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I was wondering with respect to what filtration the process $X_t=\begin{cases}\frac{1}{t}B_{1/t}\text{ if }t>0 \\ 0\text{ if }t=0 \end{cases}$ $\;\;\;$($B_t$ is a Brownian motion) is a martingale. Is it with respect to $\mathcal{F_{1/t}}$ where $\mathcal{F}_t$ is the natural filtration of the Brownian motion? Is it what we call a reverse martingale?

roi_saumon
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$X$ is not a martingale with respect to any filtration. If $X$ is a martingale, then $X^2$ is a sub-martingale, and in particular $t \mapsto \mathbb{E}[X_t^2]$ is increasing. For this $X$, we have $$\mathbb{E}[X_t^2] = \frac 1{t^2} \mathbb{E}[B_{1/t}^2] = \frac 1{t^3},$$ which is decreasing. Hence $X^2$ cannot be a sub-martingale, and therefore $X$ cannot be a martingale with respect to any filtration.

user6247850
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  • Thank you it is more clear now. I was wondering if it was possible to show directly $E[X_t|\mathcal{G}_s]=X_s$ for $t\ge s$ – roi_saumon Oct 19 '21 at 15:50
  • You would show it the same way as you do for a normal Brownian motion - Show that $X_t - X_s$ is independent of $\mathcal G_s$, then write $X_t = X_t - X_s + X_s$ before taking the conditional expectation. – user6247850 Oct 19 '21 at 15:56
  • That would work if we had $X_t-X_s$ independent on $\mathcal{G}_s$ right? So that $E[X_t-X_s|\mathcal{G}_s]=E[X_t-X_s]$. – roi_saumon Oct 19 '21 at 16:01
  • @roi_saumon Sorry, I realized that I was thinking of the process $t B_{1/t}$ instead of $\frac 1t B_{1/t}$ - I will have to rethink my earlier comment and whether or not it's a martingale. – user6247850 Oct 19 '21 at 16:08
  • very well. Also I couldn't read your comment on why $E[X_t|\mathcal{F_{1/s}}]=E[X_s]$ was wrong – roi_saumon Oct 19 '21 at 16:10
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    It was just a counterexample - if you let $t=1$ and $s = \frac 12$, $E[X_1|\mathcal F_{1/s}] = E[B_1 | \mathcal F_2] = B_1 \ne X_{\frac 12} = 2 B_2$. – user6247850 Oct 19 '21 at 16:16
  • @roi_saumon I edited my answer - as it turns out, $X$ is not a martingale with respect to any filtration. – user6247850 Oct 19 '21 at 16:34
  • Thank you. But is is a reverse martingale though? Like if we consider a reverse filtration $\mathcal{F}_s\supset\mathcal{F}_t$ if $s\le t$ then $E[X_s|\mathcal{F}_t]=X_t$ – roi_saumon Oct 19 '21 at 19:06
  • No. That would imply $X_t = E[X_0|\mathcal F_t] = E[0|\mathcal F_t] = 0$ for all $t$. – user6247850 Oct 19 '21 at 19:17
  • I see. It is just that I am confused with this post, because $B_t/t$ is the same as $t B_{1/t}$ with a change of variable no? – roi_saumon Oct 19 '21 at 19:40
  • Yes, but you are asking about $\frac 1t B_{1/t}$, not $t B_{1/t}$, as I mentioned earlier before editing my post. $t B_{1/t}$ is a martingale, but $\frac 1t B_{1/t}$ is not. – user6247850 Oct 19 '21 at 19:42
  • oh no... than I did the same mistake. I also had $tB_{1/t}$ in my mind, sorry. But then $tB_{1/t}$ you mentioned one can prove the martingale property in the same way as for the brownian motion right? But why is $X_t-X_s$ independent of $\mathcal{G}_s$? – roi_saumon Oct 19 '21 at 19:47
  • You couldn't have brought up that you were thinking of $t B_{1/t}$ when I explicitly mentioned it? Just show $X_t-X_s$ is independent of $X_u$ for all $u \le s$. Use that uncorrelated and jointly Gaussian implies independent. – user6247850 Oct 19 '21 at 19:55
  • sorry again about that. Last thing : I get that $X_t-X_s$ and $X_u$ are uncorrelated and $X$ is jointly Gaussian but if $X_t-X_s$ is independent of $X_u$ for every $u\le s$ does it really imply $X_t-X_s$ is independent of $\sigma(X_u : u\le s)$? – roi_saumon Oct 19 '21 at 20:05
  • Yes, by the monotone class theorem. – user6247850 Oct 19 '21 at 20:51
  • I was thinking again about the case $X_t=tB_{1/t}$. Then for $t>0, X_t$ is measurable with respect to $\sigma{X_s:s\le t}=\sigma{sB_{1/s}: s\le t}=\sigma{\frac{1}{u}B_u : \frac{1}{u}\le t}$ but is this equal (without the $1/u$) to $\sigma{B_u : \frac{1}{u}\le t}$? – roi_saumon Oct 21 '21 at 11:08
  • Yes. It would be better to ask new questions than to continue asking questions in the comments of this one. – user6247850 Oct 21 '21 at 14:23
  • I was trying to work the details concerning your previous comment that explains why if $X_t-X_s$ is independent of $X_u$ for every $u\le s$ then $X_t-X_s$ is independent of $\sigma(X_u:u\le s)$. For this I noticed $\sigma(X_u:u\le s)=\sigma\left(\bigcup\limits_{u\le t}\sigma(X_u)\right)$. However, it seems $\bigcup\limits_{u\le t}\sigma(X_u)$ is not a $\cap$-stable system since ${X_u\in A}\cap{X_v\in B}={X_u\in A,X_v\in B}\notin \bigcup\limits_{u\le t}\sigma(X_u)$ no? So we could not apply the monotone class lemma – roi_saumon Oct 26 '21 at 09:45