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?

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    $\begingroup$ The quick answer is no. You need $\sup\_{s\le t}\rho\_s$ to be integrable to guarantee a positive answer. $\endgroup$ – George Lowther Sep 4 '11 at 17:08

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). 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.

  • $\begingroup$ Very interesting, thanks for a complete answer. I did learn quite a lot about stochastic calculus today! I asked another question with some more specific choice for $a_t$. $\endgroup$ – Grzenio Sep 5 '11 at 9:44
  • $\begingroup$ @ George : Is there a type in your first sentence ? You meant $p$ not $\rho$ right ? PS :I will erase this comment as soon as you corrected your post or (more probably) me. $\endgroup$ – The Bridge Sep 5 '11 at 11:36
  • $\begingroup$ @The Bridge: No typo. I did mean rho. In fact, the statement about rho implies the same statement about p, by the argument further down in my answer. $\endgroup$ – George Lowther Sep 5 '11 at 12:42
  • $\begingroup$ @George : Ok, though I thought that by Hypothesis the process $\rho$ was a martingale and not only a local martingale. Am I missing something about the definition of honest or proper martingales ? I thought they were only martingales and that the vocable "proper" or "honest" had no real mathematical meaning. $\endgroup$ – The Bridge Sep 5 '11 at 13:44
  • $\begingroup$ @The Bridge: yes, "proper" (or honest) martingales are simply martingales, just emphasising the fact that you don't mean local martingales in general. So I don't think you are missing anything there. But, the property that the maximum of rho is integrable is strictly stronger than the martingale property on it's own. $\endgroup$ – George Lowther Sep 5 '11 at 13:59

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