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Letting $X_n$ be i.i.d taking the value $1$ with probability $p$, and $-1$ with probably $1-p$,how do I show that $Y_n = M_n - S_n$, where $$M_n = \max\{0, S_1, S_2,\ldots,S_n\}$$ and $$S_n = X_1+\ldots+X_n$$ is a Markov Chain? I can see that $M_n$ itself is not a Markov chain, but do not know how to show $Y_n = M_n - S_n$ is?

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  • $\begingroup$ You may verify that from one of the equivalent definitions. Please show us your work and where you're stuck, and avoid asking no-clue questions. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 13 '18 at 13:45
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You need to look at the transition probability. Observe first that since $M_n\geq S_n$, you have $Y_n\geq 0$. If $Y_n = 0$ then there is a $1/2$ probability that $Y_{n+1}=0$ and a $1/2$ probability that $Y_{n+1}=1$. if $Y_n>0$ then there is a $1/2$ probability that $Y_{n+1}=Y_n \pm 1$, this is because $M_{n+1}=M_n$ in this case. These are the transition probability for the normal one sided random walk.

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