4
$\begingroup$

I need to bound a 1D gaussian/normal (or similar) probability density function in the domain interval $[-3\sigma, 3\sigma]$ in a way that still integrates to 1. I would need something like this:

$ p(x) = \begin{cases} N(x;\mu, \sigma) &\text{if } -3\sigma \leq x \leq 3\sigma\\ 0 & \text{otherwise } \end{cases} $

This is NOT a probability density function but how could I get a bounded distribution that is similar to the gaussian case?

Thanks in advance,

Federico

  • 0
    If you just want something that looks similar, have you considered a scaled and centered beta distribution?2011-11-04

2 Answers 2

1

It seems you are not clear about what you want. To truncate any variable to a given range, you just restrict its density to that range, and divide by its integral so that integrates to 1. But if you want to generate a random variable that just "looks like" a gaussian, but has support on an interval, and its density is smooth, you can sum three (or more) uniforms. For example, if you sum three uniforms in $[-1,1]$, the result is a random variable that has support in $[-3,3]$, and its variance is $1$; you can multiply the result by $\sigma$ to get a suport $[-3 \sigma,3 \sigma]$ and standard deviation $\sigma$. The density is piecewise quadratic, it's continuous and derivable (though not infinitely differentiable, of course).

0

I don't think bump functions will help, because the Gaussian does not have compact support. I am not quite sure what you really need. If it still integrates to one, why is it not a probability density? Why not make a simple transformation of coordinates? This will still integrate to one, should be sufficiently smooth: $f(x)=\exp\left(\frac{-\tan\left(\frac{\pi}{2}\frac{x}{3\sigma}\right)^2}{2}\right)\mathbb{1}_{(x<3\sigma)}(x)$

  • 0
    @whuber, yes, I like that approach, too.2011-09-06