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Let $X_n = \sum_{m≤n} 1_{B_m} $and suppose $B_n \in F_n$. What is the Doob decomposition for $X_n$ ? I can write it down from the construction of the theorem, but is there any neat way showing the result?

Michael Hardy
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annimal
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1 Answers1

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It follows from the fact that $1_{B_k}$ is $\mathcal{F}_{n-1}$-measurable for $k \leq n-1$ that

$$\mathbb{E}(X_n \mid \mathcal{F}_{n-1}) = X_{n-1} + \mathbb{P}(B_n \mid \mathcal{F}_{n-1}).$$

Subtracting $A_n := \sum_{k=1}^n \mathbb{P}(B_k \mid \mathcal{F}_{k-1})$ on both sides yields

$$\mathbb{E}(X_n - A_n \mid \mathcal{F}_{n-1}) = X_{n-1} - A_{n-1}.$$

This shows that $M_n := X_n-A_n$ is a martingale. Consequently, the Doob decomposition is given by

$$X_n = \underbrace{\sum_{k=1}^n 1_{B_k}-\mathbb{P}(B_k \mid \mathcal{F}_{k-1})}_{\text{martingale}}+ \underbrace{\sum_{k=1}^n \mathbb{P}(B_k \mid \mathcal{F}_{k-1})}_{\text{predictable}}.$$

saz
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  • Hi, saz. I found your and Alex's answer is different. I think you are both right. Does that mean it is possible to have two answers? – Sherry Nov 21 '14 at 18:51
  • @Tuyet No, the Doob decomposition is unique (at least, if we require $M_0 = A_0=0$). – saz Nov 21 '14 at 19:59
  • Hi,saz. Can I decompose it like $$X_n = \underbrace{\sum_{k=1}^n 1_{B_k}-\mathbb{P}(B_k \mid \mathcal{F}{k-1})+\mathbb{P}(B_1 \mid \mathcal{F}{0})}{\text{martingale}}+ \underbrace{\sum{k=1}^n \mathbb{P}(B_k \mid \mathcal{F}{k-1})-\mathbb{P}(B_1 \mid \mathcal{F}{0})}_{\text{predictable}}.$$ To make the predicable part start from $0$? Thanks! – Sherry Nov 21 '14 at 23:18
  • Because I see $X_0$ is not defined here. So I make it start from $A_1=0$? Is that right?:) – Sherry Nov 21 '14 at 23:29
  • @Tuyet yes, you can do so. – saz Nov 22 '14 at 07:07