Is the expression for the distribution of $Z = -\min\{X,0\}$ which I found is correct?

Question

To begin with, if $z < 0$, then $\mathbb{P}(Z \leq z) = 0$.

We can then consider that $z\geq 0$ in order to proceed.

Here is my attempt (edit)

\begin{align*} \mathbb{P}(Z\leq z) & = \mathbb{P}(-\min\{X,0\} \leq z) = \mathbb{P}(\min\{X,0\} \geq -z)\\\\ & = \mathbb{P}((\min\{X,0\} \geq -z)\cap(X > 0)) + \mathbb{P}((\min\{X,0\} \geq -z)\cap(X\leq 0))\\\\ & = \mathbb{P}(X > 0 \geq -z) + \mathbb{P}(-z\leq X \leq 0)\\\\ & = \mathbb{P}(X > 0) + \mathbb{P}(-z \leq X \leq 0)\\\\ & = 1 - \mathbb{P}(X\leq 0) + \mathbb{P}(-z \leq X\leq 0)\\\\ & = 1 - F_{X}(0) - F_{X}(-z^{-}) + F_{X}(0)\\\\ & = 1 - F_{X}(-z^{-}) \end{align*}

Can you anyone critique my solution, so that I can tell if it is wrong or right?

Answer 1

This was an answer to a previous version of the question

Try $X$ uniformly distributed on $[-2,-1]$ and $z=3$.

Then $Z=-\min\{X,0\}$ is uniformly distributed on $[1,2]$ and $\mathbb P(Z \le 3)=1$.

Your expression seems to suggest $\mathbb{P}(Z\leq 3) = 1 - F_{X}(-3) + F_{X}(0) - F_{X}(-3^{-}) $ $=1-0+1-0=2$, which is clearly impossible.

One possible cause is that $\mathbb{P}(X > 0 \geq -z)$ need not be equal to $\mathbb{P}(X > -z)$ and is not in this example.