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Monotonic transformation to smooth the probabilities



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Thinking: Why equivalent percentage increase of A and decrease of B is not the same end result?Increase difficulty level based on probabilitiesCalculating probability based variables?Looking for peculiar vector transformationHow to find Fantasy Football Playoff ProbabilitiesProbabilities of survival vs. probabilities of being sampledSomething with a probability of 1% occurring 100 timesSicherman versus standard 2d6, playing backgammon: is there an advantage to either side?Expected Value of BillsEstimating probabilities of switching between two states










2












$begingroup$


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 $1 - mathcalF(f(x))$ different concave functions ranging from identity (blue curve), by slightly concave function (purple curve) to maximally concave function (red curve)?



enter image description here



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










share|cite|improve this question











$endgroup$











  • $begingroup$
    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.
    $endgroup$
    – Saad
    Oct 21 '18 at 9:43










  • $begingroup$
    Sad news, but thanks anyway.
    $endgroup$
    – jakes
    Oct 21 '18 at 9:48















2












$begingroup$


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 $1 - mathcalF(f(x))$ different concave functions ranging from identity (blue curve), by slightly concave function (purple curve) to maximally concave function (red curve)?



enter image description here



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










share|cite|improve this question











$endgroup$











  • $begingroup$
    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.
    $endgroup$
    – Saad
    Oct 21 '18 at 9:43










  • $begingroup$
    Sad news, but thanks anyway.
    $endgroup$
    – jakes
    Oct 21 '18 at 9:48













2












2








2


1



$begingroup$


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 $1 - mathcalF(f(x))$ different concave functions ranging from identity (blue curve), by slightly concave function (purple curve) to maximally concave function (red curve)?



enter image description here



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










share|cite|improve this question











$endgroup$




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 $1 - mathcalF(f(x))$ different concave functions ranging from identity (blue curve), by slightly concave function (purple curve) to maximally concave function (red curve)?



enter image description here



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







probability transformation monotone-functions logistic-regression






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 25 at 6:53







jakes

















asked Sep 26 '18 at 13:04









jakesjakes

298




298











  • $begingroup$
    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.
    $endgroup$
    – Saad
    Oct 21 '18 at 9:43










  • $begingroup$
    Sad news, but thanks anyway.
    $endgroup$
    – jakes
    Oct 21 '18 at 9:48
















  • $begingroup$
    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.
    $endgroup$
    – Saad
    Oct 21 '18 at 9:43










  • $begingroup$
    Sad news, but thanks anyway.
    $endgroup$
    – jakes
    Oct 21 '18 at 9:48















$begingroup$
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.
$endgroup$
– Saad
Oct 21 '18 at 9:43




$begingroup$
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.
$endgroup$
– Saad
Oct 21 '18 at 9:43












$begingroup$
Sad news, but thanks anyway.
$endgroup$
– jakes
Oct 21 '18 at 9:48




$begingroup$
Sad news, but thanks anyway.
$endgroup$
– jakes
Oct 21 '18 at 9:48










1 Answer
1






active

oldest

votes


















2





+50







$begingroup$

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:$$
mathscrF_1(f)(x) = left( lnfracMf(x) right)^a, quad mathscrF_2(f)(x) = left( left( fracMf(x) right)^b - 1 right)^a,
$$

where $a, b > 0$ are parameters. To see the rationale, consider two families of density functions:$$
f_1(x) = M exp(-(d(x, x_0))^tfrac1a), quad f_2(x) = fracM((d(x, x_0))^tfrac1a + 1)^tfrac1b,
$$

where $x_0 = (100, 50)$ and $d$ is some distance function. Note that for $d(x, y) = sqrtsmash[b](x_1 - y_1)^2 + (x_2 - y_2)^2$, $a = dfrac12$, $b = dfrac23$, $f_1$ 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 $d(x, x_0) = mathscrF_1(f_1)(x)$ and $d(x, x_0) = mathscrF_2(f_2)(x)$, 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.$$
mathscrF_1(f)(x) = left( lnfracM + cf(x) + c right)^a, quad mathscrF_2(f)(x) = left( left( fracM + cf(x) + c right)^b - 1 right)^a.
$$

