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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

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  • $\begingroup$ Why do you doubt that last equation is valid? $\endgroup$ Jan 22 at 23:39
  • $\begingroup$ Because I think: "Why would I get the same term? Is that correct?" $\endgroup$
    – Erick GR
    Jan 22 at 23:56

1 Answer 1

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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 $(E[X_1]=(1/2)(1)+(1/2)(-1)=0)$.

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