I have a set of N scores, Si, in ascending order. S1 is the smallest, S2 the next smallest, etc. Code: 1 2 3 4 S 2 3 5 8 I want to calculate a set of probabilities that are inversely proportional to the scores and that sum to "1". That is, I want S1 to have the highest probability and rest to have probabilities that are proportionately smaller. I accomplished this by first calculating a set of inverse weights, Wi, such that Code: Wi = S1 / Si I then converted those to probabilities, Pi, by Code: 1 2 3 4 Sum S 2 3 5 8 W 1.00 0.67 0.40 0.25 2.32 Then I converted those into probabilities by dividing each one by the sum of the weights, 2.32: Code: 1 2 3 4 Sum S 2 3 5 8 W 1.00 0.67 0.40 0.25 2.32 P 0.43 0.29 0.17 0.11 1.00 Now if P1 (0.43) is greater than some arbitrary value, like 0.30, then I want to set it to that value and scale the other probabilities in proportion to their relative weights, but still have them sum to 1.00. I can get the result I want by changing the equation for calculating Wi by adding an exponent: Code: Wi = (S1 / Si)^F By trial and error, I found that setting F = 0.26, I get Code: 1 2 3 4 Sum S 2 3 5 8 W 1.00 0.90 0.79 0.70 3.89 P 0.30 0.27 0.23 0.21 1.00 I would like either a way to do that directly or a closed for equation for F. When I try to do that, I get bogged down in complicated (for me) equations involving logs. I would appreciate any help.

I got this far in the search for a closed form equation for F. The equation for P1 is: Code: P1 = 1 / Sum(Wi) Sum(Wi) = (S1/S1)^F + (S1/S2)^F + (S1/S3)^F P1 = 1 / ( (S1/S1)^F + (S1/S2)^F + (S1/S3)^F ) (S1/S1)^F + (S1/S2)^F + (S1/S3)^F = 1/P1 I have come up short for a way to solve this for F. Normally, when trying to solve for an exponent, I would take the logs of both sides, but I don;t know how to get F out of that log of sums. Can anyone suggest a way to solve this equation for F ?

So you have a vector of positive numbers S and you want a second vector of positive numbers P and an positive exponent K with conditions: max(P_i) = P0 sum(P_i) = 1 S_i P_i^K = S_j P_j^K This will not always be possible. If S is constant, P0 has to be a particular value. If P_i = W S_i^-K then S_i P_i^K = W = S_j P_j^K max(P_i) = W min(S_i)^-K Thus W(K) = P0 min(S_i)^K Thus sum(P_i) = P0 min(S_i)^K sum(S_i^-K) So we want to solve for K: 1 = P0 min(S_i)^K sum(S_i^-K)for P0 = 3/10 This in general is a transcendental equation so you will want a general solver and contingency for when there is no solution or no positive solution. I get K ≈ 0.285661799704292683848547628090811554694676399989531... ≈ 2/7 Using K=2/7 (which is just and approximation), you get W = (3/10) 2^(2/7) and (3/10) 2^(2/7) ( 2^(-2/7) + 3^(-2/7) + 5^(-2/7) + 8^(-2/7) ) = 0.9999685... So using Mathematica, we can get a highly precise solution: Code: S = { 2, 3, 5, 8 }; P0 = 3/10 ; P = P0 * Min[S]^x * ( # ^ -x ) & /@ S ; ans = Solve[ Total[ P ] == 1, x, Reals] ; N[P , 79] /. ans[[1]] Code: {0.3000000000000000000000000000000000000000000000000000000000000000000000000000000, 0.267189082563102067083882226912844813277540705949340595586058500878920594709311, 0.230911198638986395604416849051029728933476685369502782406193969753574064993534, 0.201899718797911537311700924036125457788982608681156622007747529367505340297155} Replacing some of the scary syntax with the long forms gives: Code: S = { 2, 3, 5, 8 }; P0 = 3/10 ; P = P0 * Min[S]^x * Map[ Function[ s, s ^ -x ] , S] ; ans = Solve[ Total[ P ] == 1, x, Reals] ; ReplaceAll[N[P , 79] , Part[ans,1] ]