After normalization,$$
mathscrF_1(f)(x) = left( fraclndfracM + cf(x) + clndfracM + cc right)^a, quad mathscrF_2(f)(x) = left( fracleft( dfracM + cf(x) + c right)^b - 1left( dfracM + cc right)^b - 1 right)^a.
$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    thank you. great answer, deserved bounty reward indeed.
    $endgroup$
    – jakes
    Oct 8 '18 at 11:05










  • $begingroup$
    @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.
    $endgroup$
    – Saad
    Oct 19 '18 at 7:04










  • $begingroup$
    @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 $f$.
    $endgroup$
    – Saad
    Oct 21 '18 at 1:35











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2





+50







$begingroup$

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:$$
mathscrF_1(f)(x) = left( lnfracMf(x) right)^a, quad mathscrF_2(f)(x) = left( left( fracMf(x) right)^b - 1 right)^a,
$$

where $a, b > 0$ are parameters. To see the rationale, consider two families of density functions:$$
f_1(x) = M exp(-(d(x, x_0))^tfrac1a), quad f_2(x) = fracM((d(x, x_0))^tfrac1a + 1)^tfrac1b,
$$

where $x_0 = (100, 50)$ and $d$ is some distance function. Note that for $d(x, y) = sqrtsmash[b](x_1 - y_1)^2 + (x_2 - y_2)^2$, $a = dfrac12$, $b = dfrac23$, $f_1$ 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 $d(x, x_0) = mathscrF_1(f_1)(x)$ and $d(x, x_0) = mathscrF_2(f_2)(x)$, 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.$$
mathscrF_1(f)(x) = left( lnfracM + cf(x) + c right)^a, quad mathscrF_2(f)(x) = left( left( fracM + cf(x) + c right)^b - 1 right)^a.
$$

After normalization,$$
mathscrF_1(f)(x) = left( fraclndfracM + cf(x) + clndfracM + cc right)^a, quad mathscrF_2(f)(x) = left( fracleft( dfracM + cf(x) + c right)^b - 1left( dfracM + cc right)^b - 1 right)^a.
$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    thank you. great answer, deserved bounty reward indeed.
    $endgroup$
    – jakes
    Oct 8 '18 at 11:05










  • $begingroup$
    @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.
    $endgroup$
    – Saad
    Oct 19 '18 at 7:04










  • $begingroup$
    @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 $f$.
    $endgroup$
    – Saad
    Oct 21 '18 at 1:35















2





+50







$begingroup$

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:$$
mathscrF_1(f)(x) = left( lnfracMf(x) right)^a, quad mathscrF_2(f)(x) = left( left( fracMf(x) right)^b - 1 right)^a,
$$

where $a, b > 0$ are parameters. To see the rationale, consider two families of density functions:$$
f_1(x) = M exp(-(d(x, x_0))^tfrac1a), quad f_2(x) = fracM((d(x, x_0))^tfrac1a + 1)^tfrac1b,
$$

where $x_0 = (100, 50)$ and $d$ is some distance function. Note that for $d(x, y) = sqrtsmash[b](x_1 - y_1)^2 + (x_2 - y_2)^2$, $a = dfrac12$, $b = dfrac23$, $f_1$ 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 $d(x, x_0) = mathscrF_1(f_1)(x)$ and $d(x, x_0) = mathscrF_2(f_2)(x)$, 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.$$
mathscrF_1(f)(x) = left( lnfracM + cf(x) + c right)^a, quad mathscrF_2(f)(x) = left( left( fracM + cf(x) + c right)^b - 1 right)^a.
$$

After normalization,$$
mathscrF_1(f)(x) = left( fraclndfracM + cf(x) + clndfracM + cc right)^a, quad mathscrF_2(f)(x) = left( fracleft( dfracM + cf(x) + c right)^b - 1left( dfracM + cc right)^b - 1 right)^a.
$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    thank you. great answer, deserved bounty reward indeed.
    $endgroup$
    – jakes
    Oct 8 '18 at 11:05










  • $begingroup$
    @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.
    $endgroup$
    – Saad
    Oct 19 '18 at 7:04










  • $begingroup$
    @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 $f$.
    $endgroup$
    – Saad
    Oct 21 '18 at 1:35













