# Showing that this is a martingale. Using Ito formula

I have that X is an Itô-process where:

$$dX_t=u(ω,t)dt+dB_t, t∈[0,T],$$ and assume $u$ is bounded.

where:

$$L_t=exp(-\int_0^t u(w,s)dB_s - 1/2\int_0^t u(w,s)^2 ds)$$

I am supposed to show that $X_tL_t$ is a $F_t$-martingale, $F_t$ is the filtration generated by the Brownian motion.

I Have no Idea how to solve this using the Ito formula. I get stuck because i can't get rid of the drift part of the $X_t$ process. They give as a hint to use the Girsanov theorem.

• It is discouraged to hide your questions after a link. Type them in the clear.
– mlc
May 7, 2017 at 16:54

Let $(X_t)$ be an Ito process of the form $$dX_t = u(\omega, t) dt + dB_t, \ t \in [0, T]$$ where $T \in [0, \infty)$ is given and $(B_t)$ is a standard Brownian motion (for simplicity $1$ dimensional) with respect to the law $\mathbb{P}$ given a probability space $(\Omega, \mathcal{F}, \mathbb{P})$.

Define $\mathcal{F}_t$ to be the $\sigma$-algebra generated by the random variables $B_s$, $s \leq t$.

I pressume that you wanted to assume the coefficient $u$ is bounded. To have it a bit more general we could have assumed the Novikov's condition, that is, $$\mathbb{E}_{\mathbb{P}}\left[ \exp\left\{ \frac{1}{2}\int_0^T u^2(\omega, s) ds\right\} \right] < \infty$$ The Novikov condition yields that $(L_t)$ is a martingale w.r.t $(\mathcal{F}_t)$ and $\mathbb{P}$. This martingale is know as an exponential martingale and it can be shown that $$dL_t = - L_t u(\omega, t)dB_t.$$ Usually in the literature your coefficient $u(\omega, t)$ is replaced by $a(\omega, t)=-u(\omega, t)$ and so the corresponding SDE becomes $$dL_t = L_ta(\omega, t)dB_t$$ probably this is the reason (you missed the minus) why you couldn't reduce the drift term.

You wish to show that $(Z_t:=X_tL_t)$ is a martingale w.r.t $(\mathcal{F}_t)$ and $\mathbb{P}$. We proceed by applying Ito's product formula, namely $$dZ_t = X_t dL_t + L_t dX_t + dX_t dL_t$$ thus we obtain $$dZ_t = -X_t L_t u(\omega, t)dB_t + L_t \left(u(\omega, t) dt + dB_t\right) -L_tu(\omega, t)dt$$ and so $$dZ_t = L_t(1-X_tu(\omega, t))dB_t.$$

So if you show that $b(\omega, t) := L_t(\omega)(1-X_t(\omega)u(\omega, t))$ is such that $b \in L^2(\Omega \times [0, T])$ then you obtain that $$Z_t = Z_0 + \int_0^t L_s(1-X_su(\omega, s))dB_s$$ is a martingale.

Last thing, I doubt that the Girsanov theorem is a hint to solve this exercise. I would rather say that, this exercise is a simplified Girsanov with a weaker conclusion. The assumptions in this exercise are similar to the ones in the Girsanov theorem. The theorem would imply that $(X_t)$ (if assumed to have $X_0=0$) is a standard Brownian motion with respect to the new measure $\mathbb{Q}$ defined on $\mathcal{F}_T$ via $$d\mathbb{Q}(\omega) = L_T(\omega)d\mathbb{P}(\omega).$$

To sum up,

1. if $u$ is non-zero then $(X_t)$ cannot be a martingale w.r.t $(\mathcal{F}_t)$ and $\mathbb{P}$. However, if we multiply it by a `special' exponential martingale $(L_t)$ then the product $(X_tL_t)$ is martingale w.r.t $(\mathcal{F}_t)$ and $\mathbb{P}$.
2. (Girsanov Theorem) $(X_t)$ could be viewed as a martingale w.r.t $(\mathcal{F}_t)$ and $\mathbb{Q}$, where $\mathbb{Q}$ is defined on $\mathcal{F}_T$ via $$d\mathbb{Q}(\omega) = L_T(\omega)d\mathbb{P}(\omega).$$
• Thank you very much. I did have made that mistake and at the end i couldn't cut the $L_tu(w,t)dt-L_tu(w,t)dt$. Thank you again, you were very helpful. And Girsanov Hint maybe was for to check the theorem and see the similarities. May 8, 2017 at 15:47