We have a simple 1-dimensional, $N$-step random walk with steps of $\{1,-1\}$. Let $s_k\in\{1,-1\}$ be the $k$-th step and $W_k=\sum_{i=1}^k s_i$ the position after $k$ steps.
How do I prove that
1. for all $0<k_1<k_2<...<k_n\leq N$ the vectors $W_{k_1}-W_0,W_{k_2}-W_{k_1},W_{k_3}-W_{k_2},...,W_{k_n}-W_{k_{n-1}}$ are independent?
2. an increment $W_m-W_k$ for $0<k<m\leq N$ has the same distribution as $W_{m-k}$?
3. a simple random walk is a Markov chain?
4. for $0<k<m\leq N$ we get $P(W_m=b|W_k=a)=P(W_{m-k}=b-a)$ for $a,b\in\mathbb{Z}$ and assuming $P(W_k=a)>0$?
5. $s_1,...,s_N$ are independent and identically distributed?
WHAT I KNOW:
1. I don't know where to start thinking about this.
2. I would like to prove that $P(W_m-W_k=a)=P(W_{m-k}=a)$ for all $a\in\mathbb{Z}$, but how?
3. This seems very logical, but how would I prove this mathematically?
4. I was thinking about using $P(A|B)=\frac{P(A\cap B)}{P(B)}$, but how?
5. I know that for $1\leq k_1<...<k_n\leq N$ and $a_{k_1}\in\{\pm1\}$ we have $P(s_{k_1}=a_{k_1},...,s_{k_n}=a_{k_n})=2^{N-n}2^{-N}=2^{-n}$. How can I use this?
Thanks in advance!