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I need to determine wheather the process of running maximum of random walk is a martingale or not. We have random variables $\xi_i : \forall i \rightarrow P(\xi_i = 1) = P(\xi_i = - 1) = \frac{1}{2}$ also we have a random walking $S_n = \sum_{k = 1}^n \xi_i, S_0 = 0$. Denote $M_n = \max(\xi_1, ..., \xi_n)$. Is the process $M_n$ a martingale adapted to the filtration $\sigma\{\xi_1, ..., \xi_n\}$?

I tried to check a martingale property, that $\forall \ m \le n \rightarrow E(M_n \mid \sigma\{\xi_1, ..., \xi_m\}) = M_m$. As I understood, in case $\max(\xi_1, .., \xi_n) = \max(\xi_{m+1}, .., \xi_n)$ the conditional expected value $$E(\max(\xi_{m+1},...,\xi_n) \mid \sigma\{\xi_1,...,\xi_n\}) = E(\max(\xi_{m+1},...,\xi_n)) \neq M_m \rightarrow M_n$$ isn't a martingale. I understand that it's not a full solution, but some intuitive considerations. Could you please tell me whether the direction of thinking is right or not. And would be grateful for any hints.

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1 Answer 1

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$(M_n)_{n \in \mathbb{N}}$ is not a martingale since it doesn't have constant expectation. Indeed: Clearly, we have

$$\mathbb{E}(M_1) = \mathbb{E}(\xi_1)=0.$$

On the other hand,

$$\mathbb{P}(M_2 = -1) = \mathbb{P}(\xi_1=-1,\xi_2=-1) = \frac{1}{4}$$

and so

$$\mathbb{P}(M_2 = 1) = \frac{3}{4}$$

implying

$$\mathbb{E}(M_2) = (-1) \frac{1}{4} + \frac{3}{4} = \frac{1}{2} \neq 0 = \mathbb{E}(M_1).$$

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  • $\begingroup$ I'm sorry, I revealed that I didn't type my question properly, it should be $max(S_1,..S_n)$ not $max(\xi_1,...,\xi_n)$ and for such process expected value seems equal to zero. Do you know, the way I should follow to solve this task, or it would be better to edit the question? $\endgroup$
    – D F
    Dec 11, 2017 at 15:50
  • $\begingroup$ @DF Why do you think that $\mathbb{E}(\max\{S_1,S_2\})=0$?(... and the proper way would be to ask a new question... You should avoid changing the question after it has been answered.) $\endgroup$
    – saz
    Dec 11, 2017 at 15:57
  • $\begingroup$ yes, I'm wrong, it's equal to $\frac{1}{2}$ and $M_n$ isn't a martingale. Thank you again :) $\endgroup$
    – D F
    Dec 11, 2017 at 16:10

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