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.

  • $\begingroup$ Ito integral is not Gaussian, in general. It is if, say, $b_s$ is deterministic. $\endgroup$
    – ir7
    Commented Jun 14, 2020 at 22:18

1 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.

  • 1
    $\begingroup$ $\int_0^t X_ub_udW_u$ is a local martingale but not necessarily a martingale. To ensure that it's a martingale you need to show that $E\int_0^T |X_ub_u|^2du<\infty,$ which you haven't proved. $\endgroup$
    – user39119
    Commented Jun 15, 2020 at 11:18
  • $\begingroup$ @UBM Do you have any ideas how to prove that $\int_0^T |X_ub_u|^2du<\infty$? $\endgroup$
    – Helen
    Commented Jun 15, 2020 at 12:01
  • 1
    $\begingroup$ @Helen I think we can't prove that condition with the information we are given. In my opinion this approach is wrong, it takes you to a dead end. A different approach would be to say that If $a_s=− \frac{b_s^2}{2}$, then the process 𝑋 is the stochastic exponential of the process $\{\int_0^t b_sdW_s; 0 \leq t \leq T\}$, so a sufficient condition for 𝑋 to be a martingale is the Novikov condition. $\endgroup$
    – user39119
    Commented Jun 15, 2020 at 12:42
  • $\begingroup$ @UBM yes, thank you very much, I read some information about Novikov’s condition and realized that this is good advice. But the condition for submartingales is still not clear. $\endgroup$
    – Helen
    Commented Jun 15, 2020 at 13:25
  • 1
    $\begingroup$ @UBM Something must surely escape me, for I do not see immediately why $\int_0^t X_ub_udW_u$. Is assured to be a local martingale. It would be great if you could point to a reference with a page and theorem/definition number (e.g. Protter). I'm far from being an expert in stochastic calculus and I'd like to gain a better understanding of this. How does $\mathbb{E}\left[\int_0^t|X_ub_u|^2du \right]<+\infty $ ensure martingality? Does the argument use Ito isometry? $\endgroup$
    – fwd_T
    Commented Jun 15, 2020 at 17:41

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