I've got, again, a problem to determine the CDM of a random variable $Z$. In fact, we consider :
$F_X(x) = \frac{1}{2}1_{[0;1[} + (1-e^{-x})1_{[1;+\infty[}$ which is a cumulative distribution function of a random variable that we call $X$. Let : $Z=X1_{[1;+\infty[}(X)$. We want to determine the CDM of $Z$. This is what I did :
- if $x < 0$, then we have : $F_Z(x) = P(X1_{[1;+\infty[}(X) \leq x) = 0$ cause $X1_{[1;+\infty[}(X) \geq 0$.
- if $x \in [0;1[$, then : $F_Z(x) = P(X1_{[1;+\infty[}(X) \leq x) = P(X \in ]-\infty;1[)) = F(1^-) = \frac{1}{2}$
- if $x \in [1;+\infty[$, then : $F_Z(x) = P(X1_{[1;+\infty[}(X) \leq x) = P(X \in ]-\infty;1[ \cup [1;x]) = F(x) = 1-e^{-x}$.
So, I find $F_Z = F_X$, and that's a problem I think... I should not have find this... But I don't see my mistake !
Somebody could help me, again ?
Thank you !
Write $X=\frac12 (1-B) + BY$ where $B\sim\mathrm{Ber}\left(\frac12\right)$ and $Y\sim\mathrm{Exp}(1)$. Then $$ \mathsf 1_{[1,\infty)}(X) = \mathsf 1_{[1,\infty)}(X)\cdot\mathsf 1_{\{B=0\}} + \mathsf 1_{[1,\infty)}(X)\cdot\mathsf 1_{\{B=1\}} = \mathsf 1_{\{B=1\}}, $$ so $Z=X\cdot \mathsf 1_{\{B=1\}}$. It follows that $Z\stackrel d=Y$, i.e. $Z$ has $\mathrm{Exp}(1)$ distribution.