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Let $T$ be a random exponentially distributed time. $P \left(T > t \right)=e^{-t}$. Define $M$ via $M_t = 1$ if $t-T \in Q^+$, $M_t = 0$ otherwise. Where $Q^+$ being positive rationals. let $\{F_t\}$ be a filtration generated by the process $M$. We have to prove that for $t\ge s$ and and $F \in F_s$

$$E \left[ 1_{\{t-T \in Q^+\}} \right] = 0 = E \left[ M_S 1_F \right].$$

I do not see the equality can someone please help with the details.

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I apologize if my edits weren't exactly what you meant... still a bit confused on your notation for the Expected Values though. At least it'll be a good exercise on LaTeX for you ^_^. –  Nicolas Villanueva Jun 30 '11 at 20:56
    
thanks very much for the edits Nicolas. I kind of learnt from it and made some edits myself to make the problem clear. –  babu Jun 30 '11 at 21:12

1 Answer 1

The LHS is $E(M_t)=P(T\in t+Q^+)=0$ because the Lebesgue measure of $t+Q^+$ is zero and the distribution of $T$ is absolutely continuous. The RHS is nonnegative and at most $E(M_s)=0$, hence zero. The conditions that $t\ge s$ and that $F\in F_s$ are irrelevant.

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