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Let $B=(B_t)_{0\le t\le T}$ be a continuous semi-martingale and $\mathbb F=(\mathcal F_t)_{0\le t\le T}$ be its natural filtration. Denote by $\mathcal C_b(\Omega\times \mathbb R_+)$ the space of continuous bounded functions $F:\Omega\times \mathbb R_+ \to \mathbb R$, where $\Omega$ denotes the space of continuous functions defined on $[0,T]$ endowed with the uniform norm, and $\Omega\times\mathbb R_+$ is equipped with the product topology.

Now, given a sequence of $\mathbb F-$stopping times $(\tau_n)_{n\ge1}$ s.t. there exists a random variable $\tau$ satisfying

$$\lim_{n\to \infty}\mathbb E[F(B,\tau_n)]~~=~~\mathbb E[F(B,\tau)],~~ \mbox{ for all } F\in\mathcal C_b(\Omega \times \mathbb R_+).$$

My question is: could we show that $\tau$ is an $\mathbb F-$stopping time? If not, what about assuming that $B$ is Markov or even a Brownian motion? Thanks a lot for the reply!

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In the case of Brownian motion you can look at functions of the form $$g(\tau_n \wedge t) e^{i \sum \lambda_i(B_{t_i} - B_t)}$$ where $t_i > t$. Since the expectation is $$ \mathbb E(g(\tau_n \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ which converges to $$ \mathbb E(g(\tau \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ $g(\tau \wedge t)$ is independent of $\mathcal F (B_s - B_t), s>t$ which should make it $\mathcal F_t$ measurable. $${}$$ As pointed out below, that doesn't work, however the brownian case can be done completely differently by
1. continuous bounded functions of $B_{t_1},...,B_{t_n}$ are dense in $L^2(\mathcal F_{\infty}) $ because, e.g. lemma 5.3.1 of Revuz and Yor shows that linear combinations of $ sin(\sum \lambda_i B_{t_i}), cos(\sum \lambda_i B_{t_i})$ are dense. $${}$$2. Let $\epsilon > 0$. Approximate $\tau \wedge t$ to within $\frac {\epsilon} 2$ by a continuous bounded function of $B_{t_1},...,B_{t_n}$ which I'll call f. By the hypotheses $\mathbb E(f-\tau_n \wedge t)^2 \rightarrow \mathbb E(f-\tau \wedge t)^2 < \frac {\epsilon} 2$ Therefore by the triangle inequality $limsum E(t_n \wedge t - \tau \wedge t)^2 < \epsilon$. This shows that $\tau_n \wedge t \rightarrow \tau \wedge t$ in $L^2$, and therefore $\tau^t $ is a stopping time. (exercise 1.4.17 of revuz & yor). $${}$$

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  • $\begingroup$ @ michael Thanks for the reply. Yes, I get the proof for a BM. Do you have an idea for a general diffusion process? If not, could we find some stopping time $\sigma$ s.t. $(B,\tau)$ has the same law of $(B,\sigma)$? $\endgroup$ – CodeGolf Mar 30 '16 at 7:28
  • $\begingroup$ I think with a lot of diffusions (bounded, smooth parameters say) you can use time changes & girsanov to convert it to the Brownian motion problem $\endgroup$ – user83457 Mar 30 '16 at 8:58
  • $\begingroup$ @ michael I will try it. Could you specify a bit more on how to show that $\tau\wedge t$ is $\mathcal F_t$ measurable? $\endgroup$ – CodeGolf Mar 30 '16 at 9:37
  • $\begingroup$ Knowing that $\tau \wedge t$ is independent of $\mathcal F(B_s-B_t, s\ge t)$, how can we show $\tau \wedge t$ is $\mathcal F_t$ measurable? $\endgroup$ – CodeGolf Mar 30 '16 at 9:38
  • $\begingroup$ @michael's conclusion about the measurability of $\tau\wedge t$ needs some justification. Example: Toss two fair dice; let $X_i$ be the number showing on the top of die $i$. Let $Y$ be the indicator of the event that $X_1+X_2$ is even. The $\sigma$-algebra of interest is $\mathcal F=\sigma(X_1,X_2)$, and (i) $X_1$ and $X_2$ are independent, (ii) $Y$ is $\mathcal F$-measurable, (iii) $Y$ is independent of $X_2$, but (iv) $Y$ is not $\sigma(X_1)$ measurable. $\endgroup$ – John Dawkins Apr 2 '16 at 17:28

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