absolute value random walk also a markov chain Consider the random walk $S_{n}$ for $n \geq 1$. Specifically, let $X_{1},X_{2},..$ be Independent with
$$
\mathbb{P}(X_{n}=1) = p,~~\mathbb{P}(X_{n}=-1) = 1- p =:q
$$
and $S_{n} = \sum_{k=1}^{n}X_{k}$. I have read in the book of Ross, Stochastic Processes on page 166, that $|S_{n}|$ is also a markov chain. He uses Proposition 4.1.1. to prove this. Here the Proposition

My question is why 
$$
\mathbb{P}(S_{n} = i,...,|S_{j+1}| = i_{j+1},|S_{j}| = 0) = p^{\frac{n-j}{2}+\frac{i}{2}}q^{\frac{n-j}{2}-\frac{i}{2}}
$$
and
$$
\mathbb{P}(S_{n} = -i,...,|S_{j+1}| = i_{j+1},|S_{j}| = 0) = p^{\frac{n-j}{2}-\frac{i}{2}}q^{\frac{n-j}{2}+\frac{i}{2}}
$$
I know that if $S_{n} = i$ then because of the definition of $j$ $S_{j} = 0$ and $S_{j+1},...,S_{n-1} > 0$ and similar for $S_{n} = -i$ I have $S_{j+1},...,S_{n-1} < 0$. 
I hope someone can help me

The probabilities in your last two displayed equations aren't quite right. It's not
$$
\mathbb P(S_n=\pm i,\ldots,|S_{j+1}|=i_{j+1},|S_j|=0)=p^{\frac{n-j}2\pm\frac i2}q^{\frac{n-j}2\mp\frac i2}
$$
but
$$
\mathbb P(S_n=\pm i,\ldots,|S_{j+1}|=i_{j+1}\;\big|\;|S_j|=0)=p^{\frac{n-j}2\pm\frac i2}q^{\frac{n-j}2\mp\frac i2}\;,
$$
that is, given that we know that $S_j=0$, the probability for the remaining sequence to occur is given by the right-hand side. This is the case because to reach $\pm i$ in $n-j$ steps, we need to take $\frac{n-j}2\pm\frac i2$ steps in the positive direction with probability $p$ and $\frac{n-j}2\mp\frac i2$ steps in the negative direction with probability $q$: The total number of steps is $\frac{n-j}2\pm\frac i2+\frac{n-j}2\mp\frac i2=n-j$, and the distance traveled is $\frac{n-j}2\pm\frac i2-\left(\frac{n-j}2\mp\frac i2\right)=\pm i$.
