Truncated normal distribution without scaling

Question

My understanding of a truncated normal distribution $\mathcal{N}(\mu,\sigma;a,b)$ is that it results from scaling the values of a normal distribution within the bounds $[a; b]$ such that the area under the curve within these bounds becomes 1 (please correct me, if I'm wrong).

However, in a case where I want to describe a variable $z$ that results from a latent variable $y$ (with normally distributed noise $\sigma$) such that

$ z = \begin{cases} a & \text{if $y<a$}\\ b & \text{if $y>b$}\\ y & \text{otherwise} \end{cases} $

.. the truncated normal distribution in the above form would not be an appropriate distribution for $z$. Indeed, as $\sigma$ increases, the pdf within $]a;b[$ should not be scaled such that the area under the curve is 1 (hence the title "without scaling"); rather, the probability that $z$ is exactly $a$ or exactly $b$ should increase.

Now let's say, I'm not interested in the density per se, but rather in computing the probability $p(x; \epsilon)$ that the true value of whatever $z$ is measuring is $x$ ($\pm \epsilon$). Assuming that I measured $z=\mu$ and that I know the measurement noise $\sigma$, then I think that $p(x; \epsilon)$ can be described as

$ p(x; \epsilon) = \begin{cases} \int\limits_{-\infty}^{a+\epsilon}\mathcal{N}(\mu,\sigma) & \text{if $x=a$}\\ \int\limits_{b-\epsilon}^{\infty}\mathcal{N}(\mu,\sigma) & \text{if $x=b$}\\ \int\limits_{x-\epsilon}^{x+\epsilon}\mathcal{N}(\mu,\sigma) & \text{otherwise} \end{cases} $

1. Is this correct?
2. Is there a name for this form of $p(x;\epsilon)$ or even a name for the underlying probability density, which would seems like a special form of a truncated normal distribution? (-> would help researching on this)
3. Are there other probability distributions that represent are able to represent a similar scenario as outlined above?

Answer 1

The observed $z$ is left and right censored (as opposed to truncated), that is, rather than never seeing observations smaller than $a$ and larger than $b$ (truncation), you observe $z=a$ equivalent to the event $y<a$ and $z=b$ equivalent to $y>b$. The contribution to the likelihood from such observations would be $$ \begin{cases} \Phi(\frac{a-\mu}\sigma) & \text{for }z=a \\ \frac1\sigma\phi(\frac{z-\mu}\sigma) &\text{for }a< z <b\\ 1-\Phi(\frac{b-\mu}\sigma) & \text{for }z=b, \end{cases} $$ where $\phi$ and $\Phi$ is the pdf and cdf of the standard normal distribution. Note that the events that $y=a$ and $y=b$ both have probability zero when $y$ is continuously distributed so you don't need to add anything extra to the probabilities in the first and third cases above to account for the fact that $z$ can take the values $a$ or $b$ without $y$ being censored.

Another thread has a detailed discussion on this kind of likelihood involving a mix between discrete and continuous observations.