Let $X_n$ be largest number of first n throws of a fair six-sided die, show that $X = \{X_n: n\ge 1\}$ is homogeneous Markov chain
My try:
Let $Y_i$ denote the number of $i$th throw, then $X_n = \max\{Y_1,Y_2,...,Y_n\}$,
$$P(X_{n+1} = j | X_n = i, X_{n-1} = x_{n-1},...,X_1 = x_1) = P(\max\{X_n, Y_{n+1}\} = j | X_n = i, X_{n-1} = x_{n-1},...,X_1 = x_1)$$
Since $\max\{X_n,Y_{n+1}\} = j$ only depends on $X_n$ and $Y_{n+1}$ , it seems that
$$P(\max\{X_n, Y_{n+1}\} = j | X_n = i, X_{n-1} = x_{n-1},...,X_1 = x_1) = P(\max\{X_n, Y_{n+1}\} = j | X_n = i)$$
If so, why?
About homogeneity, when trying to verify $P(X_{n+1} = j | X_n = i) = P(X_2 = j | X_1 = i)$, it seems that computation is complicated since you have to check corner case $i=1 \ or \ 6.$
Any hint?