Is the rectified Gaussian distribution the same as the truncated Gaussian distribution within the interval $[0,\infty)$?


  [1]: https://en.wikipedia.org/wiki/Rectified_Gaussian_distribution

Here is the link to the [paper](http://mlg.eng.cam.ac.uk/pub/pdf/SchWinHan09.pdf). Like mentioned before, the call the product of the normal and the exponential a rectified distribution. The problem is, that i dont understand how they sample from this distribution within a Gibbs sampling procedure. So far I found out, that they used the following algebraic identity to simplify the product as follows $\mathcal{R}(x|\mu,\sigma,\alpha)\propto K\exp\left(-\frac{\left(x-\mu\right)^2}{2\sigma^2}\right)\exp\left(-\alpha x\right) \qquad \text{with}\qquad x\geq0$

 $-\left[\frac{\left(x-\mu\right)^2}{2\sigma^2}+\alpha x\right] =-\left[\frac{\left(x-\left(\mu-\sigma^2\alpha\right)\right)^2}{2\sigma^2}-\frac{1}{2}\sigma^2\alpha^2+\mu\alpha\right]$

the last two terms are dropped out since the can be regarded as a proportionalty constant. The result is a normal distribution truncated on the interval $[0,\infty]$

$\mathcal{R}(x|\mu,\sigma,\alpha)\propto K\exp\left(-\frac{\left(x-\left(\mu-\sigma^2\alpha\right)\right)^2}{2\sigma^2}\right)$

I think that this is not equivalent to the truncated distribution since the normalization is missing. But this is not a proper distribution anymore since it does not integrate to one.