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Let $(X_t)_{t\geq0}$ be a stochastic process on some probability sprace $(\Omega, \mathcal{F}, P)$. Then for $s < t$, we define the $\textit{increment}$ of the process, $X_t - X_s$ over the interval $[s, t]$. The process $(X_t)_{t\geq0}$ has $\textit{independent increments}$ if, for every set of real numbers $0 \leq t_1 < t_2 < \ldots < t_n < \infty$, the increments $$ X_{t_2} - X_{t_1}, \ X_{t_3} - X_{t_2}, \ \ldots, \ X_{t_n} - X_{t_{n-1}} $$ are independent of eachother.

During our course in stochastic processes the lecturer mentioned an equivalent way to define independent incerements of a process. For every set of real numbers $0 \leq t_1 < t_2 < \ldots < t_n < t_{n+1} < \infty$ the increment $X_{t_{n+1}}-X_{t_n}$ is independent of the random vector $(X_{t_1}, \ \ldots, X_{t_n})$. I don't see a way how to prove the eqivalence. He said this equvalence withou writing it down so I could be missing something. Any help would be appreciated.

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  • $\begingroup$ One direction looks easier than the other. Can you prove the second condition implies the first? I also note that if we have a deterministic initial condition such as $X_0=0$ then the information $X_0,X_{t_1}, X_{t_2}, ..., X_{t_n}$ is equivalent to the information $X_{t_n}-X_{t_{n-1}}, X_{t_{n-1}}-X_{t_{n-2}}, ..., X_{t_2}-X_{t_1}, X_{t_1}-X_0$. $\endgroup$
    – Michael
    Commented Apr 19 at 18:47
  • $\begingroup$ We are studying point processes, so usually $X_0=0$. $\endgroup$ Commented Apr 20 at 7:08

1 Answer 1

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The problem becomes easy if you assume $X_0=0$ surely, as you indicated in your comments. The main idea is to consider the vectors $(X_0, X_{t_1}, ..., X_{t_n})$ and $(X_0, X_{t_1}-X_0, ..., X_{t_n}-X_{t_{n-1}})$, and observe that any one of these two vectors can be used to obtain the other by simple addition or subtraction.

Fact 1: Fix positive integers $n,k$. Let $Y, W_1, ..., W_n$ be random variables. If $Y$ is independent of the random vector $(W_1, ..., W_n)$, then $Y$ is also independent of $h(W_1, ..., W_n)$ where $h:\mathbb{R}^n\rightarrow\mathbb{R}^k$ is some Borel measurable function.

Fact 2: Fix $n\geq 2$ as an integer. If $(A_1, A_2, ..., A_n)$ are random variables such that for each $i \in \{2, ..., n\}$, $A_i$ is independent of $(A_1, ..., A_{i-1})$, then $A_1, ..., A_n$ are mutually independent.

Forward direction: Suppose $(X_t)_{t\geq 0}$ has $X_0=0$ surely and also has the independent increments property. Fix $n$ as a postive integer and fix $\{t_i\}_{i=1}^{n+1}$ such that $0\leq t_1\leq ... \leq t_{n+1}$. We want to show $X_{t_{n+1}}-X_{t_n}$ is independent of $(X_0,X_{t_1}, ..., X_{t_n})$.

Since $X_0=0$ surely, we know $X_{t_{n+1}}-X_{t_n}$ is independent of $(X_0, X_{t_1}-X_0, X_{t_2}-X_{t_1}, ..., X_{t_n}-X_{t_{n-1}})$. So (by Fact 1) we know $X_{t_{n+1}}-X_{t_n}$ is independent of $h(X_0, X_{t_1}-X_0, ..., X_{t_n}-X_{t_{n-1}})$ for any measurable function $h:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1}$. Observe that \begin{align} X_0 &= 0\\ X_{t_1}&=X_0+(X_{t_1}-X_0)\\ X_{t_2}&=X_0+(X_{t_1}-X_{0}) + (X_{t_2}-X_{t_1})\\ ...\\ X_{t_n} &= X_0+(X_{t_1}-X_{0}) + (X_{t_2}-X_{t_1}) + ... + (X_{t_n}-X_{t_{n-1}}) \end{align} So $(X_0, X_{t_1}, ..., X_{t_n})=h(X_0, X_{t_1}-X_0, ..., X_{t_n}-X_{t_{n-1}})$ for the measurable function $h=(h_0, ..., h_n)$ defined by \begin{align} &h_0(a_0, ..., a_n) = a_0\\ &h_1(a_0, ..., a_n)=a_0+a_1\\ &\cdots\\ &h_n(a_0,...,a_n)=a_0+a_1+...+a_n \end{align} $\Box$

Reverse direction: Suppose $(X_t)_{t\geq 0}$ satisfies for all positive integers $n$ and all times $0\leq t_1\leq ...\leq t_{n+1}$ that $X_{t_{n+1}}-X_{t_n}$ is independent of $(X_0, ..., X_{t_n})$. We want to show it has the independent increments property. Fix $n$. By Fact 1, we know $X_{t_{n+1}}-X_{t_n}$ is independent of $h(X_0, ..., X_{t_n})$ for any measurable function $h$. Clearly there is a measurable function that maps the pure times $(X_0, ..., X_{t_n})$ to their differences $(X_1-X_0, ..., X_{t_n}-X_{t_{n-1}})$. So $X_{t_{n+1}}-X_{t_n}$ is independent of $(X_1-X_0, ..., X_{t_n}-X_{t_{n-1}})$. By Fact 2, it holds that $(X_t)_{t\geq 0}$ has the independent increment property. $\Box$


The above proof of the forward direction uses $X_0=0$ surely, while the reverse direction does not.

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