What is the distribution of min{0,X} when X follows some general normal distribution? What is the distribution of $\min\{0, X\}$ when $X$ follows some general normal distribution?
 A: Suppose $X\sim N(\mu, \sigma^2)$ with CDF $\Phi(\cdot; \mu, \sigma^2)$. Then you can calculate the probability $p$ that $X>0$, by taking 
$$ p = 1- \Phi(0; \mu, \sigma^2) $$
Then $Y:=\min\{0,X\}$ is a mixture between


*

*a point mass at $0$ with weight $p$

*and a truncated normal distribution with parameters $\mu$ and $\sigma^2$, truncated from above at $0$, with weight $(1-p)$.


This is also known as a "censored normal distribution". You may be interested in previous threads tagged both "censoring" and "normal-distribution". We have a thread on the mean and variance in the multidimensional case.
A: We have a simple answer for the distribution of $\min(0,X)$ for any random variable $X$ in terms of its distribution function.
Suppose $Y=\min(0,X)$. That is, $$Y=\begin{cases}X&,\text{ if }X\le 0\\0&,\text{ if }X>0\end{cases}$$
Trivially, a plot of $y=\min(0,x)$ for real $x$ would look like

Then distribution function (DF) of $Y$ is simply 
$$P(Y\le y)=\begin{cases}P(X\le y)&,\text{ if }y<0\\1&,\text{ if }y\ge 0\end{cases}$$
For $X\sim\mathcal N(\mu,\sigma^2)$, we get 
\begin{align}
P(Y\le y)=\begin{cases}\Phi\left(\frac{y-\mu}{\sigma}\right)&,\text{ if }y<0\\1&,\text{ if }y\ge 0\end{cases}
\end{align}
, where $\Phi$ as usual is the DF of standard normal distribution.
As already mentioned, $Y$ has a mixed distribution as it contains the mass point $0$ for $X>0$ as well as having a continuous part for $X\le 0$.
