# Independent increment vs independent sigma-algebras

Suppose we want to define a Lévy process $$\{ X_t \vert \ t \geq 0\}$$. Is it equivalent to demand independent increments i.e. $$\forall n \geq 1, \forall t_n \geq t_{n-1} \geq ...\geq t_1 \geq 0: X_{t_1}, X_{t_2}-X_{t_1},..., X_{t_n} - X_{t_{n-1}} \ \text{are independent}$$ versus demanding that $$X_t-X_s$$ is independent of the sigma-algebra generated by $$\{X_k, 0\leq k \leq s\}$$ ?

• Yes, they are equivalent. – saz May 8 at 17:00

Let $$(\Omega, \mathcal{F}, P)$$ be a probability space. Let $$X=\{X_{t}\mid t\in[0,\infty)\}$$ be a stochastic process with $$X_{0}=0$$. For each $$t\in[0,\infty)$$, define $$\sigma$$-algebras $$\mathcal{F}_{t}$$ and $$\mathcal{H}_{t}$$ by $$\mathcal{F}_{t}=\sigma\left(\bigcup_{s\in[0,t]}\sigma(X_{s})\right)$$ and $$\mathcal{H}_{t}=\sigma\left(\bigcup_{u\in(t,\infty)}\sigma(X_{u}-X_{t})\right)$$. Then the following conditions are equivalent:

(a) The process has independent increments, in the sense that: For any $$n\in\mathbb{N}$$ and any $$0=t_{0}, $$X_{t_{1}}-X_{t_{0}}$$, $$X_{t_{2}}-X_{t_{1}}$$, $$\ldots$$, $$X_{t_{n}}-X_{t_{n-1}}$$ are independent.

(b) For each $$t\in(0,\infty)$$, $$\mathcal{F}_{t}$$ and $$\mathcal{H}_{t}$$ are independent.

Firstly, we prove that $$(a)\Rightarrow(b)$$. Suppose that (a) holds.

Claim 1: For any $$0, and $$u\in(t,\infty)$$, $$\sigma(X_{t_{1}},X_{t_{2}},\ldots,X_{t_{n}})$$ and $$\sigma(X_{u}-X_{t})$$ are independent.

Proof of Claim 1: Let $$\mathcal{\mathcal{C}=}\{A\mid A=\cap_{i=1}^{n}A_{i}\mbox{ for some }A_{i}\in\sigma(X_{t_{i}}-X_{t_{i-1}})\}$$ (Here $$t_{0}=0$$ by convention). Note that $$\mathcal{C}$$ is a $$\pi$$-class (in the sense that $$A_{1}\cap A_{2}\in\mathcal{C}$$ whenever $$A_{1},A_{2}\in\mathcal{C}$$). Let $$A\in\mathcal{C}$$ and write $$A=\cap_{i=1}^{n}A_{i}$$ for some $$A_{i}\in\sigma(X_{t_{i}}-X_{t_{i-1}})$$. Let $$B\in\sigma(X_{u}-X_{t})$$. Then $$P(AB)=P(A_{1}A_{2}\ldots A_{n}B)=\prod_{i=1}^{n}P(A_{i})P(B)=P(A)P(B)$$ by observing that $$X_{t_{1}}-X_{t_{0}},X_{t_{2}}-X_{t_{1}},\ldots,X_{t_{n}}-X_{t_{n-1}},X_{u}-X_{t}$$ are independent. Let $$\mathcal{L}=\{A\in\sigma(\mathcal{C})\mid P(AB)=P(A)P(B)\}$$. It can be verified that $$\mathcal{L}$$ is a $$\lambda$$-class, in the sense that: (i) $$\Omega\in\mathcal{L}$$, (ii) $$A^{c}\in\mathcal{L}$$ whenever $$A\in\mathcal{L}$$, and (iii) For any pairwisely disjoint $$A_{1},A_{2},\ldots,\in\mathcal{L}$$, we have $$\cup_{i=1}^{\infty}A_{i}\in\mathcal{L}$$. Moreover, we have proved that $$\mathcal{C}\subseteq\mathcal{L}$$. By Dynkin's $$\pi$$-$$\lambda$$ theorem, we have $$\sigma(\mathcal{C})\subseteq\mathcal{L}$$ and hence $$\mathcal{L=}\sigma(\mathcal{C})$$. Fix $$j$$ and let $$A_{j}\in\sigma(X_{t_{j}}-X_{t_{j-1}})$$. Put $$A_{i}=\Omega$$ for any $$i\neq j$$. Then $$A_{j}=\cap_{i=1}^{n}A_{i}\in\mathcal{C}$$. Therefore, $$X_{t_{j}}-X_{t_{j-1}}$$ is $$\sigma(\mathcal{C})/\mathcal{B}(\mathbb{R})$$-measurable. Put $$j=1$$, we have: $$X_{t_{1}}$$ is $$\sigma(\mathcal{C})/\mathcal{B}$$-measurable. (Here $$\mathcal{B=\mathcal{B}}(\mathbb{R})$$). Put $$j=2$$ and observe that $$X_{t_{2}}=(X_{t_{2}}-X_{t_{1}})+X_{t_{1}}$$ which is $$\sigma(\mathcal{C})/\mathcal{B}$$-measurable. By repeating the argument, we have $$X_{t_{1}},X_{t_{2}},\ldots,X_{t_{n}}$$ are $$\sigma(\mathcal{C})$$ measurable. Hence $$\sigma(X_{t_{1}},X_{t_{2}},\ldots,X_{t_{n}})\subseteq\sigma(\mathcal{C})=\mathcal{L}$$. That is, $$\sigma(X_{t_{1}},X_{t_{2}},\ldots,X_{t_{n}})$$ and $$\sigma(X_{u}-X_{t})$$ are independent.

