Resolvent operators describe distribution of Lévy Process evaluated at independent exponential times

Question

I'm reading about resolvent operators of Lévy processes. The definition is the next:

$$U^{q}f(x)=\int_{0}^{\infty}e^{-qt}P_{t}f(x)dt,$$ where $(P_{t})_{t\geq 0}$ is the semigroup, $f$ is non negative measurable function and $q>0.$

Using Fubini Theorem we have that $$U^{q}f(x)=E_{x}(\int_{0}^{\infty}e^{-qt}f(X_{t})dt).$$

The book says the follow: From the probabilistic point of view,the resolvent operators describe the distribution of Lévy process evaluated at independent exponential times. That is, if $\tau=\tau(q)$ has an exponential law with parameter $q>0$ and is independent of $X,$ then $E_{\cdot}(f(X_{\tau}))=qU^{q}f(\cdot).$

I don't undertand the above; $$qU^{q}f(x)=E_{x}(\int_{0}^{\infty}qe^{-qt}f(X_{t})dt)$$ so the term $\int_{0}^{\infty}qe^{-qt}f(X_{t})dt$ is like the expectation of $f(X_{t})$ where $X_{t}$ is an " exponential random variable",but the fact is that the time is who has such law and not the process $X.$ I don't understand where is used the independence of $\tau$ with $X.$

What Am I understand incorrectly?

Any kind of help is thanked in advanced.

Answer 1

$$ E_x[f(X_{\tau})] = \int_0^{\infty} E_x[f(X_{\tau})|\tau=t] P(\tau=t)\,d t = \int_0^{\infty} E_x[f(X_{t})] P(\tau=t)\,d t \\= q\int_0^{\infty} e^{-qt}P_tf(x)\,dt =qU^qf(x) $$ where the independece of $\tau$ and $X$ is used to obtain the second equality.