I have a bunch of numbers taken from a normal distribution of mean $0$.

I wish to create a function that takes each of these numbers and converts it to an exponent. for example,

  • if the function's parameter is $2$, I want the number $3$std above $0$ to be converted to $2$,
  • while the number $3$std below $0$ to be converted to $1/2$ which is the inverse of the exponent $2$.
  • the number $0$ is to be converted to $1$ regardless of the parameter.

I am aware that there might be various non-identical functions which abide the rules above yet form different distributions, I was wondering if there are common/natural ways of achieving this result.

  • $\begingroup$ if the function's parameter is $2$, what do you want the number $1$ standard deviation above $0$ to be converted to? $\sqrt[3]{2}$? $\endgroup$
    – Henry
    Aug 6, 2018 at 14:56
  • $\begingroup$ That sounds like a good idea! $\endgroup$
    – Carpet4
    Aug 6, 2018 at 15:09

1 Answer 1


As you say, there are many possibilities. But if you want the function to give a power of $2$ for each $3$ standard deviations above the mean of $0$, then a possibility is with $x$ standard deviations above $0$:


and that would give $f(3)=2, f(-3)=\frac12, f(0)=1$ as requested

and for example also $f(6)=4, f(9)=8, f(1)=\sqrt[3]{2} \approx 1.259921$

Slightly more generally you could define a multi-parameter function as $$g(y, \mu, \sigma, a, b) = a^{({y-\mu})/({b \sigma})}$$ where the earlier function is the special case $f(x)=g(x,0,1,2,3)$

  • $\begingroup$ Amazing, thanks! $\endgroup$
    – Carpet4
    Aug 7, 2018 at 6:15

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