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I need to give a proof for some lemma which states the following:

If $(X_n,\mathbb{F}_n)_{n\geq 1}$ is a submartingale and $\tau$ is a stopping time, then $(X_{\tau\wedge n},\mathbb{F}_{n})_{n\geq 1}$ is also a submartingale.

Now the CBS-principle ("Can't Beat the System") is a result. I've shown that if $(X_n,\mathbb{F}_n)_{n\geq 1}$ is a submartingale, and $(C_n)_{n\geq 1}$ is predictable and satisfies that each $C_n$ is non-negative and boundend, then $\big((C\cdot X)_n,\mathbb{F}_n\big)_{n\geq 1}$ is a submartingale.

I know that I can write $(C\cdot X)_n$ as: $$(C\cdot X)_n = \sum_{i=2}^n C_i\Delta X_i$$ Since $(C\cdot X)_1 = 0$ by assumption and $\Delta$ is just the standard difference operator.

Now how do I proceed from this? I guess I have to find some relation between $(C\cdot X)_n$ and $X_{\tau\wedge n}$.

mas2
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  • Can you elaborate on this "CBS-principle"? On Google all I find is some song on YouTube... – Math1000 Dec 11 '19 at 23:27
  • Sure. If you consider $\Delta X_n = X_n - X_{n-1}$ we can interpret this as our net winnings per unit stake at time $n$ and if $C_n$ is our gambling strategy, then $(C\cdot X)_n$ is our total winnings up to time $n$. We say that the game is rigged, i.e. each subgame is favourable to the system. And as long the gambler cannot look into the future, the system will win in the long run. – mas2 Dec 12 '19 at 11:59

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Hint:

You need to pick the right $C$:

-Keep adding the increments for $i$ satisfying $\{i\le\tau\}$. Note that this event is in ${\cal F}_{i-1}$.

-Don't add any increments for larger $i$.

In other words, take $C_i = {\bf 1}_{\{i\le \tau\}}$, and then $(C\cdot X)_n = \sum_{i=2}^n {\bf 1}_{\{i\le \tau\}} (X_i -X_{i-1}) = \sum_{i=2}^{\tau \wedge n} (X_i -X_{i-1}) = X_{n \wedge \tau} -X_1$. It therefore follows that $X_{n \wedge \tau} = X_1 + (C\cdot X)_n$ is a martingale ($X_1$ is measurable with respect to ${\cal F}_n$ for all $n\ge 1$).

Fnacool
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  • Thanks! This clearly shows the relation! And the event $(i\leq \tau)$ being $\mathbb{F}_{i-1}$-measurable just follows from the "predictable"-assumption? – mas2 Dec 12 '19 at 12:15
  • Predictable means $C_i \in {\cal F}_{i-1}$: the amount you bet on the $i$-th game is a function of whatever happened before. Since ${i \le \tau} = {i-1 < \tau}= {\tau \le i-1}^c$, it follows that the process $C$, as defined, is predictable. – Fnacool Dec 12 '19 at 15:21