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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 $\textrm{max}(a,b) = \left\{ \begin{array}{l l} a & \text{if }a>b\\ b & \text{otherwise}\\ \end{array} \right.$ ?

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3 Answers 3

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$$ \Pr(Y\le y) = \begin{cases} 0 & \text{if }y<0 \\[8pt] \Pr(X\le 0) & \text{if }y=0 \\[8pt] \Pr(X\le y) & \text{if } y>0 \end{cases} $$ 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 $Z\sim\mathcal{N}(0,1)$.

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Ah yes, and indeed the pdf integrates to one, which was what was bothering me in the first place. Great! –  Jonathan Nov 9 '12 at 16:33

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 $$ p=F_X(0),\qquad f_Y=f_X\,\mathbf 1_{(0,+\infty)}. $$

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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.

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