# Rectified Gaussian distribution with cut-off point other than zero

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.

• And? What’s the problem? What have you tried? Commented Jul 8, 2019 at 18:22
• Oh I have forgotten to include what I have got, I have edited the original question Commented Jul 8, 2019 at 19:08
• For any function, or distribution for that matter, shifting with k is just: $f(x) \to f(x-k)$. Commented Jul 8, 2019 at 19:38
• But only by doing this, I would also get the cut-off point shifted, i.e. the cut-off point would now become $k$ instead of 0 Commented Jul 8, 2019 at 19:41
• I think you're a bit confused about what you want. But I think it's actually this: 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. Commented Jul 8, 2019 at 19:45

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. "

• Thanks for your help! Commented Jul 9, 2019 at 18:21