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}


\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}


\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

  • $\begingroup$ 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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.