Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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. 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

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.