Let $(S_k,k∈N_0 )$ be the symmetric random walk, that is, the process defined by
\begin{align}
S_0≔0 ; S_k =∑_{j=1}^kX_i ,k≥1
\end{align}
where the random variables $\{X_i \}_{i∈N}$ are independent and identically distributed, with
\begin{align}
P(X_i)=\dfrac{1}{2}=1-P[X_i=-1]
\end{align}
Prove that given any sequence of times $0=k_0<k_1<k_2<⋯<k_n$ , the increments $\{S_{k_i}-S_{k_{i-1}}\}_{i=1}^n$ are stationary and independent.
Then prove that $\forall k,m \in \mathbb{N}$
\begin{align}
\mathbb{E} [S_{k+m} – S_{k}] = 0
\end{align}
and
\begin{align}
Var [S_{k+m} – S_{k}] = m
\end{align}
My solution so far
So first we have these are stationary and independent.
Let
\begin{align}
Z_{a,b} = \{ S_i \neq 0, a \leq i <b, S_b=0\}
\end{align}
and
\begin{align}
Z'_{a,b} = \{ S_i – S_{a-1} \neq 0, a \leq i <b, S_b-S_{a-1}=0\}
\end{align}
For $i<j$, then $S_j-S_i$ it's a function that depends only on $X_{i+1}, \cdots , X_j$ random variables.
Therefore, $Z'_{b+1,c}$ only involves random variables $X_{b+1}, \cdots, X_c$ for any c.
Furthermore, $S_i$ is a function of $X_1, \cdots, X_i$ only. Then the event $S'_{a,b}$ involves only random variables $X_1, \cdots, X_b$
And because $X_1, X_2, \cdots $ are independent, then the events defined by $S_{a,b}$ and $S'_{b+1}$ are independent.
After that
Now I must prove
\begin{align}
\mathbb{E} [S_{k+m} – S_{k}] = 0
\end{align}
and
\begin{align}
Var [S_{k+m} – S_{k}] = m
\end{align}
So we have
\begin{align}
\mathbb{E} [S_{k+m} – S_{k}] = \mathbb{E} \left[ \sum_{j=k}^{k+m} X_j – \sum_{j=1}^{k} X_j \right] = \mathbb{E} \left[ \sum_{j=k+1}^{k+m} X_j -\sum_{j=1}^{k-1} X_j \right]
\end{align}
\begin{align}
=\sum_{j=k+1}^{k+m} \mathbb{E}[X_j] – \sum_{j=1}^{k-1} \mathbb{E}[X_j]
\end{align}
But then I struggle because I'm not sure how to get this is equal to 0
Best Answer
The independence of the increments of the rw comes from the fact that $(X_n)_{n \in \mathbb{N}}$ are IID. To see that the increments are stationary, consider that for any $m,n\in \mathbb{N}$,we have that $S_{n+m}-S_{n}$ is s.t. $$E[e^{ia(S_{n+m}-S_{n})}]=E[e^{ia(X_{n+1}+...+X_{n+m})}]=(E[e^{iaX_1}])^m$$ The distribution of the increments depends only on $m$, so they are stationary. To prove the mean and variance, use linearity of expectations $(E[X_1]=(1/2)(1)+(1/2)(-1)=0)$.