What is the distribution of $\min\{0, X\}$ when $X$ follows some general normal distribution?

  • $\begingroup$ Maybe it can help to know that $\forall x,y\in\Bbb R,\;\min(x,y)= \dfrac{x+y-|x-y|}2$. $\endgroup$ – paf Aug 2 '18 at 10:37

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 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.

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  • 1
    $\begingroup$ +1. However, censoring may be a little more refined than suggested here. You describe a "delta (truncated) normal" distribution or "zero inflated" truncated normal, rather than a censored normal. The distinction is that in the present case the zeros will be evaluated as zeros whereas with censoring the zeros are considered to be the intervals $[0, \infty).$ This isn't a superficial difference: it changes likelihood calculations for instance. $\endgroup$ – whuber Aug 2 '18 at 12:02

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

enter image description here

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 $X\le 0$.

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