Martingale definition

Question

To prove that one process is Martingale, generally we prove 3 things :

1) X is adapted.

2)$$ \mathbf{E} ( \vert X_n \vert )< \infty $$

3) $$\mathbf{E} (X_{n+1}\mid X_1,\ldots,X_n)=X_n $$

I generally prove the 1 and 3 but I have some calculus problems with 2 Here an example: $$M_{t}=B_{t}^3-\alpha\int_{0}^{t}B_{s}ds$$ How can I show the 2nd rule.(B brownian motion standard)

Thank you in advance for your help

Answer 1

Obviously,

$$|M_t| \leq |B_t|^3 + |\alpha| \int_0^t |B_s| \, ds.$$

Since $(B_t)_{t \geq 0}$ is a Brownian motion, we have in particular $B_t \sim N(0,t)$ and $B_t \sim \sqrt{t} B_1$. Consequently,

$$\mathbb{E}(|B_t|^3) <\infty \qquad \text{and} \qquad \mathbb{E}(|B_s|) = \sqrt{s} \mathbb{E}(|B_1|).$$

Using Tonelli's theorem, we get

$$\mathbb{E}(|M_t|) \leq \mathbb{E}(|B_t|^3) + \mathbb{E}(|B_1|) \, |\alpha| \int_0^t \sqrt{s} \, ds < \infty.$$

So, usually the integrability of such processes boils down to the fact that the Brownian motion has moments of arbitrary order.

Answer 2

$B_t$ is normally distributed so $E(|B_t^3|)$ is finite, so $B_t^3$ is in $L^1$.

$\int_{0}^{t}B_{s}ds$ is also normally distributed if I remember well. So it's in $L^1$. Finally, $M_t$ is in $L^1$ because $L^1$ is stable by addition as a vector space.