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Given any sequence of r.v.s $\{X_n\}$ s.t. $X_n\in L^1$ for all $n$ and history $\{\mathcal{F_n}\}$ with $X_n\in \mathcal{F_n}$ for all $n$. How can we prove that this sequence of r.v.s can be written as a sum of a supermartingale and submartingale?

Thank you!

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    Why did this flat PSQ receive upvotes? – 2017-11-20

1 Answers 1

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Hints:

  1. Define a sequence of events $A_n: = \big\{E[X_{n+1}|\mathcal F_n] \geq X_n\big\}$, and similarly define $B_n:= A_n^c = \big\{E[X_{n+1}|\mathcal F_n] < X_n \big\}$. Note that $A_n,B_n \in \mathcal F_n$ for all $n$.

  2. Define $Y_n: = \sum_{j=0}^{n-1} (X_{j+1}-X_j)1_{A_j}$ and $Z_n:=\sum_{j=0}^{n-1}(X_{j+1}-X_j)1_{B_j}$, with $X_0:=0$. Note that $X_n=Y_n+Z_n$ for all $n$.

  3. Check that $Y_n$ is a supermartingale and $Z_n$ is a submartingale, by computing $E[Y_{n+1}-Y_n|\mathcal F_n]$ and same for $Z$. Conclude.

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    If only require written as a sum of submartingale and supermartingale, the following is also a simple decomposition: $X_n=\sum_{j=0}^{n-1}(X_{j+1}-X_j)^++\sum_{j=0}^{n-1}[-(X_{j+1}-X_j)^_]$. – 2017-02-22
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    @JGWang That's even better than what I wrote above, because it really only uses the elementary fact that any real sequence is the sum of a nondecreasing sequence and a nonincreasing one. I guess my answer is an alteration of the same principle too. You should post an answer! – 2017-02-22
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    Thank you for your reply. The decomposition you offered is meaningful also. Since it is "predictable" and may be denoted by "(discrete) stochastic integrals". – 2017-02-23