Density of vertically truncated normal distribution

Question

I'm interested in the Gaussian density, truncated vertically at a threshold $\beta>0$.

$$ p(x)=\frac{1}{Z} \min\left\{\exp\Big(-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\Big),\beta\right\} \\ $$

How can I find the normalizing constant $Z$?

Answer 1

You won't be able to find this constant in a nice form. Either you have an integral experession or a numerical approximation.

I assume (like usual for a gaussian densitiy) $\Sigma \in \Bbb R ^{n\times n}$ to be positive definite and so we can conclude that $$\exp\Big(-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\Big) \le 1$$ hold.

Then for $\beta \ge 1$ we are getting a normal gaussian density for p(x) and can conclude $$Z = \sqrt{(2\pi)^n \det(\Sigma)}$$

For a shorter notation consider that with $$\begin{align*} <x,y> &= x^T\Sigma^{-1}y \\ ||x|| &= \sqrt{<x,x>}\end{align*}$$

we can write $$p(x)=\frac{1}{Z} \min\left\{\exp\Big(-\frac{1}{2}||x-\mu||^2\Big),\beta\right\}$$

For $0 < \beta < 1$ and $A:=\left\{ x \in \Bbb R^n \;\Big|\; ||x-\mu|| \ge \sqrt{-2\ln\beta}\right\}$ we have:

$$\begin{align*} Z &= \int_{\Bbb R^n} Zp(x) dx \\ &= \int_{A} \exp\Big(-\frac{1}{2}||x-\mu||^2\Big) dx + \int_{A^c} \beta dx \\ &= \sqrt{(2\pi)^n \det(\Sigma)} - \int_{A^c} \exp\Big(-\frac{1}{2}||x-\mu||^2\Big) - \beta \, dx \end{align*}$$

For a given $\Sigma$ you may be able to calculate $$\int_{A^c} \beta \, dx$$

but the expression

$$\int_{A^c} \exp\Big(-\frac{1}{2}||x-\mu||^2\Big) \, dx$$ have to be calculated numerically, even in the real case.

For some $\beta$ this can be simplified but not in the general case.