We have a given positive martingale $\rho\_t$, 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 a "dumped" process p\_t:
$$\textrm{d} p\_t = a\_t \lambda\_t \rho\_t \textrm{d} W\_t$$
where $0 \leq a\_t \leq 1$. 


(Silly) question: are we guaranteed that $p\_t$ is also a honest martingale? What would be the easiest way to prove it?



Stochastic integrals as honest martingales -- comparison criterion
No, $\rho$ need not be a proper martingale. To guarantee that $p\_t=\int\_0^ta\_sd\rho\_s$ is a martingale for all predictable $0\le a\_t\le 1$ you need the additional property that $\sup\_{s\le t}\rho\_s$ is integrable. In fact, for any cadlag martingale $\rho$, the following are equivalent.


1. $\sup\_{s\le t}\vert\rho\_s\vert$ is integrable for all $t\in\mathbb{R}^+$.
2. $\int\xi\_sd\rho\_s$ is a proper martingale for all bounded predictable processes $\xi$.


The implication (1) ⇒ (2) is a consequence of the Burkholder-Davis-Gundy inequality. This states that there exists positive constants $c < C$ such that
$$
c\mathbb{E}\left[[M]\_t^{1/2}\right]\le\mathbb{E}\left[\sup\_{s\le t}\vert M\_s\vert\right]\le C\mathbb{E}\left[[M]\_t^{1/2}\right]
$$
for all cadlag martingales M with $M\_0=0$, where $[\cdot]$ is the quadratic variation (I have written a blog post with a proof of the BDG inequality [here](http://almostsure.wordpress.com/2010/04/06/the-burkholder-davis-gundy-inequality/)). If $M=\int\xi d\rho$ for bounded predictable $\vert\xi\vert\le1$ and (1) is satisfied then
$$
\begin{align}
\mathbb{E}\left[\sup\_{s\le t}\vert M\_s\vert\right]&\le C\mathbb{E}\left[\left(\int\_0^t\xi^2d[\rho]\_s\right)^{1/2}\right]\\\\&\le C\mathbb{E}\left[[\rho]\_t^{1/2}\right]\\\\&\le Cc^{-1}\mathbb{E}\left[\sup\_{s\le t}\vert\rho\_s-\rho\_0\vert\right]\\\\& < \infty.
\end{align}
$$
This is enough to guarantee that the local martingale M is a proper martingale.
Conversely, to prove (2) ⇒ (1) then, assuming that $\sup\_{s\le t}\rho\_s$ is not integrable, you just need to find a bounded predictable process $\xi$ such that $\int\_0^t\xi d\rho$ is not a martingale. Supposing that $\rho$ is continuous and writing $\rho^\ast\_t=\sup\_{s\le t}\rho\_s$, you can use the identity
$$
\int\_0^t f^\prime(\rho^\ast\_s)d\rho\_s=f(\rho^\ast\_t)-f(\rho\_0)+f^\prime(\rho^\ast\_t)(\rho\_t-\rho^\ast\_t).
$$
This holds for all absolutely continuous functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$. For example, choosing $f^\prime(x)=\sum\_{n\in\mathbb{Z}}(-1)^n1\_{\{n\le x < n+1\}}$ then $f(x)=\int\_0^x f^\prime(y)dy$ will satisfy $0\le f\le1$. So, up to a bounded value, the right hand side of the above identity has absolute value $\vert\rho\_t-\rho^\ast\_t\vert$, which is not integrable, so is not a martingale. The case where $\rho$ is not continuous can be handled similarly, but is a bit messier.


