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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.

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@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
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1 Answer

up vote 3 down vote accepted

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. $$ M_{\tau\wedge t}=\lim\_{n\to\infty}\int\_0^{\tau\wedge t}\xi^{(n)}\_s\\,d\rho\_s =\lim\_{n\to\infty}\frac1n\sum\_{k=1}^n\rho\_{\tau\_{k/n}\wedge\tau\wedge t}. $$ 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.

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Thanks for your reply, its really convenient that this process IS a martingale in my model! I am trying to understand the proof and everything seems straightforward apart from the last paragraph, where you use the fact that the limit in the last equation preserves the uniform integrability property. Could you provide some (citeable) reference? Btw, is there a reference for the whole proof that I could cite? –  Grzenio Sep 12 '11 at 9:07
    
There's the following which is not really citeable and was written by myself, so not an independent statement either, but it has proofs. planetmath.org/encyclopedia/… –  George Lowther Sep 12 '11 at 10:25
    
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
    
That's correct. Also, the definition of a martingales only involves finite time intervals. If a process is a martingale over all finite time intervals then it is a martingale. You shouldn't worry about what happens at infinity. –  George Lowther Sep 14 '11 at 21:00
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