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# Markov Chain watch

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1. Let X_i, i=1,..,n be a sequence of independent and identically distributed random variables. Let S_n=X_1+...+X_n, with S_0=0. Define the sequence K_n (n>=0) by K_(n+1)=max(K_n + K_(n+1), 0), with K_0=0. I'd like to show that K_n has the same probability distribution as R_n=maximum[over 0<=m<=n](S_m).
2. I think your definition of K_(n + 1) should be max(K_n + X_(n + 1), 0). The symbol below means "has the same distribution as".

K_1
= max(X_1, 0)
= R_1

K_2
= max(max(X_1, 0) + X_2, 0)
= max(max(X_1 + X_2, X_2), 0)
= max(X_1 + X_2, X_2, 0)
max(X_1 + X_2, X_1, 0)
= R_2

K_3
= max(max(X_1 + X_2, X_2, 0) + X_3, 0)
= max(max(X_1 + X_2 + X_3, X_2 + X_3, X_3), 0)
= max(X_1 + X_2 + X_3, X_2 + X_3, X_3, 0)
max(X_1 + X_2 + X_3, X_1 + X_2, X_1, 0)
= R_3

etc

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Updated: November 16, 2005
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