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?
Thanks in advance,
Federico
 A: 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).
A: 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: $f(x)=\exp\left(\frac{-\tan\left(\frac{\pi}{2}\frac{x}{3\sigma}\right)^2}{2}\right)\mathbb{1}_{(x<3\sigma)}(x)$
