# Stochastic integrals as honest martingales — exponential damping

We have a given positive martingale ρt, with the dynamics: $$\textrm{d}\rho_t = \lambda_t \rho_t \textrm{d}W_t$$ where $W_t$ is a standard Brownian motion. Now we have an "exponentially dampened" process $p_t$: $$p_t=\int_0^t \exp(-\int_0^s r_u \textrm{d} u)\textrm{d} \rho_s$$ where $r_u \geq 0$. If needed I could add stronger assumptions for $r_u$, e.g.:

(1) $r_u>0$

(2) $\exp(-\int_0^t r_u \textrm{d} u) \rightarrow 0$ as $t\rightarrow \infty$

I know (from the answer to my previous question Stochastic integrals as honest martingales -- comparison criterion) that in generality $p_t$ is not guaranteed to be a martingale, but is it the case in one of these specific cases? If not I would appreciate a (simple) counterexample.

• @Douglas, you're right, cheers! – Grzenio Sep 6 '11 at 7:12
• @Grzenio : I think there is a little problem with your integrand because as written you have $p_t=exp(−\int_0^t r_u.du).ρ_t$, shouldn't it be $p_t=\int_0^texp(−\int_s^t r_u.du).dρ_s$ ? Regards – The Bridge Sep 6 '11 at 11:53
• @Grzenio (and myself): By the way in both case you don't have necessarily a martingale, in your case because $p_t=exp(-\int_0^tr_u.du).\rho_t$ obviously need not to be a martingale in general and in the case I suggested it is the same, if you have $p_t=\int_0^t exp(−\int_s^tr_u.du).dρ_s$ then you are not assured that is a martingale (or a local martingale) because of the dependence of the integrand in $t$ (apply Itô-Leibniz rule or Stochastic-Fubini to see that a non nul drift appears in the general case). – The Bridge Sep 6 '11 at 13:29
• Hi @TheBridge, thanks for pointing me to the typo, of course there should not be $t$ dependence in the integrand! – Grzenio Sep 6 '11 at 14:37
• @ Grzenio : May be you could have a look at Novikov's criterion it is usually used to get a true martingale from a local martingale but it might be applied here or be useful to impose a condition over the $r$ process. Regards – The Bridge Sep 6 '11 at 14:42

Yes, in this case it is true that $p$ is a proper martingale! Note that your integrand $\exp\left(-\int_0^tr_u du\right)$ is an adapted, continuous, and decreasing process bounded by 1. So, the following statement implies that $p$ is a martingale.
Let $\rho$ be a cadlag martingale and $\xi$ be an adapted left-continuous and decreasing process with $0\le\xi\le1$. Then, $M=\int\xi d\rho$ is a martingale.
As stochastic integration with respect to a bounded predictable integrand preserves the local martingale property, $M$ must at least be a local martingale. Then, it is standard that a local martingale $M$ is a proper martingale if and only if the set $$\left\{M_{\tau\wedge t}\colon\tau{\rm\ is\ a\ stopping\ time}\right\}$$ is uniformly integrable, for each fixed $t\in\mathbb{R}^+$. Processes satisfying this uniform integrability property are sometimes said to be of class (DL), which is a restriction of the class (D) property to finite time intervals $[0,t]$.
Let's show that $M$ is of class (DL). Without loss of generality, by subtracting $\rho_0$ from $\rho$ if necesary, we can assume that $\rho_0=0$. As $\rho$ is a martingale, it is of class (DL), so the set $$S=\left\{\rho_{\tau\wedge t}\colon\tau{\rm\ is\ a\ stopping\ time}\right\}$$ is uniformly integrable, for fixed $t\in\mathbb{R}^+$. As $\xi$ is decreasing and left continuous, we can define random times $\tau_x=\inf\left\{t\in\mathbb{R}^+\colon\xi_t < x\right\}$ for $0\le x\le1$. These are stopping times, either by applying the debut theorem or using the fact that $\{\tau_x < s\}=\{\xi_s < x\}$ (strictly speaking, this requires right-continuity of the underlying filtration but, as the martingale property of adapted processes is unchanged by replacing the filtration by its right-continuous version, this is not important). For each positive integer $n$, the process $$\xi^{(n)}_s=\frac1n\sum_{k=1}^n1_{[0,\tau_{k/n}]}$$ satisfies $\vert\xi^{(n)}-\xi\vert\le\frac1n$. So, for any stopping time $\tau$, bounded convergence gives the following limit in probability. $$M_{\tau\wedge t}=\lim_{n\to\infty}\int_0^{\tau\wedge t}\xi^{(n)}_sd\rho_s =\lim_{n\to\infty}\frac1n\sum_{k=1}^n\rho_{\tau_{k/n}\wedge\tau\wedge t}.$$ As $\tau_{k/n}\wedge\tau$ is a stopping time, this expresses $M_{\tau\wedge t}$ as a limit in probability of convex combinations of elements of $S$. As taking the convex hull and closure in probability of a set of random variables preserves the uniform integrability property, this means that the set of all $M_{\tau\wedge t}$ for stopping times $\tau$ is uniformly integrable. So, $M$ is of class (DL) and is a proper martingale.
• Just to double check, this construction works if $\xi_S$ doesn't go to zero at infinity? Then the stopping times $\tau_x$ can become infinite, but it shouldn't matter - the indicator function would in these cases be always one. Is that correct? – Grzenio Sep 14 '11 at 16:12