pdf of a clamped gaussian random variable Having a Gaussian random variable X distributed as $\mathcal{N}(\mu, \sigma^2)$, is it possible to determine the probability distribution function of $Y = \textrm{max}(X,0)$, having 
$\textrm{max}(a,b) = \left\{
  \begin{array}{l l}
    a & \text{if }a>b\\
    b & \text{otherwise}\\
  \end{array} \right.$ ? 
 A: Let $f_X$ and $F_X$ denote the PDF and the CDF of $X$. Then the distribution of the random variable $Y$ has an atom at $0$ of mass $p$ and a density $f_Y$ on $(0,+\infty)$, where 
$$
p=F_X(0),\qquad f_Y=f_X\,\mathbf 1_{(0,+\infty)}.
$$
A: Let's find the cumulative distribution function $F_Y(y)$ of $Y$. If $y\lt 0$, then $F_Y(y)=\Pr(Y\le y)=0$.
If $y=0$, then $F_Y(y)=\Pr(Y\le 0)=\Pr(X\le 0)$. If we let $F_X(x)$ be the cumulative distribution function of our normal, this is just $F_X(0)$. You may want to express this number in terms of the standard normal.
For $y\gt 0$, $F_Y(y)=F_Y(0)+\Pr(0\lt X\le y)$. This simplifies to $F_X(y)$. 
We end up with a point mass at $0$, with the rest familiar. 
A: $$
\Pr(Y\le y) = \begin{cases} 0 & \text{if }y<0 \\[8pt] \Pr(X\le 0) & \text{if }y=0 \\[8pt] \Pr(X\le y) & \text{if } y>0  \end{cases}
$$
The third piece in this piecewise definition you presumably already know about; it's the CDF of $X$.  The second piece is the value of that same CDF at one point.
If it helps, notice that
$$
\Pr(X\le 0) = \Pr\left( \frac{X-\mu}{\sigma} \le \frac{0-\mu}{\sigma} \right) = \Pr\left( Z \le \frac{-\mu}{\sigma} \right)
$$
Where $Z\sim\mathcal{N}(0,1)$.
