Under which conditions the given random process is martingale and under which submartingale?

Question

Let $a_t $ be adapted to the filtration random process $a_t: P\{\int _0^T|a_t|dt < \infty \} = 1 $ and $ b_t \in M_T^2. \quad$ Under which conditions the random process $$X_t = exp\{\int _0^ta_sds+\int _0^tb_sdW_s\} \; t \in [0, T]\,$$ is martingale and under which submartingale ?
As I understand, this is a famous example of "exponential martingale" and the answer is:
The process will be martingale for $ a_s = -\frac {b_s^2}{ 2 } $.
But I don’t understand how to prove it. And what conditions will be for submartingale?
My attempt to prove was:
Let's try to find conditions when $E(\frac{X_t}{X_s}|\mathcal F_s)= 1$ .

$E(\frac{X_t}{X_s}|\mathcal F_s)=exp\{\int _s^ta_sds\} E(exp\{\int _s^tb_sdW_s\}) $
Also, I understand that $\int _s^tb_sdW_s$ has Gaussian distribution.
But I do not know what to do next. I would be grateful for any help.

Answer 1

One can approach this using the Ito lemma. Let $I_t=\int_0^t a_u du+\int_0^tb_udW_u, (\forall) t\in [0;T]$. Then, by definition we have that: $$ dI_t=a_t+b_tdW_t. $$ Using Ito lemma applied to $f(I_t)$, where $f(x)=e^x$, we get: $$ dX_t=d\left(e^{I_t}\right)=\underbrace{e^{I_t}}_{X_t}dI_t + \frac{1}{2}e^{I_t}d\langle I \rangle_t, $$ where $\langle I \rangle_t$ is the quadratic variation of $(I_t)_{t\geq 0}$. This quadratic variation can be obtained using the rules of stochastic calculus: $$ d\langle I \rangle_t =(b_t)^2 dt. $$ Therefore, $$ dX_t=X_tdI_t+\frac{1}{2}X_t(b_t)^2dt=\left(a_t+\frac{b_t^2}{2}\right)dt+X_tb_tdW_t. $$ This is really just a shorthand notation for: $$ X_t=X_0+\int_0^t \left(a_u+\frac{b_u^2}{2}\right)du+\int_0^t X_ub_udW_u. $$ But since the last term of the above formula is a stochastic integral (which is a martingale), we have that: $$ \mathbb{E}\left[X_t\right]=\mathbb{E}\left[X_0\right]+\mathbb{E}\left[\int_0^t\left(a_u+\frac{b_u^2}{2}\right)du\right]. $$ To ensure the martingality of $(X_t)_{t\geq 0}$, a necessary condition is: $$ \mathbb{E}\left[\int_0^t\left(a_u+\frac{b_u^2}{2}\right)du\right] = 0. $$ This is somewhat different than what you have written above, as the integral $$ \int_0^t\left(a_u+\frac{b_u^2}{2}\right)du $$ is a random variable. Your condition is sufficient, but not necessary.

Since the submartingale condition is $$ \mathbb{E}\left[X_t|\mathcal{F}_s\right]\geq X_s, \text{for }s\leq t $$ (assuming the filtration is indeed $\left(\mathcal{F}_t\right)_{t\geq 0}$), then the sufficient condition for $(X_t)_{t\geq 0}$ to be a submartingale should be straightforward to see now.