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.
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You can not be sure that $(X_n+Y_n)$ is adapted to either filtrations. – Stefan Hansen Nov 05 '12 at 07:20
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I try to construct an example... – Mike Nov 05 '12 at 07:28
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Is there any example can illustrate that? – Mike Nov 05 '12 at 07:36
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- construct $X_i, Y_i, i=1,...5$ so that you can tell what X and Y are individually, e.g., X integral $Y_i \in (\frac 14 ,\frac 34)$, alternately be willing to assume both $F_i, G_i$ are subfields of your field. 2. Choose $Y_1,...,Y_5$ equivalent to $X_5,...,Y_1$, scale the X's small, make them decreasing, get yourself a submartingale where $Y_1, Y_2$ determines $X_4,X_5$.
– mike Nov 05 '12 at 11:45
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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.
foamster
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Note that if $(Z_n,H_n)_{n \geq 0}$ is a martingale/sub-martingale/super-martingale, we have that $(Z_n,F_n)$ is also a martingale/sub-martingale/super-martingale, where $F_n = \sigma(Z_0,\cdots,Z_n)$, because $\mathbb{E}[\mathbb{E}[Z_{n+1}|H_n]|F_n] = \mathbb{E}[Z_{n+1}|F_n]$.
Now if $Z_n$ are iid., then we have $\mathbb{E}[Z_{n+1}|F_n] = \mathbb{E}[Z_{n+1}]=\mathbb{E}[Z_{n}]$. Thus the martingale(/sub-martingale/super-martingale) condition becomes $\mathbb{E}[Z_{n}] =(/\geq/\leq) Z_n, \forall n\geq 0$, respectively. Which implies $Z_n$ is a constant sequence. Thus a non constant iid. sequence cannot be a martingale/sub-martingale/super-martingale.
This coupled with @foamster's solution gives an example without resorting to martingale convergence theorems.
