sum of submartingales is not submartingale?

Please give an example satisfying: submartingale $(X_n)\; w.r.t\; \mathcal{(F_n)}, (Y_n)\; w.r.t \;\mathcal{(G_n)}$. But $(X_n + Y_n)$ is not a submartingale $w.r.t$ any filtration.

Thanks

Solution 1:

Let $Z_i$ be i.i.d random variables with $\mathbb{E} Z_1 = 0$. Let $X_n =- \sum_1^n Z_i$ with $\mathcal F_n = \sigma(Z_1,\cdots,Z_n)$ and let $Y_n = \sum_1^{n+1} Z_i$ with $\mathcal G_n = \sigma(Z_1,\cdots,Z_{n+1} )$. Notice that both $(X_n)$ and $(Y_n)$ are submartingales with respect to the filtrations they generate (even stronger they are martingales). To show a contradiction, that $(X_n + Y_n)$ is a submartingale with respect to a filtration $(\mathcal H_n)$. Our construction gives $X_ n + Y_ n = Z_{n+1}$, so $(Z_{n+1})$ is a submartingale with respect to $(\mathcal H_n)$. Since $\sup_n \mathbb E[Z_{n+1}] = 0 < \infty$ we can apply the martingale convergence theorem (for submartingales) and conclude that $$Z_n \overset{a.s.} \to Z_\infty,$$ for some random variable $Z_\infty$. Thus we would have $$Z_i \text{ are i.i.d with $\mathbb E Z_i = 0$ } \implies Z_i \text{ converge a.s.} \qquad \qquad (*).$$ (*) is false, for example consider $\Omega = \{ 0,1\}^\infty$ and $Z_i = Ber(\frac 12)$. The $Z_i$ do not converge almost surely to any random variable since Bernouli random variables satisy $\{\omega : Z_i \text{ converge} \}$ has measure zero.