# $M_t$ martingale wrt natural filtration, still a martingale wrt completed filtration?

Let the independent processes $M=(M_t, t\ge 0)$ and $X=(X_t, t\ge 0)$ be defined on the probability space $(\Omega, \mathcal F, P)$.

Assume $M$ is a martingale with respect to its natural filtration $\mathcal F^M_t$.

Denote $\mathcal F^X_t$ as the natural filtration of $X$.

Will $M$, due to the independence of the two processes, still be a martingale with respect to the filtration $\mathcal F^M_t\vee \mathcal F^X_\infty$?

If the above is true, and $X$ is a positive increasing and pathwise continuous process with $X_0=0$ a.s., would the process $M(X)=(M_{X_t}, t\ge 0)$ be well defined on the same probability space? In that case would $M(X)$ also be a martingale with respect to the natural filtration of $M(X)$?

I just saw that there is a similar question with answer: Martingale preservation under independent enlargement of filtration

• @ByronSchmuland I simplified the question slightly. Aug 8, 2017 at 14:01
• Looks good! I will think about your problem.
– user940
Aug 8, 2017 at 16:01

Fix $t>s$ and let ${\cal C}=\{A_M \cap A_X: A_M\in{\cal F}^M_s, A_X\in{\cal F}^X_\infty \}.$

The martingale property for $M$ means that for $A_M\in{\cal F}^M_s$, we have $$\mathbb{E}\left((M_t-M_s){\bf 1}_{A_M}\right)=0.\tag1$$

Thus, for any set in $\cal C$ the independence of $M$ and $X$ gives
$$\mathbb{E}\left((M_t-M_s){\bf 1}_{A_M \cap A_X}\right) =\mathbb{E}\left((M_t-M_s){\bf 1}_{A_M}{\bf 1}_{A_X}\right)= \mathbb{E}\left((M_t-M_s){\bf 1}_{A_M}\right) \mathbb{E}({\bf 1}_{A_X})=0.$$

Since $\cal C$ is a $\pi$-system that generates ${\cal F}^M_s\vee{\cal F}^X_\infty$, the Monotone Class Theorem says that $$\mathbb{E}\left((M_t-M_s){\bf 1}_{A}\right)=0\tag2$$ for every $A\in {\cal F}^M_s\vee{\cal F}^X_\infty$. This means that $(M_t)$ is a martingale with respect to the filtration $({\cal F}^M_t\vee{\cal F}^X_\infty).$

• Thank you for your answer! A small question, how could one assure $\mathcal C$ not to be empty? Aug 8, 2017 at 21:33
• ${\cal F}^M_s\cup{\cal F}^X_\infty\subseteq{\cal C}$
– user940
Aug 8, 2017 at 21:40
• Ok! Thank you for the fast reply. In connection to the old question which I "ment to" state (but changed): can a probability space always be constructed in such a way that any two stochastic processes defined on that probability space follows any wanted distributions. For example, can I always construct a probability space for which $M$ is any mean corrected levy process and $X$ an integrated CIR och OU process or any other pathwise continuous increasing process, $M$ and $X$ again being independent? Aug 8, 2017 at 21:51
• As you probably can guess, i'm rather new to this field and still an undergrad. :-) Aug 8, 2017 at 22:06
• Yes, you just need to take the product space equipped with the product measure. This is what you meant to do in the first version of your post, but some of your notation was not correct.
– user940
Aug 8, 2017 at 23:00