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Jun 7, 2017 at 0:04 vote accept matematicaActiva
May 29, 2017 at 20:29 comment added Anthony Quas Aren't those quantities numerically equal? I think the distribution of $T_0(\omega)$ is exactly $\nu$.
May 29, 2017 at 18:11 comment added matematicaActiva In this case, I have $\int \log^{+}\left( \delta_{1}(T)+\delta_{2}(T) \right)\nu (dT) <\infty$, but I don`t have $$\int \log^{+}\left( \delta_{1}(T_{0}(\omega))+\delta_{2}(T_{0}(\omega)) \right)\mathbb{P} (d\omega) <\infty,$$ which is what I should have to use your argument, I do not know if that is easy to infer.
May 29, 2017 at 17:52 history answered Anthony Quas CC BY-SA 3.0