I'm trying to understand what is a Markov chain through One Thousand Exercises in Probability by Geoffrey Grimmet and David Stirzaker. I'm on exercise 6.1.2 of this book.
I know that a $X_n$ is a Markov chain if $$P(X_{n+1}=j|X_1=i_1,...,X_n=i_n)=P(X_{n+1}=j)$$
I'm trying to find out if the largest number $X_n$ shown up to the nth roll is a Markov chain?
with $Y_n$, the outcome of the nth throw, $X_{n+1}=$max{$X_n,Y_{n+1}$}, I don't understand why (solution given by the book):
$$ p_{s,i} = \left\{ \begin{array}{ll} 0 & \mbox{if} \ \ \ j<i\ \\ \frac{1}{6}i & \mbox{if j=i}\\ \frac{1}{6} & \mbox{if}\ j>i \end{array} \right. $$
for $1 \le i, j\le 6$ what probability is that? the one of $X_{n+1}$? How is it obtained, I understand the first line but neither the second nor the last one.
the equation above leads similarily according to the book to:
$$ p_{s,i}(n) = \left\{ \begin{array}{ll} 0 & \mbox{if} \ \ \ j<i\ \\ (\frac{1}{6}i)^n & \mbox{if j=i}\\ \end{array} \right. $$
The Markov chain in question has the following states: $$\{1,2,3,4,5,6\}.$$ Let $p_{i,j}^n$ denote the probability that the chain goes to state $j$ assuming that it is in state $i$ before the $(n+1)^{th}$ roll:
$$p_{i,j}^n=P(X_{n+1}=j|X_n=i).$$
This conditional probability is $0$ if $i>j$ since the state changes only if the next roll results in a larger number than the maximum achieved so far.
$i=j$ means that the chain remains in state $i$. This event takes place if the result of the next roll is less or equal than $i$. There are $i=\#\{1,2...,i\}$ such possibilities of equal probability: $\frac16$. This is why
$$p_{i,j}^n=P(X_{n+1}=j|X_n=i)=\frac i6, \text{ if } i=j.$$
There is only one possibility of probabilty $\frac16$ to get to the state $j>i$ from state $i$, namely the result of the roll has to be exactly $j$. This is why
$$p_{i,j}^n=P(X_{n+1}=j|X_n=i)=\frac 16, \text{ if } i<j.$$
I am sorry but I don't understand what $$ p_{s,i}(n) $$ is.