Is the infinite sum of the increments of some stochastic process from some point on equal to the process at that time

Question

Let $\{ X_n \}, n \in \mathbb N_0$ be a stochastic process, define $Y_{n+1} := X_{n+1} - X_n$ and $Y_0 = 0$, i.e. the increments of the process $X$. Then do we have $$ X_n = \sum_{k \ge n} Y_k \in \sigma(Y_n, Y_{n+1}, \ldots). $$ This seems to be intuitively true as the increments cancel out. First for the finite sums we have $\sum_{i=k}^n Y_i = X_k - X_n$, so that $\lim_{m\to \infty} \sum_{k=n}^m Y_k = \lim_{m\to\infty} (X_k - X_m)$, but I do not see that the last term $X_m$ vanishes in some sense?

Answer 1

No, in general, this does not hold true. Simply consider the (deterministic) process

$$X_n := \begin{cases} 0, & \text{$n$ even} \\ 1, & \text{$n$ odd} \end{cases}$$ then

$$\sum_{k=1}^m Y_k = \begin{cases} 0, & \text{$m$ is even}, \\ 1, & \text{$m$ is odd} \end{cases}$$

which means that the partial sum $S_m := \sum_{k=1}^m Y_k$ does not converge as $m \to \infty$, i.e. $\sum_{k=1}^{\infty} Y_k$ is not even well-defined.

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It is not difficult to show the following statement:

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of random variables and set $Y_{n+1} := X_{n+1}-X_n$. Then $$X_n =- \sum_{k \geq n} Y_k \tag{1}$$ for some (all) $n \in \mathbb{N}$ if and only if $\lim_{k \to \infty} X_k = 0$.

Proof: Since $$\sum_{k=n}^m Y_k = X_m-X_n$$ we have $$\left| X_n + \sum_{k=n}^m Y_k \right| = |X_m|.$$ Consequently, the left-hand side converges to $0$ as $m \to \infty$ if, and only if, the right-hand side converges to $0$, and this is exactly what we wanted to prove.