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I am trying to determine for what values of a, b the process $X_t=e^{aW_t+bt}, t \ge 0$ is a martingale with respect to $F_t^{W}$. Here $W_t$ is a brownian motion.

I know I need to show that $\mathbb E(X_t|F_s^{W})=X_s$, but I am not sure how to compute $\mathbb E(e^{aW_t+bt}|F_s^{W}).$ Any help would be appreciated.

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    $\begingroup$ Have you tried decomposing $W_t = W_s + (W_t-W_s)$? It's the only trick I know, but it works. $\endgroup$ Apr 19, 2017 at 22:58
  • $\begingroup$ How do I deal with the expectation of the exponential though? $\endgroup$
    – user75514
    Apr 19, 2017 at 23:00
  • $\begingroup$ The brownian motion $W_t$ has a normal distribution. Have you done any exercise about deriving the moment generating function of normal before? $\endgroup$
    – BGM
    Apr 20, 2017 at 4:27
  • $\begingroup$ Yes, I am familiar with MGFs. So I can substitute the MGF for $W_t$ for the expression inside the conditional expectation, but I am not sure how to tease the rest out. $\endgroup$
    – user75514
    Apr 20, 2017 at 20:57

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Answer here Determine when the process $\exp(\lambda t + \sigma W(t))$ is a sub- or a supermartingale

I will briefly write a solution.

Let $X \sim \mathcal{N}(0, 1)$: $$\mathbb{E}e^{\alpha X} = \int\limits_\mathbb{R}e^{\alpha z}\dfrac{1}{\sqrt{2\pi}}e^{-z^2/2} dz = \dfrac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}}e^{\alpha z-z^2/2}dz = $$ $$ = \dfrac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}}\exp\left({-\dfrac{1}{2}((z - \alpha)^2 - \alpha^2)}\right)dz =$$ $$ = e^{\alpha^2/2}\int\limits_{\mathbb{R}}\dfrac{1}{\sqrt{2\pi}}\exp\left({-\dfrac{1}{2}(z - \alpha)^2}\right)dz = e^{\alpha^2/2}.$$

And now check that it is martingals:

  1. $\mathbb{E}|e^{\alpha W_t+\beta t}| = \text{/ exp > 0 /} = \mathbb{E}e^{\alpha W_t+\beta t} = e^{\frac{\alpha^2t}{2} + \beta t} < +\infty$.

  2. $\mathbb{E}(X_t|\mathcal{F}_u) = \mathbb{E}(e^{\alpha W_t+\beta t}|\mathcal{F}_u) = \mathbb{E}(e^{\alpha W_u + \beta t}e^{\alpha (W_t - W_u)}|\mathcal{F}_u) =_* e^{\alpha W_u + \beta t}\mathbb{E}(e^{\alpha (W_t - W_u)}|\mathcal{F}_u) =_{**} e^{\alpha W_u + \beta t}\mathbb{E}(e^{\alpha (W_t - W_u)}) =_{***} e^{\alpha W_u + \beta t}\mathbb{E}(e^{\alpha\sqrt{t-u} \frac{W_t - W_u}{\sqrt{t-u}}}) = e^{\alpha W_u + \beta t}e^{\frac{\alpha^2(t-u)}{2}} = e^{\alpha W_u + \beta u}e^{\beta (t - u)}e^{\frac{\alpha^2(t-u)}{2}}.$

Exponent's degree: $$\beta(t - u) + \dfrac{\alpha^2(t- u)}{2} = \left(t - u\right)\left(\beta + \dfrac{\alpha^2}{2}\right).$$ $$\beta + \dfrac{\alpha^2}{2} = 0 \Leftrightarrow \text{martingal}.$$

(*): $\mathbb{E}(\xi\eta|\mathcal{F}_{\xi}) = \xi\mathbb{E}(\eta|\mathcal{F}_{\xi})$,

(**): $\mathbb{E}(\xi|\mathcal{F}) = \mathbb{E}\xi$ if $\xi$ independ $\mathcal{F}$.

(***): $\frac{W_t - W_u}{\sqrt{t-u}} \sim \mathcal{N}(0, 1).$

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