I need to compute $\Bbb{E}(\tau^{X} \ \Bbb{1}_{\{\tau^{X}<+\infty\}})$ where:
$1) $ $\tau^y$ is a r.v. representing the time spent by a particle until it "jumps", ( $ y \in R_{\geq 0} $ is the starting position of the particle ). Its law is given by
$$ P(\tau^y > t)= e^{y(e^{-t}-1)}$$ Observe that $ P(\tau^y =+\infty ) = e^{-y}$ . I've also computed (for other purposes,but I think it may help) the law and expectation of $\hat{\tau}^y:=\tau^y \Bbb{1}_{\{\tau^{y}<+\infty\}} $: $$P(\hat{\tau}^y>t)=e^{-y}(e^{ye^{-t}}-1)\ ;\ \Bbb{E}(\hat{\tau}^y)=e^{-y}\int_0^y\frac{e^s-1}{s}ds $$ $2)$ $ X $ is a R.V. with values in $R_{\geq 0} $ with $$ F_X(a) = \begin{cases} 0 \ \ \ \ \ \ if\ \ a<0\\ e^{-2}\ \ \ \ \ if\ \ a\in[0,1]\\ e^{2(a-2)}\ \ \ if\ \ a\in(1,2)\\ 1 \ \ \ \ \ \ if\ \ a\geq 2 \end{cases} $$
Note that $P(X=0)=e^{-2}$, so it has an atom in $0$
I thought about conditional expectation, conditioning on $X$ , but I'm not sure how to handle it.
@Did
A few questions about your answer:
1) Which property did you use to get that result? Could you lend me useful bibliography where that property appears?
2) When computing $u(s)$ did you "compute" the "density" of $X$ ? ( I put the "" because I'm not sure it DO have density, since $F_X$ has a jump at 0; and also I'm wondering if a minus is missing .