Assuming that $\{ W ( t ) | t \geq 0 \}$ is a Brownian motion, I'm trying to check whether the process $$X ( t ) = W ( t ) + 4 t$$ is martingale with respect to filter $\mathcal{F}_t$. For this we should check if $\mathbb{E}[X(t)|\mathcal{F}_s] = X(s)$.

My try: \begin{align*} \mathbb{E}[W(t) + 4t|\mathcal{F}_s] &= \mathbb{E}[(W(t) - W(s) + W(s)) + 4t]\\ &= \mathbb{E}[(W(t) - W(s))] + \mathbb{E}[W(s)] + \mathbb{E}[4t]\\ &= t - s + W(s) + 4t\\ &= W(s) - s + 5t\\ \end{align*} Then since $W(s) - s + 5t \neq W(s) + 4s$, this is not martingale.

First, I was wondering if the approach is correct. Secondly, a good number of online examples are using Itô's formula to check for martingale. I was wondering if there is a measure by which we choose what way to solve it or it is just to see if we can find the expectation easily or not.

  • 2
    $\begingroup$ Yes, your approach is correct (and in my oppinion much better than applying Itô's formula which is straight overkill). When you have to check whether a process is a martingale, you should always check first whether the process has constant expectation. If it does not have constant expectation (as in this case), the process cannot be a martingale, and you are done. $\endgroup$ – saz Mar 4 '19 at 6:39
  • $\begingroup$ For one thing $\mathbb{E}[(W(t)-W(s))]=t-s$ is not correct, it should be zero $\endgroup$ – Blade Mar 6 '19 at 0:17
  • $\begingroup$ Ah right, sorry, I missed that. There are also three "$\mid \mathcal{F}_s$" missing, as I see now. $\endgroup$ – saz Mar 6 '19 at 7:09

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