# Bound 1D gaussian domain in the interval $[-3\sigma, 3\sigma]$ so it still is a probability density function

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?

Federico

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All you have to do is to divide $p(x)$ over by $\int_{-3 \sigma}^{3 \sigma} p(x) \mathrm{d} x$. –  Sasha Sep 5 '11 at 13:51
Agree with Shasha. Unless you want to demand that your function should vanish smoothly, then you'll need something else... –  valdo Sep 5 '11 at 13:59
Thanks Sasha and valdo. What I actually need is that it vanished smoothly; I haven't commented on it in the question... –  Federico Sep 5 '11 at 14:09
Do you know about bump functions? –  cardinal Sep 5 '11 at 14:44
If you just want something that looks similar, have you considered a scaled and centered beta distribution? –  guy Nov 4 '11 at 23:09

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).
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: $f(x)=\exp\left(\frac{-\tan\left(\frac{\pi}{2}\frac{x}{3\sigma}\right)^2}{2}\right)\mathbb{1}_{(x<3\sigma)}(x)$
The point of my comment was to consider transforming the Gaussian into a bump function by using the appropriate mapping from $(-\infty, \infty)$ to $(-3\sigma, 3\sigma)$. –  cardinal Sep 5 '11 at 18:32
@cardinal Another natural approach is to mollify the truncated distribution. This means define a "bump function" $\psi$ supported on $[-3\sigma, 3\sigma]$, equal to $1$ on $[-3\sigma+\varepsilon, 3\sigma-\varepsilon']$ for arbitrarily small positive $\varepsilon$, $\varepsilon'$. Normalizing the function $x \to \psi(x) \exp(-(x/\sigma)^2/2)$ does the trick. –  whuber Sep 6 '11 at 16:39