# Martingale property for two stochastic processes

Let $(\Omega,F,P)$ be a probability space with filtration $\left\{F_{t}\right\}_{t\geq 0}$ generated by one dimensional Brownian motion $(B_{t})_{t\geq 0}$ defined on $(\Omega,F,P)$, assume that $\theta$ is a real constant and

$$Z_t := \exp \left(\theta B_t - \frac{1}{2}\theta^2 t\right)$$

Check the martingale property with respect to filtration $\left\{F_{t}\right\}_{t\geq 0}$ for the following processes:

I. $(B_{t})_{t\geq 0}$

II. $(Z_{t})_{t\geq 0}$

My attempt:

I. We have to show: $E[B_t | \mathcal{F}_s] = B_s$, so

$E[B_t | \mathcal{F}_s] = E[B_t - B_s + B_s | \mathcal{F}_s] = B_s + E[B_t - B_s | \mathcal{F}_s] = B_s + 0 = B_s$

My question: Why is $E[B_t - B_s | \mathcal{F}_s] = 0$?

II. We have to show $E[Z_t | \mathcal{F}_s] = Z_s$, so

$$E[Z_t | \mathcal{F}_s] = E[e^{\theta (B_t - B_s) + \theta B_s}e^{e^-\frac{1}{2}\theta ^2(t-s) - \frac{1}{2} \theta^2s}| \mathcal{F}_s]== E[Z_s e^{\theta (B_t - B_s)}| \mathcal{F}_s]e^{-\frac{1}{2}\theta^2 (t-s)}$$

$$= Z_s E[e^{\theta (B_t - B_s)}| \mathcal{F}_s]e^{-\frac{1}{2}(t-s)\theta^2}= Z_s e^{\frac{1}{2}(t-s) \theta^2}e^{-\frac{1}{2}(t-s) \theta^2}=Z_s$$

My question: Why is $E[e^{\theta (B_t - B_s)}| \mathcal{F}_s] = e^{\frac{1}{2}(t-s) \theta^2}$ ?

• $B_t-B_s|\mathcal{F}_s$ has normal distribution with mean $0$ and variance $t-s$ – gt6989b Mar 14 '16 at 14:32
• But then why is $E[e^{B_t}] = e^{\frac{t}{2}}$ and why is the $\theta$ squared? Thanks btw! – Ptru Mar 14 '16 at 15:14
• $e^{B_t}$ has lognormal distribution – gt6989b Mar 14 '16 at 16:16
• You can calculate $E[e^{\theta B_t}]$ via Ito's formula. – Xuxu Oct 6 '16 at 0:23

Assume that $F_{s}=\sigma\left(B_{i}\ \colon\ 0\leq i\leq s\right)$.

Then for every $0\leq s\leq t$ random variable $B_{t}-B_{s}$ is independent of $F_{s}$ and random variable $B_{s}$ is $F_{s}$ - measurable.

We also know that: $B_{u}\sim \mathcal{N}(0,u)$

Check basic properties of conditional expectation: link

I. $$\mathbb{E}\left[B_{t}|F_{s}\right]=\mathbb{E}\left[B_{t}-B_{s}+B_{s}|F_{s}\right]=\mathbb{E}\left[B_{t}-B_{s}|F_{s}\right]+\mathbb{E}\left[B_{s}|F_{s}\right]$$

$$=\mathbb{E}\left[B_{t}-B_{s}\right]+B_{s}=\mathbb{E}\left[B_{t}\right]-\mathbb{E}\left[B_{s}\right]+B_{s}=0+0+B_{s}=B_{s}$$

II.

$$\mathbb{E}\left[Z_{t}|F_{s}\right]=\mathbb{E}\left[e^{\theta B_{t}-\frac{1}{2}\theta^{2}t}|F_{s}\right]=e^{-\frac{1}{2}\theta^{2}t}\mathbb{E}\left[e^{\theta B_{t}}|F_{s}\right]=e^{-\frac{1}{2}\theta^{2}t}\mathbb{E}\left[e^{\theta (B_{t}-B_{s})+\theta B_{s}}|F_{s}\right]$$

$$=e^{\theta B_{s}-\frac{1}{2}\theta^{2}t}\mathbb{E}\left[e^{\theta (B_{t}-B_{s})}|F_{s}\right]=e^{\theta B_{s}-\frac{1}{2}\theta^{2}t}\mathbb{E}\left[e^{\theta (B_{t}-B_{s})}\right]$$

$$=e^{\theta B_{s}-\frac{1}{2}\theta^{2}t}\cdot e^{\frac{1}{2}\theta^{2}(t-s)}=e^{\theta B_{s}-\frac{1}{2}\theta^{2}s}=Z_{s}$$

since

$$B_{t}-B_{s}\sim\mathcal{N}(0,t-s)$$

and

$$\mathbb{E}\left[e^{\theta (B_{t}-B_{s})}\right]=\int_{\Omega}e^{\theta (B_{t}-B_{s})}d\mathbb{P}=\int_{\mathbb{R}}e^{\theta x}\frac{1}{\sqrt{2\pi (t-s)}}e^{-\frac{x^{2}}{2(t-s)}}dx$$

$$=\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi (t-s)}}exp\left(\frac{-\left(x-\theta(t-s)\right)^{2}+\theta^{2}(t-s)^{2}}{2(t-s)}\right)dx$$ $$=e^{\frac{1}{2}\theta^{2}(t-s)}\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi (t-s)}}exp\left(\frac{-\left(x-\theta(t-s)\right)^{2}}{2(t-s)}\right)dx$$

$$=e^{\frac{1}{2}\theta^{2}(t-s)}\cdot 1= e^{\frac{1}{2}\theta^{2}(t-s)}$$

because

$$f(x)=\frac{1}{\sqrt{2\pi (t-s)}}exp\left(\frac{-\left(x-\theta(t-s)\right)^{2}}{2(t-s)}\right)$$

is probability density function of random variable with normal distribution of the form: $$\mathcal{N}\left(\theta(t-s),t-s\right)$$

so

$$\int_{\mathbb{R}}f(x)\ dx=1$$