2





+50







2





+50



2




+50



$begingroup$

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:$$
mathscrF_1(f)(x) = left( lnfracMf(x) right)^a, quad mathscrF_2(f)(x) = left( left( fracMf(x) right)^b - 1 right)^a,
$$

where $a, b > 0$ are parameters. To see the rationale, consider two families of density functions:$$
f_1(x) = M exp(-(d(x, x_0))^tfrac1a), quad f_2(x) = fracM((d(x, x_0))^tfrac1a + 1)^tfrac1b,
$$

where $x_0 = (100, 50)$ and $d$ is some distance function. Note that for $d(x, y) = sqrtsmash[b](x_1 - y_1)^2 + (x_2 - y_2)^2$, $a = dfrac12$, $b = dfrac23$, $f_1$ 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 $d(x, x_0) = mathscrF_1(f_1)(x)$ and $d(x, x_0) = mathscrF_2(f_2)(x)$, 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.$$
mathscrF_1(f)(x) = left( lnfracM + cf(x) + c right)^a, quad mathscrF_2(f)(x) = left( left( fracM + cf(x) + c right)^b - 1 right)^a.
$$

After normalization,$$
mathscrF_1(f)(x) = left( fraclndfracM + cf(x) + clndfracM + cc right)^a, quad mathscrF_2(f)(x) = left( fracleft( dfracM + cf(x) + c right)^b - 1left( dfracM + cc right)^b - 1 right)^a.
$$






share|cite|improve this answer











$endgroup$



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:$$
mathscrF_1(f)(x) = left( lnfracMf(x) right)^a, quad mathscrF_2(f)(x) = left( left( fracMf(x) right)^b - 1 right)^a,
$$

where $a, b > 0$ are parameters. To see the rationale, consider two families of density functions:$$
f_1(x) = M exp(-(d(x, x_0))^tfrac1a), quad f_2(x) = fracM((d(x, x_0))^tfrac1a + 1)^tfrac1b,
$$

where $x_0 = (100, 50)$ and $d$ is some distance function. Note that for $d(x, y) = sqrtsmash[b](x_1 - y_1)^2 + (x_2 - y_2)^2$, $a = dfrac12$, $b = dfrac23$, $f_1$ 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 $d(x, x_0) = mathscrF_1(f_1)(x)$ and $d(x, x_0) = mathscrF_2(f_2)(x)$, 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.$$
mathscrF_1(f)(x) = left( lnfracM + cf(x) + c right)^a, quad mathscrF_2(f)(x) = left( left( fracM + cf(x) + c right)^b - 1 right)^a.
$$

After normalization,$$
mathscrF_1(f)(x) = left( fraclndfracM + cf(x) + clndfracM + cc right)^a, quad mathscrF_2(f)(x) = left( fracleft( dfracM + cf(x) + c right)^b - 1left( dfracM + cc right)^b - 1 right)^a.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 8 '18 at 0:03

























answered Oct 3 '18 at 6:23









SaadSaad

20.7k92452




20.7k92452











  • $begingroup$
    thank you. great answer, deserved bounty reward indeed.
    $endgroup$
    – jakes
    Oct 8 '18 at 11:05










  • $begingroup$
    @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.
    $endgroup$
    – Saad
    Oct 19 '18 at 7:04










  • $begingroup$
    @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 $f$.
    $endgroup$
    – Saad
    Oct 21 '18 at 1:35
















  • $begingroup$
    thank you. great answer, deserved bounty reward indeed.
    $endgroup$
    – jakes
    Oct 8 '18 at 11:05










  • $begingroup$
    @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.
    $endgroup$
    – Saad
    Oct 19 '18 at 7:04










  • $begingroup$
    @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 $f$.
    $endgroup$
    – Saad
    Oct 21 '18 at 1:35















$begingroup$
thank you. great answer, deserved bounty reward indeed.
$endgroup$
– jakes
Oct 8 '18 at 11:05




$begingroup$
thank you. great answer, deserved bounty reward indeed.
$endgroup$
– jakes
Oct 8 '18 at 11:05












$begingroup$
@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.
$endgroup$
– Saad
Oct 19 '18 at 7:04




$begingroup$
@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.
$endgroup$
– Saad
Oct 19 '18 at 7:04












$begingroup$
@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 $f$.
$endgroup$
– Saad
Oct 21 '18 at 1:35




$begingroup$
@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 $f$.
$endgroup$
– Saad
Oct 21 '18 at 1:35

















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