# What does truncated distribution mean?

In a research article about sensitivity analysis of an ordinary differential equation model of a dynamic system, the author provided the distribution of a model parameter as Normal distribution (mean=1e-4, std=3e-5) truncated to the range [0.5e-4 1.5e-4]. He then uses samples from this truncated distribution for simulations of the model. What does it mean to have a truncated distribution and sample from this truncated distribution?

I could come up with two ways of doing this:

• Sample from a Normal distribution but ignore all the random values falling outside the specified range before simulations.
• Somehow obtain a special “Truncated Normal” distribution and get samples from it.

Are these valid and equivalent approaches?

I believe in the first case, if one were to plot the experimental cdf/pdf of the sample, it would not look like a Normal distribution because the curves do not extend to $\pm\infty$.

So, to truncate the $N(\mu, \sigma^{2})$ distribution to an interval $(a,b)$ would be to generate a random variable that has density
$$p_{a,b}(x) = \frac{ \phi_{\mu, \sigma^{2}}(x) }{ \int_{a}^{b} \phi_{\mu, \sigma^{2}}(y) dy } \cdot \mathcal{I} \{ x \in (a,b) \}$$
where $\phi_{\mu, \sigma^{2}}(x)$ is the $N(\mu, \sigma^2)$ density. You could sample from this density in a number of ways. One way (the simplest way I can think of) to do this would be to generate $N(\mu, \sigma^2)$ values and throw out the ones that fall outside of the $(a,b)$ interval, as you mentioned. So, yes, those two bullets you listed would accomplish the same goal. Also, you are right that the empirical density (or histogram) of variables from this distribution would not extend to $\pm \infty$. It would be restricted to $(a,b)$, of course.