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

Here is the link to the paper. In this paper, they call the product of the normal and the exponential a rectified distribution. The problem is, that I don’t 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

\begin{align} \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] \end{align}

the last two terms are dropped out since the can be regarded as a proportionality 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. How do I sample from this distribution and why can it be used in the gibbs sampler if it is not a proper density?

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

Here is the link to the paper. In this paper, they call the product of the normal and the exponential a rectified distribution. The problem is, that I don’t 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

\begin{align} \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] \end{align}

the last two terms are dropped out since they can be regarded as a proportionality 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 any more since it does not integrate to one. How do I sample from this distribution and why can it be used in the Gibbs sampler if it is not a proper density?

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

Here is the link to the paper. In this paper, they call the product of the normal and the exponential a rectified distribution. The problem is, that I don’t 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 proportionality 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. How do I sample from this distribution and why can it be used in the gibbs sampler if it is not a proper density?

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

Here is the link to the paper. In this paper, they call the product of the normal and the exponential a rectified distribution. The problem is, that I don’t 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

\begin{align} \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] \end{align}

the last two terms are dropped out since the can be regarded as a proportionality 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. How do I sample from this distribution and why can it be used in the gibbs sampler if it is not a proper density?

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

Here is the link to the paper. Like mentioned before, the call the product of the normal and the exponential a rectified distribution. The problem is, that I don’t 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 proportionality 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.

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

Here is the link to the paper. In this paper, they call the product of the normal and the exponential a rectified distribution. The problem is, that I don’t 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 proportionality 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. How do I sample from this distribution and why can it be used in the gibbs sampler if it is not a proper density?