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I know that if $(X_n)$ and $(Y_n)$ are martingales with respect to their natural filtration then $(X_n +Y_n)$ does not have to be a martingale with respect to its natural filtration. However, I did not manage to come up with an example. I would really appreciate if one could point out an example to me.

user007
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  • Well, we have $|X_n+Y_n|{L^{1}}\leq |X_n|{L^{1}}+|Y_n|_{L^{1}}<+\infty$ and $X_n, Y_n\sim \mathcal{F}_n$. Also $E(X_n+Y_n|\mathcal{F}_n)=X_n+Y_n$. Then the sum of two martingales is martingale. – A. P. May 09 '24 at 05:22
  • It is meant like here in the last part of the sentence of the second to last paragraph https://math.stackexchange.com/questions/22367/sum-and-product-of-martingale-processes – user007 May 09 '24 at 05:23
  • Ok, I misinterpreted the usual question. Yes, a priori, when we consider a martingale $X_p$ with its filtration $F_{p}^{X}$ and $Y_q$ with its own filtration $F_{q}^{Y}$, the sum $X_p+Y_p$ with respect to the common filtration $F_{r}^{X+Y}$ may not be a martingale. The way I would construct an example is by adding some noise to one of the martingales, something like $X_n = X_{n-1} + Z_n$ and $Y_n = Y_{n-1} + Z_n + W_n$, with the initialization of the iteration at $0$ and suitable choices of the increments. – A. P. May 09 '24 at 06:00

1 Answers1

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Let $(X_n)$ and $(Y_n)$ be two simple random walks defined on the same probability space, generated by a sequence of i.i.d. fair coin tosses $\omega_1,\omega_2,\dots $ with $\omega_i\sim U\{-1,1\}$ such that $$\begin{align*} X_n &= \omega_1 +\omega_2 +\dots + \omega_n \\ Y_n &= \omega_2 + \omega_1 + \dots +\omega_n \end{align*}$$

Then $(X_n)$ and $(Y_n)$ are martingales with respect to their natural filtrations, but $$\mathbb{E}[X_2+Y_2|X_1+Y_1]=\mathbb{E}[2(\omega_1+\omega_2)|\omega_1+\omega_2]=X_1+Y_1+\omega_1+\omega_2\neq X_1+Y_1$$ so the sum is not a martingale with respect to its natural filtration.

Joseph Basford
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