Rectified Gaussian distribution with cut-off point other than zero

Question

I am trying to figure out the pdf of Rectified Gaussian distribution with cut-off point other than zero, or actually, a shifted Rectified Gaussian distribution with cut-off point at zero.

Here is what I have got so far, let $X$ denote rectified Gaussian distribution with $\mu,\sigma^2$

$$f(x) = \Phi(-\frac{\mu}{\sigma})\delta(x)+\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}U(x)$$ where $\delta(x)$ is the Dirac delta function and $U(x)$ is the unit step function.

$\delta(x)=\begin{cases} +\infty, & \text{if}\ x=0 \\ 0, & \text{otherwise} \end{cases}$ and $U(x)=\begin{cases} 0, & \text{if}\ x\leq0 \\ 1, & \text{otherwise} \end{cases}$

I am trying to get it to have a cut of point at $k$ instead of 0, what I am thinking is that in order to have a cut of point at $k$, should I just change $\delta(x)$ and $U(x)$ to $$\delta(x)=\begin{cases} +\infty, & \text{if}\ x=k \\ 0, & \text{otherwise} \end{cases},\text{and}$$

$$U(x)=\begin{cases} 0, & \text{if}\ x\leq k \\ 1, & \text{otherwise} \end{cases}.$$

But I am unsure if they are the correct way to do it.

Answer 1

Let's write the density of a standard normal as $\phi(x)$ and the cumulative distribution as $\Phi(x)$, then the rectified Gaussian as above can be written as

$$f(x;\mu,\sigma) = \Phi\left(-\frac{\mu}{\sigma}\right)\delta(x)+\phi\left(\frac{x-\mu}{\sigma}\right)U(x)$$

and the generalization you're looking for as

$$f(x;\mu,\sigma,k) = \Phi\left(\frac{k-\mu}{\sigma}\right)\delta(x-k)+\phi\left(\frac{x-\mu}{\sigma}\right)U(x-k) \; .$$

It should be clear from the description I gave in my comment: "the rectified gaussian is what you obtain when you take an arbitrary gaussian, take all the mass left of a point k and concentrate it at the point k. Keep the rest of the mass on the right like it was. "