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A local martingale is a martingale iff it is in the class DL.

The condition: for every $t\in[0,\infty)$

$$E[\sup\limits_{0\leq s\leq t} |M_s|]<\infty\tag1$$

guarantees a local martingale $M$ is a martingale by ensuring it satisfies the condition for being in the class DL. Moreover, by Burkholder-Davis-Gundy, this means: for every $t\in[0,\infty)$,

$$E[\langle M\rangle^{1/2}_t]<\infty$$

My question is: do there exist (simple?) examples of continuous martingales where these two conditions are violated?

What I am really asking is that these are known to be sufficient conditions to guarantee local martingales are martingales, why are they not necessary?

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  • $\begingroup$ I misunderstood the question and thus my answer does not answer the question. Therefore, I erased it. $\endgroup$ – ofer zeitouni Jul 17 '14 at 13:36
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There exist indeed a uniformly integrable martingale $X$ whose max $\sup_{k\in \mathbb{N}} |X_k|$ is not integrable. A simple example in discrete time can be found here http://www.math.fsu.edu/~nichols/martingalezoo.pdf (see the last 2 lines of point 5).

A related construction is found in example 4.1 of the paper by A.S. Cherny "Some particular problems of martingale theory", which produces a uniformly integrable martingale $(X_n)_{n\in \mathbb{N}}$ and a bounded process $(H_n)_{n\in \mathbb{N}}$ such that the stochastic integral $M_n:=\sum_{k=0}^n H_k (X_{k+1}-X_k)$ is a martingale which is not uniformly integrable (by the Burkholder-Davis-Gundy inequality this implies that $\sup_{k\in \mathbb{N}} |X_k|$ is not integrable)

A second (completely different) example in continuous time is sketched in Exercise 3.15 of the 2nd Chapter of Revuz-Yor's Book "Continuous Martingales and Brownian Motion, 3rd ed"

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  • $\begingroup$ I just ran into yet another example of a uniformly integrable martingale which does not satisfy condition (1), this one with time index $[0,\infty)$: see section 4 in arxiv.org/pdf/1508.07564.pdf . This paper also discusses in detail the relation between local martingales and uniformly integrable martingales $\endgroup$ – pietro siorpaes Sep 19 '18 at 9:43
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I guess the following is a counter-example: take $B$ a standard two-dimensional Brownian motion and $$ M_t = \log ( \Vert B_t \Vert ) $$ $M$ is a local martingale (compute the SDE satisfied by $M$, rembering that $\Vert B \Vert$ is a Bessel process) , but it does not hold that $$ E( \langle M \rangle_t ) < \infty $$ nor that $$ E( \sup_{0 \le s \le t} |M_s|) < \infty. $$ (to be checked; I am not 100% sure about this last statement).

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    $\begingroup$ I'm confused: does $B_t$ start at 0? Then surely $M_t$ is not continuous? And aren't we looking for a martingale, not a local martingale? $\endgroup$ – Nate Eldredge Jul 22 '14 at 2:35
  • $\begingroup$ You could let $B$ start somewhere else than $0$, but then I agree there is always a possibility that $B$ comes back at $0$ at some point. There are similar examples in higher dimensions, where $B$ is transient, but OK, this does not guarantee that $M$ is strictly speaking continuous. Regarding your second comment, we are looking for a local martingale, not a martingale. $\endgroup$ – Olivier Leveque Jul 22 '14 at 7:28
  • $\begingroup$ No, I was looking for a martingale, not a local martingale.I am saying the conditions given are sufficient conditions for martingales. I want to see why they are not necessary conditions. $\endgroup$ – lost1 Feb 20 '15 at 15:50

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