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?


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 '15 at 19:03
  • $\begingroup$ @Math1000 Yeah, no problem. I'll go to sleep :) $\endgroup$ – saz Jul 1 '15 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 '15 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 '15 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 '15 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.