Show that the following process is a martingale without Ito's formula

Question

I'm trying to do the following exercise from Peres's Brownian Motion Book.

Exercise 2.13. Let $(B_t)_{t \geq 0}$ be a linear Brownian motion.

1. Show that $\exp(\sigma B_t - \sigma^2 t/2)$ is a martingale for any $\sigma>0$.

2. By taking derivatives $\frac{\partial}{\partial^n}$ at zero, derive that the following processes are martingales:

   • $B_t^2-t$,
   • $B_t^3 - tB_t$,
   • $B_t^4- 6t B_t^2 + 3t^2$

The following I understand how to do using Ito's formula, but I was wondering if it's possible to do without. Part a) is straightforward,

so we have that

$$\exp(\sigma B_t-\sigma^2t/2) $$

is a martingale.

Now when we take the derivative of this with respect to sigma, before taking the limit, we still have a martingale, as it's a linear combination of a martingale.

If we write

$$ A_{\sigma}(t) = \frac{1}{\sigma} ( \exp(\sigma B_t-\sigma^2t/2) - 1 ) $$

then for any $\sigma >0$ this is a martingale. So for $B \in \mathcal F_s$

$$ \mathbb E A_\sigma(t) 1_B=E A_\sigma(s) 1_B $$

If I'm correct. To show that the derivative function is then a Martingale, it suffices to show that we can take a limit of $\sigma $ to zero and then move the limit inside the integral.

I would like to show, but don't know how, that $$\mathbb E ( |A_\sigma (t)-A(t)|) $$ goes to zero, using something like bounded convergence. So if we have that this limit is zero, then we must have that $$ \mathbb E A(t)1_B = \lim_{\sigma \to 0} \mathbb E A_\sigma(t) 1_B= \lim_{\sigma \to 0} E A_\sigma(s) 1_B = \mathbb E A(s)1_B$$

Does anyone know if it's possible? Many thanks

Answer 1

Fix $t \geq 0$. By the mean value theorem, we have

$$|e^x-1| \leq |x| e^{|x|} \qquad \text{for all $x \in \mathbb{R}$.} \tag{1}$$

For $\sigma \in (-1,1)$ we get

$$\begin{align*} A_{\sigma}(t) \stackrel{\text{def}}{=} \frac{1}{\sigma} \left( e^{\sigma B(t)-\sigma^2 t/2}-1 \right) &\stackrel{(1)}{\leq} \left( |B(t)|+ \frac{t}{2} \right) e^{|B(t)| + t/2} =: X(t). \end{align*}$$

Since $B(t)$ is Gaussian, it holds that $X(t) \in L^1(\mathbb{P})$. As

$$A(t) := \frac{d}{d\sigma} e^{\sigma B(t)-\sigma^2 t/2} \bigg|_{\sigma=0} = B(t)$$

we have

$$|A_{\sigma}(t)-A(t)| \leq X(t)+|B(t)|$$

for any $\sigma \in (-1,1)$ and the right-hand side is an integrable dominating function. On the other hand, the expression on the left-hand side converges to $0$ as $\sigma \to 0$, and therefore we may apply the dominated convergence theorem to conclude that

$$\lim_{\sigma \to 0} \mathbb{E}(|A_{\sigma}(t)-A(t)|)=0.$$