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Given two Martingale processes $(X_t)$ and $(Y_t)$, are their sum $(X_t+Y_t)$ and their product $(X_t \times Y_t)$ also Martingale?

If not, will the two $(X_t)$ and $(Y_t)$ being independent grant their sum and product Martingale? Thanks!

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Are you considering $t\in\mathbb N$ or $t\in\mathbb R_+$? – Morning Feb 16 '11 at 15:32
@Morning: can be both. Will they make difference? – steveO Feb 16 '11 at 16:05
@steveO: I guess you should make it precise what it means by $X_t+Y_t$ to be a martingale. With respect to the same filtration? In the discrete case, the martingale structure is relatively simple and you can prove/disprove easily. – Morning Feb 16 '11 at 16:54
I think it best to look at the martingale property with respect to a given filtration. This is the most common in my experience, and fits in naturally with the view of the filtration as representing the observable information up to each time. If you just look at the martingale property with respect to the filtration generated by the individual processes then you are, in some sense, missing some observable events and you cant expext it to behave so well. Maybe there is a good reason in your case, but I'm just pointing out what I consider the best approach is in general. – George Lowther Feb 16 '11 at 21:09
@George: Your comment seems like an excellent motivation to look at what happens for different filtrations... And this is exactly what Vershik, Tsirelson, Émery and others are studying. Keywords here are immersion, cosiness, standardness, hiding a drift, innovations... See Tsirelson's survey at the ICM'98 or Émery and Schachermayer's paper for example. – Did Feb 17 '11 at 21:48
up vote 13 down vote accepted

The product of two independent martingales is a martingale--or rather it is or it is not, depending on the precise formulation of the hypothesis! When it is, one says that the martingales are orthogonal. This is explained, for example, by Alexander Cherny in the chapter Some Particular Problems of Martingale Theory of the Shiryaev Festschrift.

And yes, the sum of two independent martingales is a martingale but, here again, it might be wise to state the result with some care and, first of all, as mentioned by steveO in a comment, to specify the filtration(s) one is considering.

The trivial version is that if $X$ and $Y$ are two martingales (independent or not) with respect to a given filtration $\mathcal{G}$, then the sum $X+Y$ is also a martingale with respect to $\mathcal{G}$. But what happens if one assumes that $X$ is a martingale with respect to its own filtration $\mathcal{F}^X$ and that $Y$ is a martingale with respect to its own filtration $\mathcal{F}^Y$?

(Recall that the filtration $\mathcal{F}^Z$ of a process $Z$ is defined by $\mathcal{F}^Z_n=\sigma(\{Z_k;k\le n\})$ for every $n$.)

Then, if $X$ and $Y$ are independent, $X+Y$ is a martingale with respect to its own filtration $\mathcal{F}^{X+Y}$ but, first, the proof, while not terribly difficult, requires to be careful (and uses more than the linearity of conditional expectations), and, second, for non independent martingales, this becomes horribly wrong.

To get an idea of the problem, consider a given integrable random variable $\xi$ and two $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$ and try to find conditions guaranteeing that $E(\xi|\mathcal{A}\vee\mathcal{B})=E(\xi|\mathcal{A})$. Is $\xi$ independent of $\mathcal{B}$ enough? No, one has to assume that $\mathcal{B}$ is independent of $\sigma(X)\vee\mathcal{A}$.

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is this related to the orthomartingales? (The reference that I am aware of is Multiparameter processes by Khoshnevisan.) I briefly looked over Cherny's paper. It seems not. – Morning Feb 16 '11 at 16:58
@Morning: Not in ways that I would be aware of. – Did Feb 17 '11 at 21:49
@did Could you please provide an easy example of such two martingales "...non independent martingales, this becomes horribly wrong". I tried but couldn't come up with anything. – grozhd Dec 13 '12 at 19:35
@grozhd Could you please start with the indication on exactly this, which I took pain to add? – Did Dec 13 '12 at 19:40

The sum is a martingale by linearity of conditional expectation.

The product is not a martingale in general - if $X_t$ is a martingale, then $X_t \times X_t = X_t^2$ is a submartingale by Jensen's inequality.

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Thanks! I mean if the sum is Martingale wrt itself. When proving $E(X_{n+1} + Y_{n+1} | X_i + Y_i, i=1,...,n) = X_n + Y_n$, will the left hand side be represented in terms of $E(X_{n+1} | X_i, i=1,...,n)$ and $E(Y_{n+1} | Y_i, i=1,...,n)$? If yes, how to separate the condition X_i + Y_i, i=1,...,n into X_i i=1,...,n and Y_i, i=1,...,n – steveO Feb 16 '11 at 16:20

I am wondering whether we can write some simple proof without much knowledge of measure theory. I also face this problem in my introductory level course. Suppose that $X_t$ and $Y_t$, $t\in \mathbb{R}$ are two independent martingales. Then can we just say $\mathbb{E}(X_t+Y_t|X_u+Y_u,u\le s)=\mathbb{E}(X_t|X_u+Y_u,u\le s)+\mathbb{E}(Y_t|X_u+Y_u,u\le s)=\mathbb{E}(X_t|X_u,u\le s)+\mathbb{E}(Y_t|Y_u,u\le s)=X_s+Y_s$.

And for the product, can we say $\mathbb{E}(X_tY_t|X_uY_u,u\le s)=\mathbb{E}(X_t|X_uY_u,u\le s)\mathbb{E}(Y_t|X_uY_u,u\le s)=\mathbb{E}(X_t|X_u,u\le s)\mathbb{E}(Y_t|Y_u,u\le s)=X_sY_s$.

Then both the sum and the product are martingales.

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This suggested edit probably should have been a comment. – Martin Sleziak Mar 19 '13 at 13:13

Regarding the product of two MGs, for square integrable Martingales $X_t$ and $Y_t$ , you have something called the cross variation process $<X,Y>_t$, this being zero (as) would imply $XY$ is a MG. (See Karatzas Shreve, section 1.5).

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Welcome to Math.SE. When posting a new Answer to an older (by four years) Question, it is less than helpful to provide such a short comment without pointing how it adds to the material presented in existing Answers. – hardmath Feb 14 at 4:01
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. - From Review – Ben S. Feb 14 at 5:15

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