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Let $A$ be a set of values $\{a_1,a_2,a_3,a_4,a_5\}$ where $a_1 = 2$, $a_2 = 1$, $a_3 = 4$, $a_4 = 1$ and $a_5 = 2$, so, the $avg(A) = 2$.

I'm looking for a normalization where the values below the average would be increased while the values over the average would be reduced, with the condition that the normalized version of a greater number would also be greater than the normalized version of a smaller one.

I could just tranform to the logarithmic scale but the reduction is too much. My attempt was: $\frac{a_i}{log_{avg(A)}a_i}$. The problem is that it does not work if the value of $a_i$ is 1, for example. Also, $\frac{2}{log_{2}2} = 2$ while $\frac{3}{log_{2}3} = 1.89$, violating the condition.

Is there any (good) way to achieve what I stated in the 2nd paragraph?

user206695
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  • As far as I understood, you also need to keep average value unchanged. In this case you may just take segment $[a_{min}, a_{max}]$ and scale it linearly with a coefficient appropriate for you. – Andrei Rykhalski Feb 11 '15 at 15:37

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You can compute $a_{avg}$, the average of all your numbers. Then replace $a_i$ with $a_{avg} + \frac 1n (a_i-a_{avg})$. Choose $n$ to your liking. The average will not change, nor will values that start at the average.

Ross Millikan
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  • Thanks! The only thing is that I would like to reduce more if the number is greater, but I'm going to work over that :). – user206695 Feb 11 '15 at 16:25
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    You can use different $n$s, one if $a_i \gt a_{avg}$ and another if it is less. The average will move, but points that started at the average will not. – Ross Millikan Feb 11 '15 at 16:27
  • Thanks again! What I meant was that the decrease in 100 should be greater than the decrease in 80! The ideal would be a "self-defined" n according to the number, for example. – user206695 Feb 11 '15 at 17:18
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    That will already happen. If we choose $n=2$ and $a_{avg}=60$ you will take $100$ to $80$ and $80$ to $70$. If that isn't enough, try $a_i \to a_{avg}+\sqrt{a_i-a_{avg}}$. There are lots of functions. You need to think clearly about what you want to do. – Ross Millikan Feb 11 '15 at 17:23
  • I see that. I'm going to give a further look on that. But actually the proportion of the decrease was pretty good with $\frac{a_i}{log_{avg(A)}a_i}$. The only problem is that I was not able to think a way to use this proportion and hold the propertie I described. Thank you very much! – user206695 Feb 11 '15 at 18:05