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I have pairs of values e.g.

$A_{1} = -10$ , $B_{1} = 4$

$A_{2} = 6$ , $B_{2} = 4$

$A_{3} = 80$ , $B_{3} = 79$

I want to create a probability, $P(A) + P(B) = 1$ , such that the smaller values of A or B have higher probability. For example:

$P(A_{1}) = 0.8$ , $P(B_{1}) = 0.2$

$P(A_{2}) = 0.35$ , $P(B_{2}) = 0.65$

$P(A_{3}) = 0.47$ , $P(B_{3}) = 0.53$

I have broken my head trying to generate this type of proability but I haven't been able. I thought something like:

$P(A_{1}) = \frac{|A|}{|A+B|}$

But I am afraid this doesn't yield the expected results, specially when there are negative and positive values like in $A_{1}$ and $B_{1}$.

I would appreciate some suggestion on how I can generate such probabilities.

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  • $\begingroup$ When you write "e.g." and then "For example:" is there a hidden functional relation betweeen the two sets of values ? Or are these two sets not linked at all ? $\endgroup$ – keepAlive May 3 '17 at 10:19
  • $\begingroup$ @Tnerual there is no relation at all. I made up the probabilities as an illustrative example. The point of the question is in fact what you mention, creating a link between the values and the probabilities such that smaller values of A or B have higher probability. $\endgroup$ – adrian1121 May 3 '17 at 10:22
  • $\begingroup$ I am curious to know what you think of my answer since it does yield the expected results ? $\endgroup$ – keepAlive May 3 '17 at 16:27
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There are many choices, but here's an approach that will let you design many solutions easily. Break this task into two steps.

One scheme is to take some transformation $t$ of the values and then scale to $1$: $P(A) = \frac{t(A)}{t(A)+t(B)}$.

Note that $t$ should be positive and decreasing; this will make sure that smaller values get higher probability and that probabilities remain between $0$ and $1$.

For example either of $t(x) = \exp(a+bx)/(1+\exp(a+bx))$, or $t(x)=a+b.\exp(-cx)$
would all work when all the parameters ($a,b,c$) are positive. Any continuous monotonic decreasing survivor function on the real line would work ($t(x)=1-F(x)$ for any cdf $F$), as would $\log(1/F(x))$. Any number of other such functions might be used.

So consider $t(x) = \log(1/\Phi(\frac{x}{5}))$ (where $\Phi$ is the standard normal cdf). (A variation on one of the functions I mentioned before.)

Then for $A=-10$, $B=4$ we would have $P(A) = \frac{\log(1/\Phi(\frac{-10}{5}))}{\log(1/\Phi(\frac{-10}{5}))+\log(1/\Phi(\frac{4}{5}))} = \frac{3.783}{ 3.783+0.2381}=0.941$ (and $P(B)=0.059$).

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  • $\begingroup$ Thank you for your answer. I see your point! However, I am afraid this trannsformation is not what I am looking for: In the last line of your answer you derive $P(A) = 0.3125$. But notice that I want to assign higher probability to the smaller value (A), so intuitively $A$ should yield higher probability than $B$'s. Thank you $\endgroup$ – adrian1121 May 3 '17 at 13:03
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    $\begingroup$ 1. You cannot write a function that returns a probability from inputs that are not probabilities without some form of transformation of those values. Your own function $\frac{|A|}{|A+B|}$ involves transformation as does the other answer here. But the point of my answer is to try to illustrate approaches to tailoring a solution that will do exactly what you want. $\:$ 2. Sorry, I was thrown off by your example function with absolute values and thought you intended "closer to 0" by "smaller". I've fixed that issue now $\endgroup$ – Glen_b -Reinstate Monica May 3 '17 at 17:05
  • $\begingroup$ Thank you very much for your answer. This is a clever and straightforward approach which allows for some freedom when modeling the transformation. Upvote and accept! Thanks. $\endgroup$ – adrian1121 May 4 '17 at 8:36
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What about using a monotonic transformation which maps elements from $\mathbb{R}$ to $\mathbb{R}_{>0}$: like a power transformation, doing it in the base you want (I mean instead of using the constant $e ≈ 2.71828...$):

If one has

$A_{1} = -10$ , $B_{1} = 4$

$A_{2} = 6$ , $B_{2} = 4$

$A_{3} = 80$ , $B_{3} = 79$

and one uses the following base, say $e'= 0.905723664263907$ (lower than $1$ to make the function be monotonic decreasing on $\mathbb{R}$), one gets

$e'^{A_{1}}=2.6918003852647$, $e'^{B_1}=0.672950096316179$

$e'^{A_{2}}=0.552044756836908$, $e'^{B_2}=0.672950096316179$

$e'^{A_{3}}=0.000362791574496337$, $e'^{B_3}=0.000400554373050618$

finally one gets the following weights

$\frac{e'^{A_{1}}}{e'^{A_{1}}+e'^{B_{1}}} = 0.8$

$\frac{e'^{A_{2}}}{e'^{A_{2}}+e'^{B_{2}}} = 0.450650675$

$\frac{e'^{A_{3}}}{e'^{A_{3}}+e'^{B_{3}}} = 0.475264951$

As can be seen above, I chose $e'$ so as to get $0.8$ for the weight of $A_1$. But as you surely know, it could have been chosen relatively to the desired weight associated to $A_2$, $A_3$, none of those, any optimization/equalization criterion or simply setting $e'=\exp(-1)$.

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  • $\begingroup$ Thank you very much for your answer! Your idea is smart and straightforward! Upvote! It worked perfectly! However, I accept the answer from @Glen_b as his approach allows more freedom and control over the transformation. Thanks again though. Adrian. $\endgroup$ – adrian1121 May 4 '17 at 8:35

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