True mean of a truncated distribution?

Question

I use C++ GSL library to generate random numbers now. The numbers obey a distribution, (e.g. normal or lognormal distribution). This library requires the input of expected value ${\mu}$ (i.e. mean) and std. deviation $\sigma$. I'd like sample random numbers within an interval $(a, b)$. So, when the generated numbers is out of the interval, then that number is discarded. For instance, I'd like to sample numbers having a normal distribution with $\mu$ = 5, $\sigma$ = 0.5, $a$ = 0.5, and $b$ = 10. The $\mu$ must be in the interval $(a, b)$.

The question is: someone said the input $\mu$ and $\sigma$ are not true $\mu$ and $\sigma$ of the sampled numbers, because the pdf $f(x)$ is truncated. But I think, since the input $\mu$ is within the $(a, b)$, and $(a, b)$ is wide enough, they might not be so different.

Is it a big difference between the input and true $\mu$ and $\sigma$? Is it anyway to analytically derive the true values of them?

Maybe, the true $\mu$ is $E(x) = \int_a^bxf(x)dx$? am I right?

Answer 1

Suppose that $G$ is an absolutely continuous distribution, perhaps with support on all real numbers, or some other interval, and $F$ is a truncated distribution with the same pdf, except that it is restricted to have support on $[a,b]$. To write it out in notation,

$$ f(x)= \begin{cases} C g(x) & x\in[a,b] \\ 0 &\text{otherwise} \end{cases} $$

for some constant $C$ which renormalizes the distribution.

When $G$ is an absolutely continuous distribution, we can write the CDF of $F$ as $$ F(x) = \frac{G(x)-G(a)}{G(b) - G(a)} $$

and by LOTUS, the distribution $F$ has expectation

$$ \mu=\frac{\int_a^b t d G(t)}{G(b)-G(a)} $$

and variance

$$ \sigma^2 =\frac{\int_a^b t^2 d G(t)}{G(b)-G(a)} - \mu^2. $$

Answer 2

For truncated distributions, it is customary to parameterise them in terms of the original parameters, ie those of the untruncated distribution. For instance, the truncated normal distribution$$\mathcal N_-^+(\mu,\sigma^2,a,b)$$restricts the untruncated normal distribution$$\mathcal N_(\mu,\sigma^2)$$to the interval $(a,b)$, meaning that the untruncated normal $\mathcal N_(\mu,\sigma^2)$ density is restricted to the $(a,b)$ interval and renormalised$$f_-^+(x;\mu,\sigma^2,a,b)=\varphi(x;\mu,\sigma^2)\Big/\int_a^b\varphi(\xi;\mu,\sigma^2)\,\text d\xi\tag{1}$$Using the original mean-variance parameterisation is easier than using the truncated mean-variance parameterisation as

1. The definition of the distribution is straightforward, as shown by (1). By contrast, if given the truncated mean-variance parameterisation, one would need to revert back to the untruncated mean-variance parameterisation to write (1).
2. In most situations, there exist no analytical formula for the mean and/or variance of the truncated distributions. Take for instance a truncated multivariate Normal distribution.

Note however that the truncated normal distribution$$\mathcal N_-^+(\mu,\sigma^2,a,b)$$ allows for$$\sigma^2<0$$when $a$ and $b$ are both finite. Meaning that $$\exp\{(x-\mu)^2/2\kappa\}\Big/\int_a^b \exp\{(\xi-\mu)^2/2\kappa\}\,\text d\xi$$ is well-defined for all $\kappa\in\mathbb R$.

Note also that simulating from the untruncated version until the outcome stands in $(a,b)$ may prove a very inefficient way of simulating from the truncated version.