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I have $X_1,...,X_n \sim i.i.d.(0,1)$ and $S_n=X_1+...+X_n$,

I define trunctation $X_i^{(v)}=X_i\textbf1_{\{|X_i|\leq v\}}-E[X_i\textbf1_{\{|X_i|\leq v\}}]$ and $S_k^{(v)}=\Sigma_{i=1}^k X_i \textbf1_{\{|X_i|\leq v\}}-E \left[ \Sigma_{i=1}^k X_i \textbf1_{\{|X_i|\leq v\}} \right]$. So i obtain random walk $\{ S_i^{(v)} \}_{i=1}^n$ which increments are bounded by $2v$. In my paper there is random walk is mean zero. Thanks to this two properties it's a martingale.

I wonder if there is any definition of martingale I don't know. In my opinion i should have filtration $F_n = \sigma(X_1,...,X_n)$ and then checked the definition.

Secondly there is the same conclusion with $\left\{ S_n - S_n^{(v)} \right\}_{i=1}^n$. In paper I have it's mean zero P-integrable random walk therefore it's a martingale.

Can someone please explain me or tell definition I need to understand this.

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1 Answer 1

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The typical definition of a martingale is that $$\mathbb E[X_{n+1} \mid \mathcal F_n] = X_n$$ where $X_n$ is adapted to the filtration $\mathcal F_n$. This is equivalent to $$\mathbb E[X_{n+1} - X_n \mid \mathcal F_n] = 0$$ since $X_n \in \mathcal F_n$. Note that it is common to take $\mathcal F_k = \sigma(X_1, \dots, X_k)$.

Any stochastic process with mean-zero (hence, integrable) increments is indeed a martingale with respect to the natural filtration $\mathcal F_k = \sigma(X_1, \dots, X_k)$.

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  • $\begingroup$ Okay I understand. However I will have to think about it for a second because I don't see why the truncated sum has mean 0. $\endgroup$
    – nodis6
    Commented Jun 15, 2022 at 8:11
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    $\begingroup$ Note that anything minus its own expectation necessarily has mean 0. $\endgroup$ Commented Jun 15, 2022 at 10:50
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    $\begingroup$ Oh, it's the most usefull comment i've read. Thank you very much for help. $\endgroup$
    – nodis6
    Commented Jun 15, 2022 at 10:54

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