Let $B$ be a standard Brownian motion and further let
I know that for $\lambda \in \mathbb R$, we have that $(\exp(\lambda B_{t} -\frac{\lambda^{2}}{2}t))_{t\geq 0}$ is a martingale, but how can I use it to prove that the following are martingales:
$B_{t}^{2}-t,\; \; B_{t}^{3}-3tB_{t}, \; \; B_{t}^{4}-6tB_{t}^2+3t^{2}\; \; (*)$. By differentiating wrt $\lambda $ and then setting $\lambda = 0$, we use that for $s < t$:
$\mathbb E [\exp(\lambda B_{t} -\frac{\lambda^{2}}{2}t)\lvert \mathcal{F}_{s}]=\exp(\lambda B_{s} -\frac{\lambda^{2}}{2}s)$
Now differentiating both sides wrt $\lambda$, and setting $\lambda = 0$ we get:
$\mathbb E[B_{t}\lvert \mathcal{F}_{s}]=B_{s}$ [Side question: What can of justification do I need to bring the differentiation operator of $\lambda$ into the conditional expectation?]
Ok, so now we have the idea to show that $(B_{t})$ is a martingale, but how can we use it to show that $(*)$ are all martingales?
It's exercise from Steele book, Stochastic calculus and financial applications. So firstly note that we can expand $M_t = \exp{(\lambda B_t - (\lambda^2/2)t)}$ as Taylor series in $\lambda$. We have $$ M_t = \sum_{i=0}^{\infty} \frac{(\lambda(B_t - (\lambda/2)t))^{i}}{i!}. $$ Now note that the above expression can be written as $$ M_t = \sum_{i=0}^{\infty} \lambda^{i} H_i(t,B_t), $$ where the $H_i(t,B_t)$ are polynomials. Now for $0 \leq s \leq t$ by martingale property of $M_t$ we have $$ \mathbb{E}[M_t|\mathcal{F_s}] = M_s, $$ so by series representation $$ \mathbb{E} \big[\sum_{i=0}^{\infty} \lambda^{i} H_i(t,B_t)|\mathcal{F_s} \big] = \sum_{i=0}^{\infty} \lambda^{i} H_i(s,B_s). $$ Now one have to justify that expectation and summation above can be interchanged. Then by arbitrariness of $\lambda$ we have that to have equality of coefficients, and this give us $\forall i \geq 0$ $$ \mathbb{E} [H_i(t,B_t)|\mathcal{F_s}] = H_i(s,B_s). $$ To prove that $B^2_t-t,B^3_t−3tB_t,B^4_t−6tB^2_t+3t^2$ are martingales one have to find $H_2, H_3, H_4$ which is the easiest part of this exercise.