Prove that this process is a martingale

Question

Let $\{ B_t \}_{t \in [0,T]}$ be a standard Brownian motion and denote its natural filtration $\mathbb{F} := \{ \mathcal{F}_t \}_{t \in [0,T]}$ and let $\{U_t\}_{t \in [0,T]}$ be a uniformly bounded and $\mathbb{F}$ adapted process. We consider the process \begin{equation} M_t := exp \left( - \int_0^t U_s dB_s - \frac{1}{2} \int_0^t U_s^2 ds \right). \end{equation}

We want to prove that $\{ M_t \}_{t \in [0,T]}$ is an $\mathbb{F}$-martingale given the assumption that $\int_0^T E\left( M_s^2 \right) ds < \infty$.

MY PROBLEM:

My reflex in this type of context is to define the process I am looking at as a function of time and other processes. Then, I can apply Ito's Lemma and I should be left with just the Ito integral and I can just invoke its martingale property to complete the proof. Just to give you an example, take a simpler, though similar process: \begin{align} N_t := exp \left(\alpha B_t - \frac{1}{2} \alpha^2 t \right) \\ g(t,x) := exp( \alpha x - \alpha^2 t/2) \rightarrow \begin{cases} \partial_x g(t,x) = \alpha g(t,x) \\ \partial_x^2 g(t,x) = \alpha^2 g(t,x) \\ \partial_t g(t,x) = \frac{-\alpha^2}{2} g(x,t) \end{cases} \\ \rightarrow dg(t,x) = \alpha g(t,x) dB_t + \frac{1}{2} \alpha^2 g(t,x) dt + \frac{-1}{2} \alpha^2 g(t,x) dt = \alpha g(t,x) dB_t + 0 \\ \leftrightarrow N_t - N_0 = \alpha \int_0^t N_t dB_t. \end{align}

If $N_t$ respects $\int_0^T E \left( N_t^2 \right) ds < \infty$, we have that $N_t - N_0$ is equal to an Ito integral, hence it is a martingale with respect to the natural filtration. My problem is that I have no clue how to extend that idea to the first process above. Frankly, I'm not sure how I would define my function g(t,x) or take analogous derivatives for $M$.

In case someone has the time to take a stab at it, it might be useful to point out I'm an economist and not a mathematician. I apologize in advance if my work is a little sloppy or if I missed something obvious. I'm just trying to learn the basics right now. Thanks in advance and any help will be appreciated.

Additional details

A similar question was asked here and works by invoking the Novikov condition. If we can show that \begin{equation} E \left( exp \left( \frac{1}{2} \int_0^T U_s^2 ds \right) \right) < \infty \end{equation} is true, then $\{ M_t \}_{t \in [0,T]}$ is an $\mathbb{F}$-martingale. However, given where I took this problem, this would require me to dig in a subsequent chapter to complete it and the problems are chapter-specific.

Answer 1

I found a solution to this problem here.

In case someone ever comes across this same problem, the trick lies in using the right Ito process to apply Ito's Lemma. I was stuck trying to work with a function g(x,t), but I could have picked: $ Z_t := -\int_0^t Us dB_s - \frac{1}{2} \int_0^t U_s^2 ds$ and g(x) = e^x more simply. This readily implies $Z_t = g'(Z_t) = g''(Z_t)$ and we can use Ito's Lemma in its simplest form: \begin{align} g(Z_t) - g(Z_0) = \int_0^t g'(Z_t) dZ_t + \frac{1}{2}\int_0^t g''(Z_t) d \left< Z \right>_t \\ dM_t = - U_t M_t dB_t. \end{align}

and that's obviously a martingale. And there I was wondering how the hell should I take derivatives w.r.t to t in there and reading through papers and notes of all sorts for hours... At least I learned something.