If $X$ has CDF $F$, how can I find the CDF of $U= \max \{0,X \}$? If $X$ has CDF $F$, how can I find the CDF of $U=\max\{0,X\}$? Obviously the suport of $U$ consists solely of nonnegative values. 
Am I right then in thinking  that for $u=0, F_U (u)=F_X(0)$ and for  $u > 0$, $F_U (u)
 =F_X(u)$?
Thank you.
 A: $F_{U}\left(u\right)=P\left\{ \max\left\{ X,0\right\} \leq u\right\} =P\left\{ \omega\in\Omega\mid X\left(\omega\right)\leq u\wedge0\leq u\right\} $ and
$F_{X}\left(u\right)=P\left\{ X\leq u\right\} =P\left\{ \omega\in\Omega\mid X\left(\omega\right)\leq u\right\} $
If $u<0$ then $\left\{ \omega\in\Omega\mid X\left(\omega\right)\leq u\wedge0\leq u\right\} =\emptyset$
so that in that case $F_{U}\left(u\right)=0$
If $u\geq0$ then $\left\{ \omega\in\Omega\mid X\left(\omega\right)\leq u\wedge0\leq u\right\} =\left\{ \omega\in\Omega\mid X\left(\omega\right)\leq u\right\} $
so that in that case $F_{U}\left(u\right)=F_{X}\left(u\right)$.
A: Clearly, for $u<0$, $F_U(u) = \mathbb{P}(\max\{0,X\} \leq u)= 0$.  Now, for $u\geq 0$,
\begin{align}
F_U(u) &= \mathbb{P}(\max\{0,X\} \leq u) \\
&=\mathbb{P}(X<0)\mathbb{P}(\max\{0,X\} \leq u|X<0) + \mathbb{P}(X\geq 0)\mathbb{P}(\max\{0,X\} \leq u|X\geq0)\\
&= \mathbb{P}(X<0) +  \mathbb{P}(X\geq 0)  \mathbb{P}(X \leq u |X\geq0)\\
&=\mathbb{P}(X<0) + \mathbb{P}(X\geq 0)  \frac{\mathbb{P}(0\leq X \leq u)}{\mathbb{P}(X \geq0)}\\
&=\mathbb{P}(X<0) + \mathbb{P}(0\leq X \leq u) \\
&\overset{(a)}=\mathbb{P}(\{X<0\}\cup\{0\leq X \leq u\})\\
&=\mathbb{P}(X\leq u)\\
&=F_X(u),
\end{align}
where $(a)$ follows since the events $\{X<0\}$, $\{0\leq X \leq u\}$ are mutually exclusive (for $u \geq 0$). Sumarizing,
\begin{equation}
F_U(u) = \begin{cases}
0 & u<0\\
F_X(u) & u \geq 0
\end{cases}
\end{equation}
