# Determine the conditions for Brownian motion to be a Martingale

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.

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

Hint

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

Also

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

and

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

• $\beta=-1$ and $\alpha , \gamma$ are arbitary – math Jan 2 '17 at 17:30
• 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. – math_apprentice Jan 2 '17 at 17:35
• What is your problem? – math Jan 2 '17 at 17:38
• $X_t= W_t^2+\alpha W_t- t +\gamma$ is a martingale for every $\alpha$ and $\gamma$ – math Jan 2 '17 at 17:40
• How do you know that is a Martingale? – math_apprentice Jan 2 '17 at 17:41