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Consider the process $dX_t=W_t dt+0 dW_t$, alternatively $X_t=\int_0^t W_s ds$. $W_t$ is Brownian motion. I read a proof that $X_t$ is a martingale that simply states "Because the diffusion of $dX_t$ is 0, $X_t$ is not a martingale."

By definition, a stochastic process $X_t$ adapted to a filtration $\{F_t\}$ is a martingale iff $$ E(|X_t|) <\infty, t \geq 0 \tag{1} $$ and $$E(X_t|{\cal F}_s)=X_s, 0\leq s<t \tag{2}$$

Question: What exactly about conditions (1) and (2) establishes that if a random process has 0 diffusion, it is not a martingale?

I am asking because I see the 0-diffusion condition used often for this purpose, but in the above example, of a process which is still random even though it has a zero diffusion, I don't get it.

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    $\begingroup$ Full answer here. $\endgroup$ – Did Oct 26 '17 at 22:25
  • $\begingroup$ A constant process has no diffusion term, but is also a martingale. $\endgroup$ – Theoretical Economist Oct 26 '17 at 23:05
  • $\begingroup$ Thanks Did for showing the question was asked before. What do you think of Theoretical Economist's counterexample? $\endgroup$ – Lars Ericson Oct 27 '17 at 4:01
  • $\begingroup$ Also Did, not to pick nits, but not all martingales are local martingales, the other question you referenced is particular to local martingales. $\endgroup$ – Lars Ericson Oct 27 '17 at 4:15

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