# A question on independence of increments

How could I prove the following? Let $X=(X_t)_{t \in[0,1]}$ be a real-valued stochastic process on a probability space $(\Omega,F,P)$ with $X_0=0$ a.s Show that the following statement as are equivalent

1 $X$ has independent increments ie. $\forall n \in \mathbb{N}$ and for every choice of $0 \leq t_0 < t_1...< t_n \leq 1$ we have that $X_{t_i}-X_{t_{i-1}}$ where $i=1,2,..n$ are independent

2 For every $0\leq t<u$, $X_u-X_t$ is independent of $F_t^{X}=\sigma(X_s,s\leq t)$

How could I go about proving this,can someone give me a hint? I have been trying to use the definition of independence of Rv's ie X, Y are independent if the $\sigma$-algebra generated by the are independent but I am stuck.

• (2) $\Rightarrow$ (1): Recall that (real-valued) random variabeles $Z_1,\ldots,Z_n$ are independent if, and only if, $$\mathbb{E} \exp \left( \imath \, \sum_{j=1}^n \xi_j Z_j \right) = \prod_{j=1}^n \mathbb{E}\exp(\imath \, \xi_j Z_j)$$ for all $\xi_1,\ldots,\xi_n \in \mathbb{R}$. Now fix $0 \leq t_0 < \ldots \leq t_n$. By assumption, $X_{t_n}-X_{t_{n-1}}$ is independent from $$\sum_{j=1}^{n-1} \xi_j (X_{t_j}-X_{t_{j-1}}).$$ Consequently, \begin{align*} \mathbb{E}\exp \left( \imath \, \sum_{j=1}^n \xi_j (X_{t_j}-X_{t_{j-1}}) \right) &= \mathbb{E} \left[ \exp \left( \imath \, \sum_{j=1}^{n-1} \xi_j (X_{t_j}-X_{t_{j-1}}) \right) \exp(\imath \, \xi_n (X_{t_n}-X_{t_{n-1}}) \right] \\ &= \mathbb{E} \exp \left( \imath \, \sum_{j=1}^{n-1} \xi_j (X_{t_j}-X_{t_{j-1}}) \right) \cdot \mathbb{E}\exp(\imath \, \xi_n (X_{t_n}-X_{t_{n-1}})). \end{align*} Can you take it from here?
• (1) $\Rightarrow$ (2): Note that \begin{align*} \mathcal{F}_t^X &= \sigma \left( \bigcup_{n \in \mathbb{N}} \bigcup_{u_1<\ldots<u_n \leq t} \sigma(X_{u_1},\ldots,X_{u_n}) \right) \\ &= \sigma \left( \bigcup_{n \in \mathbb{N}} \bigcup_{u_1<\ldots<u_n \leq t} \sigma(X_{u_1},\ldots,X_{u_n}-X_{u_{n-1}}) \right), \end{align*} i.e. $$\mathcal{D} := \bigcup_{n \in \mathbb{N}} \bigcup_{u_1<\ldots<u_n \leq t} \sigma(X_{u_1},\ldots,X_{u_n}-X_{u_{n-1}})$$ is a $\cap$-stable generator of $\mathcal{F}_t^X$. Conclude that the independence of $(X_{u_1},\ldots,X_{u_n}-X_{u_{n-1}})$ and $X_u-X_t$ for any sequence $0 \leq u_1 < \ldots < u_n \leq t$ already implies the claim.
• Thank you Saz. In the first part of the proof ($(2)\implies (1)$ , we just proceed iteratively and say that $X_{t_{n-1}}-X_{t_{n-2}}$ is independent of \sum_{j=1}^{n-2}t_j(X_{t_{j}}-X_{t_{j-1}} and so on and we get $(1)$. I really cannot follow your proof for the second implication i.e $(1) \implies (2)$ .I mean $\sigma(X_s)$ is the smallest $\sigma$-algebra which makes X_s measurable. I see that you take countable unions but we have uncountably many $X_s$ such that $s\leq t \leq 1$ ,I cannot follow why $\mathcal{F}_{t}^X=\sigma(X_s,s\leq t)$ is equivalent the the multiple union exp u used. – user3503589 Oct 19 '14 at 13:18
• @user3503589 Just do it step by step. First equality: First of all, note that we can choose $n=1$, $u_1 = s$ for any $s \leq t$. This shows that $X_s$, $s \leq t$, is msb. with respect to the $\sigma$-algebra on the right-hand side, i.e. $\mathcal{F}_t^X$ "$\subseteq$" the right-hand side. On the other hand, for any $n$ and $u_1<\ldots<u_n \leq t$, we know that $X_{u_1},\ldots,X_{u_n}$ are $\mathcal{F}_t^X$-measurable. Hence, the $\sigma$-algebra on the right-hand side is contained in $\mathcal{F}_t^X$. Hence, they are equal. So far, so good? (Mind that the union is not countable.) – saz Oct 19 '14 at 13:26
• @user3503589 Great. So for the second equality, it suffices to show $$\sigma(X_{u_1},\ldots,X_{u_n}) = \sigma(X_{u_1},\ldots,X_{u_n}-X_{u_{n-1}}).$$ Use that $$X_{u_j} = \sum_{k=1}^j (X_{u_k}-X_{u_{k-1}})+X_1$$ in order to get "$\subseteq$". "$\supseteq$" is easy. – saz Oct 19 '14 at 13:33
• ($X_1$ in the last equation should read $X_{u_1}$, sorry.) – saz Oct 19 '14 at 13:50