# Is every right continuous local martingale of finite variation constant?

I was reading a chapter in Dellacherie and Meyer. Suppose we have right continuous adapted processes $A$, $A'$ of finite variation. Both are null at zero and the difference is a local martingale.

I know the following lemma:

A continuous local martingale of finite variation is constant.

In Dellacherie and Meyer they conclude that in the above situation $A=A'$. So can continuity in the above lemma be relaxed to right continuity?

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Do you know the proof of the lemma? What wouldn't work if the local martingale is only right-continuous? –  Davide Giraudo Sep 23 '12 at 8:47
@DavideGiraudo I know the proof, but there we are using that every continuous adapted process is locally bounded, which is not true for RCLL and therefore of course not true for RC processes. –  user20869 Sep 23 '12 at 9:08
A right continuous version of a compensated Poisson process provides an exemple of a right continuous FV martingale which is not constant. You need predictability of the difference to conclude that A = A'. –  Alexandre C. Jun 9 at 19:12

I think that the most general result is the following:

Every predictable local martingale of finite variation is a constant
(up to indistinguishability)


But in general a right-continuous adapted process $X=(X_t)_{t\geq 0}$ is not predictable (only $X_{-}$ is).

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Do you have a reference or a proof for this result? –  Davide Giraudo Sep 23 '12 at 12:05
@DavideGiraudo: See Jacod and Shiryaev's Limit Theorems for Stochastic Processes p. 32. –  Stefan Hansen Sep 23 '12 at 12:51
It was missing that $A,A'$ are predictable. Thanks for your answer. I didn't know this result! –  user20869 Sep 23 '12 at 15:24

How about the following example? I've stripped it down as much as possible.

Let $\Omega = \{H, T\}$ be a probability space consisting of a single fair coin flip (i.e. $\mathcal{F} = 2^\Omega$, $P(A) = \frac{1}{2} |A|$). Define the filtration $$\mathcal{F}_t = \begin{cases} \{\emptyset, \Omega\}, & t < 1 \\ \mathcal{F}, & t \ge 1. \end{cases}$$ This satisfies the usual conditions of completeness and right continuity. Now take $$M_t(\omega) = \begin{cases}0, & t < 1 \\ 1, & t \ge 1,\, \omega = H \\ -1, & t \ge -1,\, \omega = T. \end{cases}$$ In words, $M_t$ says "flip a coin at time 1". $M_t$ is an adapted, right continuous (indeed RCLL) martingale of finite variation, but it is not constant.

This would appear to contradict your statement from Dellacherie and Mayer, so you may want to check that you have read it correctly.

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Thank you for your example. As I wrote above, it was missing, that $A,A'$ are predictable. Then you could apply the Theorem stated by Stefan. –  user20869 Sep 23 '12 at 15:26