Let $F(x) = \frac{1}{2} 1_{[0,1[} + (1-e^{-x})1_{[1,\infty[}$ the cumulative distribution function of $X$.
Let $Z=X 1_{]1,\infty[}$. I have to calculate the cumulative distribution function of $Z$. Then, I have to prove that $Z$ can be written than $Z=BU+(1-B)V$ with $B,U,V$ are random variables independant, $U$ discrete and $V$ continue.
I wrote $F(x) = \begin{cases} 0 & \text{ if } x<0 \\ 1/2 & \text{ if } 0 \leq x < 1 \\ 1-e^{-x} & \text{ if } x\geq 1 \end{cases}$
Then $P(Z \leq x) = \begin{cases} 1-e^{-x} & \text{ if } x > 1 \\ 0 & \text{ otherwise } \end{cases}$.
But I'm not sure it's the cumulative distribution function of $Z$. Someone could help me ? Thank you in advance.
Notice that $Z:=X\mathbf 1_{X\in\,]0,\infty[}$ means that:$${\begin{split}\{Z\leqslant z<0\}&=\{\}\\ \{0\leqslant Z\leqslant z\leqslant 1\}&=\{X\leqslant 1\} &\qquad = \{Z=0\}\\\{1<Z\leqslant z\} &= \{1<X\leqslant z\}\end{split}}$$
Also recall that CDF is the Cummulative Distribution Function.