long as $f$ really depends on a certain distance function. But since there seems to be only empirical information about $f$ based on sampling, it's unlikely to determine the range. – Saad Oct 21 '18 at 9:43
• Sad news, but thanks anyway. – jakes Oct 21 '18 at 9:48

Assume that the density $$f$$ is positive on $$[0, 100]^2$$ and there exists a constant $$M > 0$$ such that $$f(x) \leqslant M$$ for any $$x \in [0, 100]^2$$. Two families of prototypical transformations are given first:$$\mathscr{F}_1(f)(x) = \left( \ln\frac{M}{f(x)} \right)^a, \quad \mathscr{F}_2(f)(x) = \left( \left( \frac{M}{f(x)} \right)^b - 1 \right)^a,$$ where $$a, b > 0$$ are parameters. To see the rationale, consider two families of density functions:$$f_1(x) = M \exp(-(d(x, x_0))^{\tfrac{1}{a}}), \quad f_2(x) = \frac{M}{((d(x, x_0))^{\tfrac{1}{a}} + 1)^{\tfrac{1}{b}}},$$ where $$x_0 = (100, 50)$$ and $$d$$ is some distance function. Note that for $$d(x, y) = \sqrt{\smash[b]{(x_1 - y_1)^2 + (x_2 - y_2)^2}}$$, $$a = \dfrac{1}{2}$$, $$b = \dfrac{2}{3}$$, $$f_1$$ and $$f_2$$ are the density functions of bivariate normal distribution and that of bivariate Cauchy distribution, respectively. Solving $$d(x, x_0)$$ from $$f_1$$ and $$f_2$$ yields $$d(x, x_0) = \mathscr{F}_1(f_1)(x)$$ and $$d(x, x_0) = \mathscr{F}_2(f_2)(x)$$, respectively.
Now, since it may be unpractical to assume that $$f$$ is positive, or $$f$$ may be almost $$0$$ at points far from $$x_0$$, then it is useful to introduce another parameter $$c > 0$$, i.e.$$\mathscr{F}_1(f)(x) = \left( \ln\frac{M + c}{f(x) + c} \right)^a, \quad \mathscr{F}_2(f)(x) = \left( \left( \frac{M + c}{f(x) + c} \right)^b - 1 \right)^a.$$ After normalization,$$\mathscr{F}_1(f)(x) = \left( \frac{\ln\dfrac{M + c}{f(x) + c}}{\ln\dfrac{M + c}{c}} \right)^a, \quad \mathscr{F}_2(f)(x) = \left( \frac{\left( \dfrac{M + c}{f(x) + c} \right)^b - 1}{\left( \dfrac{M + c}{c} \right)^b - 1} \right)^a.$$
• @jakes Since $\mathscr F(f)$ are transformations to recover the “distance,” it would be inevitably decreasing to some sense. But considering $1-\mathscr F(f)$ with suitable $a,b,c$ should give what you need. – Saad Oct 19 '18 at 7:04
• @jakes Assume that $f$ depends only on the distance $d$. Since $d$ itself changes linearly with respect to $d$, transforming $f(d)$ to $d$ would offset the variation of $f$. – Saad Oct 21 '18 at 1:35