Let $W_T$ denote normalised univariate Brownian motion and let

$X_t = W_t^2 + \alpha W_t + \beta t + \gamma$

where $\alpha, \beta$ and $\gamma$ are constants. Determine conditions on these constants such that $X_t$ is a Martingale.

  • $\begingroup$ you can use ito's lemma or compute the conditional expectation directly. $\endgroup$ – Gordon Jan 2 '17 at 16:39
  • $\begingroup$ @Gordon. Thanks. Please could you expand on this a little? I'm completely new to this. $\endgroup$ – math_apprentice Jan 2 '17 at 16:46
  • $\begingroup$ @Gordon is it simply just determining for which values of $\alpha,\beta,\gamma$ the $dt$ term is equal to zero? $\endgroup$ – math_apprentice Jan 2 '17 at 17:09
  • $\begingroup$ Yes. That is correct. $\endgroup$ – Gordon Jan 2 '17 at 17:16
  • $\begingroup$ @Gordon isn't that true for any values of $\alpha,\beta,\gamma$ as the second derivative $dt$ are zero for all terms? $\endgroup$ – math_apprentice Jan 2 '17 at 17:23


Let $s<t$. $$E[W_t^2|F_s]=E[(W_t-W_s)^2|F_s]+2 E[(W_t-W_s)W_s|F_s]+E[W_s^2|F_s]$$ so $$E[W_t^2|F_s]=E[(W_t-W_s)^2]+2W_s E[(W_t-W_s)]+W_s^2=t-s+W_s^2$$


$$E[\alpha W_t|F_s]=\alpha E[W_t-W_s|F_s]+\alpha E[ W_s|F_s]=\alpha E[W_t-W_s]+\alpha W_s=\alpha W_s$$


$$E[\beta t+\gamma |F_s]=\beta t +\gamma$$

  • $\begingroup$ $\beta=-1$ and $\alpha , \gamma $ are arbitary $\endgroup$ – math Jan 2 '17 at 17:30
  • $\begingroup$ Thanks @math. I'm not sure I completely follow this argument so please can you explain a little more? I follow the maths I'm just not sure why, for instance, $t-s+W_s^2$ means that term is a Martingale in that form. $\endgroup$ – math_apprentice Jan 2 '17 at 17:35
  • $\begingroup$ What is your problem? $\endgroup$ – math Jan 2 '17 at 17:38
  • $\begingroup$ $X_t= W_t^2+\alpha W_t- t +\gamma $ is a martingale for every $\alpha$ and $\gamma$ $\endgroup$ – math Jan 2 '17 at 17:40
  • $\begingroup$ How do you know that is a Martingale? $\endgroup$ – math_apprentice Jan 2 '17 at 17:41

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