If $X$ has CDF $F$, how can I find the CDF of $U= \max \{0,X \}$?

Question

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.

Answer 1

$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 $F_{U}\left(u\right)=F_{X}\left(u\right)$.

Answer 2

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}