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I'm currently trying to distribute 100% (1) over 20 points.
So far the standardized gaussian normal distribution did some good work with a standard deviation of 2 and a mean of 10, which distributes roughly 1 (0.999..) over my 20 points: Looks like this.

Now, I can just increase the deviation a lot but that doesn't add up to 1 anymore and stops being an equal distribution: 1 is bigger than 20 and the sum is 0.9~.

Is there a way to stretch and move the normal distribution while keeping the sum at 1?

1 Answers 1

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  • Translation of mean location does nothing to the sum.
  • For the variance we need to scale as follows $\sigma^2 \mathcal{N}(0,t) =\mathcal{N}(0,\sigma^2 t)$

What does the last statement mean? Well, the normal distribution if multiplied by a scalar will automatically adjust the variance to yield a normal distribution so the sum is maintained (this is pretty nice property of a normal distribution)

From the comments it seems you want a truncated normal distribution

So if you want a normal distribution that sums to 1 between a predefined interval then you can use $$ f(x; \mu, \sigma, a, b) = \frac{\frac{1}{\sigma}N\left(\frac{x-\mu}{\sigma}\right)}{\Phi\left(\frac{b-\mu}{\sigma}\right) - \Phi\left(\frac{a-\mu}{\sigma}\right)} $$ where the interval is $-\infty \leq a < b \leq \infty$ and $\Phi(x)$ is the cumulative normal distribution.

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    What I mean is that I want to keep the interval the same, e.g. over 20 points. If I change the deviation/mean the sum is indeed maintained, just not for [1;20], which is what I'm trying to do. For example look at the second graph, how would you display this while keeping the sum at 1? – 2017-01-11
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    Yeah that's what I mean. So it's $$ \frac{pdf}{cdf - cdf} $$ right? This works pretty good but for whatever reason doesn't seem [standardized (a=1, b=20, deviation=2, mean=10)](http://puu.sh/tiMhU/abd374eeb1.png). I can't find something on google for standardized truncated distributions, am I doing it wrong or am I misunderstanding the formula? – 2017-01-11
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    Well, I just standardized it myself, f(x)/sum of all f(x), its kind of a hack but whatever. Thanks for your help! – 2017-01-12
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    Sorry for the delay - but if you use the parameters above with a normally distributed value then the sum should converge to one. I mean integrating the function above over $[a,b]$ should yield the denominator. But yes the sum you did should normalize to one by definition but it seems strange. – 2017-01-12