Using Itô’s Lemma to Analyze Exponential Processes
Summary
The document asks when an exponential process driven by an integrated drift and a Brownian stochastic integral is a martingale or submartingale. Its answer applies Itô’s lemma to the exponential and identifies a drift term involving the original drift and half the squared diffusion coefficient. Setting that term to zero gives the familiar candidate condition for a martingale; a nonnegative drift suggests a submartingale condition when the process is integrable.
The explanation usefully distinguishes a random drift integral from its expectation, but its claim that the stochastic integral is a martingale needs an integrability qualification. In addition, a zero expected integrated drift alone is not enough to prove the conditional martingale property. Establishing a true martingale may require suitable integrability conditions, such as an exponential-martingale criterion. The discussion provides an Itô-calculus outline rather than a complete statement of the assumptions needed for each conclusion.
Key ideas
- Itô’s lemma produces a drift term involving the process, the drift coefficient, and half the squared diffusion coefficient.
- Setting the instantaneous drift to zero is the standard candidate condition for the exponential process to be a martingale.
- A nonnegative drift can support a submartingale conclusion when the process satisfies the necessary integrability conditions.
- The expected value of an integrated random drift does not by itself establish the conditional martingale property.
- The argument requires additional assumptions to ensure the stochastic integral and exponential process are true martingales.
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Full text
# Under which conditions the given random process is martingale and under which submartingale?
# Under which conditions the given random process is martingale and under which submartingale?
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 by fwd_T (score 2, accepted)
https://quant.stackexchange.com/a/54911
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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