# Increments of random walk are stationary and independent

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 $$0=k_0 , the increments $$\{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 and \begin{align} Z'_{a,b} = \{ S_i - S_{a-1} \neq 0, a \leq i

For $$i, then $$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

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

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