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](https://mathoverflow.net/questions/74521/stochastic-integrals-as-honest-martingales-comparison-criterion/74540#74540)) 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.



Stochastic integrals as honest martingales — exponential damping
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.


