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In our homework assignment, we were supposed to find an example showing that the assumption of right-continuity in the statement of the stopping theorem cannot be omitted in general (cf. http://www.math.ethz.ch/education/bachelor/lectures/fs2012/math/bmsc/bmsc_fs12_04.pdf exercise 4-2 c)).

In the hint it said: For a standard exponentially distributed random variable T, consider the process $ M = (M_t)_{t \geq 0} $ given by $ M_t = (T \wedge t ) + 1_{ \{ t \leq T \} } $ together with the $P$-augmentation of the filtration generated by the process $ (T \wedge t) $.

Moreover, we were told that we should try to prove that $ M_t = E[T | \widetilde{\mathcal{F}}_t] $, were $ \widetilde{\mathcal{F}}_t $ denotes the $P$-augmentation of the sigma algebra $ \mathcal{F}_t = \sigma (T \wedge s ; s \leq t) $.

Well, I know that once if proven $ M_t = E[T | \widetilde{\mathcal{F}}_t] $, it follows that $ M $ is a uniformly integrable $ \widetilde{\mathcal{F}}_t $-martingale. Also, I can show that $ T $ is a $ \widetilde{\mathcal{F}}_t $-stopping time. Therefore, if the assumption of right-continuity were not necessary, the stopped process $ M^T $ with $ M_t^T = (T \wedge t ) + 1_{ \{ T \wedge t \leq T \} } = (T \wedge t ) + 1 $ would also be a uniformly integrable martingale (by the stopping theorem). Then, the difference $ N_t := M_t^T - M_t = 1_{ \{ t > T \}} $ would also be a uniformly integrable martingale. But $ E[N_0] = 0 $ whereas $ E[N_{\infty}] = E[1] = 1 $, a contradiction.

Can anybody help me prove $ M_t = E[T | \widetilde{\mathcal{F}}_t] $? Or would anybody happen to know a different counterexample?

Thanks a lot!

Regards, Si

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the basic idea is $T 1_{(T<t)}$ is $F_t$ measurable, $(T > t)$ is an atom in $F_t$ and $E(T \vert T > t) = T +1$ by the memoryless property of the exponential. –  mike May 2 '12 at 10:56
    
@mike: Thanks a lot! Unfortunately, I still don't understand: do you follow from the memoryless property that $ E[T 1_{ \{ t < T \} } | \mathcal{F}_t] = (1 + T) 1_{ \{ t < T \} } $ ? (Ps: I don't have to hand this exercise in, it was due long time ago (as you can see in the link above)) –  Mad Si May 2 '12 at 13:04
    
$T1_{(t<T)}$ is is $F_t$ measurable ( in fact it is $M_t 1_{(t<T)}$ which are 2 $F_t$ measurable functions) so $\mathbb E (T1_{(t<T)} \vert F_t) = T1_(t<T)$. $\mathbb E (T1_{(T>t)} \vert F_t)= \mathbb E(T \vert T>t) 1_{(T>t)}$, which only depends on $T>t$ being an atom in $F_t$, and then u can evaluate $\mathbb E(T \vert T>t)$ using the fact that it is exponential. Discussion of filtrations generated by stopping time in Quantitative Risk Management, McNeil , Frey,Embrechts, –  mike May 2 '12 at 13:49
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