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 independent.
I did the following: $S_{k_1}-S_{k_{0}}= \sum_{k=1}^{k1}X_i, S_{k_2}-S_{k_{1}}= \sum_{i=k_1+1}^{k_2} X_i,..., S_{k_n}-S_{k_{n-1}}=\sum_{i=k_{n-1}+1}^{k_n}X_i$
Now I'm asked to prove rigorously that these are independent. My understanding is that they are sums of iid random variables and no term contains the same $X_i$. However this is not sufficient. I need some mathematical way to prove they are independent. Can someone please help?
For simplicity, let us assume that $0 = k_0 < k_1 < k_2$, the general case for $n$ is similar. You want to prove that, $$ S_{k_1} - S_{k_0} , S_{k_2} - S_{k_1} $$ are independent random variables. By definition, of indepedence, this means that if you choose two integers $p$ and $q$ then, $$ P( S_{k_1} - S_{k_0} = p , S_{k_2} - S_{k_1} = q ) = P( S_{k_1} - S_{k_0} = p )P( S_{k_2} - S_{k_1} = q ) $$
Now, $S_{k_0} = S_0 = 0$, so that $P( S_{k_1} - S_{k_0} = p ) = P(S_{k_1} = p)$ and that, $$ P( S_{k_1} - S_{k_0} = p , S_{k_2} - S_{k_1} = q ) = P( S_{k_1} = p , S_{k_2} - S_{k_1} = q )$$ Recall that $S_{k_1} = X_1 + ... + X_{k_1}$ and $S_{k_2} = X_1 + ... + X_{k_1} + X_{k_1+1} + ... + X_{k_2}$. Therefore, $$ P( S_{k_1} = p , S_{k_2} - S_{k_1} = q ) = P(X_1 + ... + X_{k_1} = p, X_{k_1+1} + ... + X_{k_2} = q)$$ Since the sequence $(X_j)_{j\geq 1}$ is independent it means, we can rewrite this probability, $$ P(X_1 + ... + X_{k_1} = p, X_{k_1+1} + ... + X_{k_2} = q) = P(X_1 + ... + X_{k_1} = p)P(X_{k_1+1} + ... + X_{k_2} = q) $$ Now you should be able to check that the condition you want to check gives you the consistent answer with independence.