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I'm looking for a way to adjust the probability distribution of a uniform random function I'm using in a program.

I want to find some discrete probability distribution that includes a parameter for "adjusting" how uniform the distribution is relative to the "center" event. For example, if that parameter, lets say s=0, then the probability of 5 events is exactly equal at 20% each. If I were to change that parameter to s=1, the function would now adjust the probability depending on the distance from the center. So, the probability would change to:

  • Event 1: 16%
  • Event 2: 18%
  • Event 3: 20%
  • Event 4: 22%
  • Event 5: 24%

And adjusting s higher would yield increasing probability above the "center" event and decreasing probability below the center event, depending on distance from the center - all while retaining a total probability of 100%. Statistics is not my strong point, so I'm just hoping someone with some experience might understand what I'm looking for :)

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Consider the function $p_s(x)=\frac{2-3s}{10}+\frac{s}{10}x$ with $s\in[-1,1]$.

Notice that $$\sum_{i=1}^{5}p_s(i)=\sum_{i=1}^{5}\left(\frac{2-3s}{10}+\frac{s}{10}i\right)=1-\frac{15}{10}s+\frac{15}{10}s=1$$ and the restriction on $s$ forces all of $p_s(1)$ through $p_s(5)$ to be at least zero and at most one. Thus, $p_s$ is a probability distribution over the integers $1$ through $5$.

The example you give above is the special case when $s=\frac{2}{10}$: $p_{s=2/10}(x)=0.14+0.02x$.

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  • $\begingroup$ Excellent! Thank you very much - just what I was looking for. I realize I was a bit vague in my description, but you nailed it. If you don't mind me picking your brain a bit, I rearranged the equation into p(x)=2/10+s(x-3)/10. If i were to want a more exponential change in distribution, how would I modify it? I tried (x-3)^3, but that throws off the summation to 1... $\endgroup$
    – 7200rpm
    Commented Mar 14, 2013 at 2:28
  • $\begingroup$ Try this: Let $$p_1(x)=\frac{(x-3)^3+8}{\sum_{j=1}^{5}\left((j-3)^3+8\right)}$$ and $$p_{-1}(x)=p_1(6-x)$$ and $$p_s(x)=s\times p_1(x)+(1-s)\times p_{-1}(x)$$ with $s\in[0,1]$. When $s=1/2$, $p_s$ is uniform over $1,\ldots,5$. $\endgroup$
    – assumednormal
    Commented Mar 14, 2013 at 3:31
  • $\begingroup$ If you have a function $f(x)$ such that $f(1),\ldots,f(5)$ are all finite, a few simple steps will turn this function into a probability distribution over the integers $1,\ldots,5$. (1) Define $g(x)=f(x)-\textrm{min}(f(1),\ldots,f(5))$. (2) Define $p(x)=g(x)/\sum_{i=1}^{5}g(i)$. Now $p(x)$ is a probability distribution over the integers $1,\ldots,5$ and it has a shape similar to the original function $f(x)$. $\endgroup$
    – assumednormal
    Commented Mar 14, 2013 at 3:41
  • $\begingroup$ Thanks again! Super helpful, sometimes it can be difficult to find a function based only on vague sense of the kind of output you want. $\endgroup$
    – 7200rpm
    Commented Mar 14, 2013 at 4:04

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