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Let $M$ a càdlàg martingale square integrable.

My question is simple : Does a continuous version of M exist in every situation (like a jump in a rational time) ? or which condition(s) could I set on $M$ to get a continuous version ?

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    Er...I don't see how you think that there could possibly always be a continuous version. Say you have a sequence of iid exponential random variables $\tau_k$ and $t_n=\sum_{k=1}^n \tau_k$, then define $X_t=\sum_{n: t_n \leq t} 2B_n-1$ where $B_n$ are iid Bernoulli(1/2) variables. (This is like a Poisson process except that instead of always jumping up by 1, you independently choose whether to jump up or down by 1 with equal probabilities.) Then $X_t$ is a cadlag square integrable martingale, but its jumps are "intrinsic", there is no getting rid of them. – 2017-02-17
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    I ask this question because during the class my prof has spoken about the martingale representation theorem and has written at the end of the theorem and without explanation, that a continuous version of the martingale $M$ exists. So my question is if the martingale is cadlag and by the mean of this theorem could we find a continuous version of $M$... (Maybe my question was set too general and look like idiot in a first read) – 2017-02-17
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    Can you state the theorem you're referring to more specifically? The one I'm finding is about measurability with respect to a Brownian motion's filtration, which can't possibly hold for a jump process (and as I showed in my example, there are jump processes that are martingales and "nice"). – 2017-02-17
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    Let a probability space $(\Omega,\mathcal F,P)$, let $B=(B_t)_{t\geq 0}$ an $\mathcal F_t$ brownian motion valued in $\mathbb R^d$, $B_0=0$. We assume $\mathcal F$ is the natural filtration of $B$ and potentially complete and right continuous. Let $M=(M_t)_{t\geq 0}$ an $\mathcal F_t$-martingale square integrable. Then it exits a "unique" process $H=(H_t)_{t\geq 0}$ valued in $\mathbb R^d$, predictible, s.t. $\mathbb E(\int^t_0\mid H_s\mid^2ds)<+\infty$, s.t a.s : $$M_t=E(M_0)+\int^t_0H_sdB_s, t\geq 0$$ **In specific, $M$ admits a continuous version (no proof)** – 2017-02-17
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    Yeah, the key assumption here that is *not* trivial is that $M$ is a $\mathcal{F}_t$ martingale. Note that $\mathcal{F}_t$ is not the natural filtration of $M_t$, but rather the natural filtration of $B_t$. So we're saying "given a process which is filtered by the natural filtration of a *different* process, here is a relationship between the two". This relationship is very similar to the statement "if $X$ is a random variable and $Y$ is $\sigma(X)$ measurable then there exists a Borel function $\phi$ such that $Y(\omega)=\phi(X(\omega))$". – 2017-02-17
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    As for why there's a continuous version under these circumstances, that's not *that* hard to understand in view of properties of the Ito integral. – 2017-02-17
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    Hence in a filtration only generated by a brownian, every cadlag martingale have a continuous version (because of the Ito integral property) ? – 2017-02-17
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    Right...but this constraint is so strong that it is not as useful as it sounds. – 2017-02-17
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    why ? what could be the problem ? – 2017-02-17
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    Eh, it's easier to see from the conclusion of the theorem perhaps: there is a massive class of processes that violate this assumption, so this is not a good tool for "classification of processes". – 2017-02-17

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