Let $X_n$, $n\geqslant 1$ be iid Rademacher random variables, i.e. $X_1$ takes values $\pm 1$ each with probability $\frac12$. Define $S_0=0$ and $S_n=\sum_{i=0}^n X_i$, and $\mathcal F_n = \sigma(S_0, \ldots, S_n)$. Then it is straightforward to show that $|S_n|$ is a submartingale with respect to $\{F_n\}$, so it has a (unique) Doob decomposition $$|S_n| = M_n + A_n, $$ where $M_n$ is a martingale and $A_n$ is a nondecreasing predictable process. Explicitly, \begin{align} A_n &= \sum_{k=1}^n \mathbb E\left[|S_k|-|S_{k-1}| \mid \mathcal F_{k-1}\right]\\ M_n &= |S_n| - A_n. \end{align} I'm trying to find a simpler expression for $M_n$, of the form $M_n = (H\cdot S)_n$ where $H$ is a predictable process and $\cdot$ denotes martingale transform, but I got bogged down in the algebra. Any hints, either as to how to approach this or what $H$ might look like?


1 Answer 1


We have to consider three cases separately:

  • $S_{k-1}(\omega)<0$: Then $S_k(\omega) \leq 0$ and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)| = -S_k(\omega)+ S_{k-1}(\omega) = - X_k(\omega).$$
  • $S_{k-1}(\omega)=0$: Then $|S_k(\omega)|=1$ and therefore $$|S_k(\omega)|-|S_{k-1}(\omega)|=1.$$
  • $S_{k-1}(\omega)>0$: Then $S_k(\omega) \geq 0$ and therefore $$|S_k(\omega)| - |S_{k-1}(\omega)| = S_k(\omega)-S_{k-1}(\omega) =X_k(\omega).$$

This shows

$$|S_k|-|S_{k-1}| = -X_k 1_{\{S_{k-1} <0\}}+ 1_{\{S_{k-1}=0\}}+ X_k 1_{\{S_{k-1}>0\}}. \tag{1}$$

Using this identity and the independence of the random variables, we get

$$\mathbb{E}(|S_k|-|S_{k-1}| \mid \mathcal{F}_{k-1}) = 1_{\{S_{k-1}=0\}}.$$


$$M_n = |S_n| - \sum_{k=1}^n 1_{\{S_{k-1}=0\}} = \sum_{k=1}^n (|S_k|-|S_{k-1}|-1_{\{S_{k-1}=0\}}).$$

By $(1)$, this implies

\begin{align} M_n &= \sum_{k=1}^n (-X_k 1_{\{S_{k-1}<0\}}+X_k 1_{\{S_{k-1}>0\}})\\ &= \sum_{k=1}^n (- 1_{\{S_{k-1}<0\}}+ 1_{\{S_{k-1}>0\}})X_k\\ &= \sum_{k=1}^n (- 1_{\{S_{k-1}<0\}}+ 1_{\{S_{k-1}>0\}})(S_k-S_{k-1})\\ \\ &= (H \bullet S)_n\end{align}


$$H_k := - 1_{\{S_{k-1}<0\}}+ 1_{\{S_{k-1}>0\}}.$$

  • $\begingroup$ Thanks for the answer, I'm at work atm so I'll have to digest it later :) $\endgroup$
    – Math1000
    Jul 1, 2015 at 19:03
  • $\begingroup$ @Math1000 Yeah, no problem. I'll go to sleep :) $\endgroup$
    – saz
    Jul 1, 2015 at 19:07
  • $\begingroup$ It all makes sense. I should have thought to write $|S_n|$ as a telescoping sum of $|S_k|-|S_{k-1}|$. Thanks again! $\endgroup$
    – Math1000
    Jul 1, 2015 at 23:45
  • $\begingroup$ I added a little detail in the last part to make it a bit more explicit, hope you don't mind @saz $\endgroup$
    – Math1000
    Jul 2, 2015 at 13:49
  • 2
    $\begingroup$ +1. One sometimes write the result as $$H_k=\mathrm{sign}(S_{k-1}),$$ with the inconvenience that this forces to define the sign of $0$ as being $0$, but the advantage (which can be debated...) of mimicking Tanaka's formula. $\endgroup$
    – Did
    Jul 24, 2015 at 15:59

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