# Find $(A_{t})_{t\geq 0}$ such that $(B_{t}^4+A_{t})_{t\geq 0}$ is a martingale for Brownian motion $(B_t)_{t \geq 0}$

Let $$(B_{t})_{t\geq 0}$$ be a Brownian motion and $$(\mathcal{F}_{t})_{t\geq 0}$$ its canonical filtration. Find a Stochastic Process $$(A_{t})_{t\geq 0}$$ such that $$(B_{t}^{4}+A_{t})_{t\geq 0}$$ is a Martingale.

I don't want to settle for a single $$(A_{t})_{t\geq 0}$$ process, I would like to be able to find a family of processes.

My attempt: Let $$X_{t}:=B_{t}^{4}+A_{t}$$ and $$t>s$$, then we have \begin{align} \mathbb{E}[X_{t}-X_{s}|\mathcal{F}_{s}]&= \mathbb{E}[B_{t}^{4}+A_{t}-B_{s}^{4}-A_{s}|\mathcal{F}_{s}]\\ &=3(t-s)^{2}+6(t-s)B_{s}^{2}+\mathbb{E}[A_{t}-A_{s}|\mathcal{F}_{s}] \end{align} If $$(X_{t})_{t\geq 0}$$ is a martingale, then we should have $$0=3(t-s)^{2}+6(t-s)B_{s}^{2}+\mathbb{E}[A_{t}-A_{s}|\mathcal{F}_{s}]$$ and and we should also have that $$(A_{t})_{t\geq 0}$$ is adapted to filtration $$(\mathcal{F}_{t})_{t\geq 0}$$. Therefore, we have $$A_{s}=3(t-s)^{2}+6(t-s)B_{s}^{2}+\mathbb{E}[A_{t}|\mathcal{F}_{s}] \tag{\bigstar}$$

Therefore, it all comes down to solving equation ($$\bigstar$$).

• How about Ito's formula? You are not allowed to use it ? – Paresseux Nguyen Nov 19 at 3:26
• @ParesseuxNguyen No, this exercise is in a section where Ito's formula is not covered. – Diego Fonseca Nov 19 at 3:27
• Are there any conditions ruling out the trivial solutions $A_t = -B_t^4 + c$ for any $c \in \mathbb{R}$? – user6247850 Nov 19 at 3:40

(I take it that you are looking for a non-trivial solution - otherwise you could simply take $$A_t := -B_t^4$$, as was pointed out in a comment.)

Using that $$(B_t)_{t \geq 0}$$ has independent and stationary increments and the fact that $$B_t \sim N(0,t)$$, it follows that

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

is a martingale for any $$\lambda \in \mathbb{R}$$. Equivalently,

$$\int_F M_s^{(\lambda)} \, d\mathbb{P} = \int_F M_t^{(\lambda)} \, d\mathbb{P} \tag{1}$$

holds for all $$s \leq t$$ and $$F \in \mathcal{F}_s$$. Now the idea is to differentiate $$(1)$$ with respect to $$\lambda$$. For the first derivative, we get

$$\int_F (B_s-\lambda s) M_s^{(\lambda)} \, d\mathbb{P} = \int_F (B_t-\lambda t) M_t^{(\lambda)} \, d\mathbb{P}. \tag{2}$$

Since this holds in particular for $$\lambda=0$$, we see that $$\int_F B_t \, d\mathbb{P} = \int_F B_s \, d\mathbb{P}$$ for all $$s \leq t$$ and $$F \in \mathcal{F}_s$$, which means that $$(B_t)_{t \geq 0}$$ is a martingale (no surprise here; we know this already). By calculating higher derivatives, we can increase the "power" of the martingale. E.g. by differentiating $$(2)$$ once more again with respect to $$\lambda$$, we find that

$$\int_F ((B_s-\lambda s)^2-s) M_s^{(\lambda)} \, d\mathbb{P} = \int_F ((B_t-\lambda t)^2-t) M_t^{(\lambda)} \, d\mathbb{P}.$$

Putting $$\lambda=0$$, we see that $$(B_t^2-t)_{t \geq 0}$$ is a martingale. To get a martingale involving $$B_t^4$$ we need two differentiate two more times. Doing so yields

$$\int_F (B_s-\lambda s)((B_s-\lambda s)^2-3s) M_s^{(\lambda)} \, d\mathbb{P} = \int_F (B_t-\lambda t)((B_t-\lambda t)^2-3t) M_t^{(\lambda)} \, d\mathbb{P}$$

(which shows, for $$\lambda=0$$, that $$(B_t^3-tB_t)_{t \geq 0}$$ is a martingale) and

\begin{align*} &\int_F \left[((B_s-\lambda s)^2-s)((B_s-\lambda s)^2-3s)-2s(B_s-\lambda s)^2\right] M_s^{(\lambda)} \, d\mathbb{P} \\ &= \int_F\left[((B_t-\lambda t)^2-t)((B_t-\lambda t)^2-3t)-2t(B_t-\lambda t)^2 \right] M_t^{(\lambda)} \, d\mathbb{P},\end{align*}

which shows, for $$\lambda=0$$, that

$$(B_t^2-t)(B_t^2-3t)-2t B_t^2=B_t^4-6tB_t+3t^2$$

is a martingale.

Remark: There is a close connection to Hermite polynomials, see e.g. this question.