Why does $\left\{\left|X-\sum_{i=1}^mX_i\right|\le\delta\right\}$ depend only on $X_i,i\ge m+1$, if all $X_i$ are independent? Assumptions
Let $(X_i)_{i\in\mathbb{N}}$ a sequence of independent real-valued random variables and $$B_{m,n}:=\left\{\left|X^{(i)}-X^{(n)}\right|\le2\delta \text{ for all }i\in[n,m-1]\text{ and }\left|X^{(m)}-X^{(n)}\right|>2\delta\right\}$$ where $$X^{(n)}:=\sum_{i=1}^nX_i$$
and $\delta>0$. Moreover, let $X^{(n)}$ converge stochastically to some $X$.

Problem
I've read that


*

*$B_{m,n}$ depends only on $X_{n+1},\ldots,X_m$ and

*$\left\{\left|X-X^{(m)}\right|\le\delta\right\}$ depends only on $X_i,i\ge m+1$



My Ideas
I absolutely don't understand why (1) and (2) are true. Especially (2) seems to be counter-intuitively. Shouldn't $\left\{\left|X-X^{(m)}\right|\le\delta\right\}$ depend only on $X_1,\ldots,X_m$?
 A: *

*By definition, $$|X^{(i)}-X^{(n)}| = \left|\sum_{j=n+1}^i X_j \right|$$ is measurable with respect to $\sigma(X_{n+1},\ldots,X_i) \subseteq \sigma(X_{n+1},\ldots,X_{m})$ for all $i \in  \{n,\ldots,m\}$. Therefore, $$B_{m,n} = \bigcap_{i=n}^{m-1} \{|X^{(i)}-X^{(n)}| \leq 2\delta\} \cap |X^{(m)}-X^{(n)}|>2\delta\} \in \sigma(X_{n+1},\ldots,X_m).$$

*Fix $m \in \mathbb{N}$. Since $X^{(n)} \to X$ in probability, there exists a subsequence $(X^{(n_k)})_k$ such that $X^{(n_k)} \to X$ almost surely. Without loss of generality, we may assume that $n_k = k$ for all $k \in \{1,\ldots,m\}$. Hence, $$X = \lim_{k \to \infty} X^{(n_k)} = X^{(m)}+ \underbrace{\lim_{k \to \infty} \sum_{j=m+1}^{n_k} X_j}_{=:Y}.$$ Since $Y$ is measurable with respect to $\sigma(X_{i}; i \geq m+1)$, we get that $Y=X-X^{(m)}$ is measurable with respect to $\sigma(X_i; i \geq m+1)$. As $(X_i)_{i \in \mathbb{N}}$ is a sequence of independent random variables, we conclude that $X-X^{(m)}$ is independent from $\sigma(X_i; i \leq m)$.