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Claim 2: For any $$t\in(0,\infty)$$ and $$u\in(t,\infty)$$, $$\mathcal{F}_{t}$$ and $$\sigma(X_{u}-X_{t})$$ are independent.

Proof of Claim 2: For each finite subset $$I\subseteq(0,t]$$ satisfying $$t\in I$$, we write $$\mathcal{C}_{I}=\sigma\left(\cup_{t\in I}\sigma(X_{t})\right)$$. Let $$\mathcal{C}=\cup\{\mathcal{C}_{I}\mid I\subseteq(0,t]\mbox{ is a finite subset satisfying }t\in I\}$$. Observe that $$\mathcal{C}\subseteq\mathcal{F}_{t}$$ and $$\mathcal{C}$$ is a $$\pi$$-class. (For, let $$A_{1},A_{2}\in\mathcal{C}$$, then $$A_{1}\in\mathcal{C}_{I_{1}}$$ and $$A_{2}\in\mathcal{C}_{I_{2}}$$ for some finite subsets $$I_{1}$$ and $$I_{2}$$ of $$(0,t]$$ satisfying $$t\in I_{1}$$ and $$t\in I_{2}$$. Take $$I=I_{1}\cup I_{2}$$, then $$I$$ is a finite subset of $$(0,t]$$, $$t\in I$$ and $$A_{1}\cap A_{2}\in\mathcal{C}_{I}\subseteq\mathcal{C}$$.) By Claim 1, $$\mathcal{C}$$ and $$\sigma(X_{u}-X_{t})$$ are independent. By Dynkin's theorm again, we have that: $$\sigma(\mathcal{C})$$ and $$\sigma(X_{u}-X_{t})$$ are independent. Clearly, for any $$s\in(0,t]$$, $$X_{s}$$ is $$\sigma(\mathcal{C})$$-measurable, so $$\mathcal{F}_{t}\subseteq\sigma(\mathcal{C})$$. On the other hand, for each finite subset $$I\subseteq(0,t]$$ with $$t\in I$$, we clearly have $$\mathcal{C}_{I}\subseteq\mathcal{F}_{t}$$. It follows that $$\sigma(\mathcal{C})\subseteq\mathcal{F}_{t}$$. That is, $$\sigma(\mathcal{C}) = \mathcal{F}_t$$.

Claim 3: For any $$t\in(0,\infty)$$ and $$u_{1},u_{2},\ldots,u_{n}$$ with $$t, $$\mathcal{F}_{t}$$ and $$\sigma\left(X_{u_{1}}-X_{t},X_{u_{2}}-X_{t},\ldots,X_{u_{n}}-X_{t}\right)$$ are independent.

Proof of Claim 3: Let $$\mathcal{C}=\{B\mid B=\cap_{i=1}^{n}B_{i}$$,$$B_{i}\in\sigma(X_{u_{i}}-X_{u_{i-1}})\}$$, where $$u_{0}=t$$ by convention. Clearly $$\mathcal{C}$$ is a $$\pi$$-class. Let assert that $$\mathcal{F}_{t}$$ and $$\mathcal{C}$$ are independent. Let $$A\in\mathcal{F}_{t}$$ and $$B\in\mathcal{C}$$. Write $$B=\cap_{i=1}^{n}B_{i}$$, where $$B_{i}\in\sigma(X_{u_{i}}-X_{u_{i-1}})$$. Observe that $$AB_{1}B_{2}\ldots B_{n-1}\in\mathcal{F}_{u_{n-1}}$$ and $$B_{n}\in\sigma\left(X_{u_{n}}-X_{u_{n-1}}\right)$$. Since $$\mathcal{F}_{u_{n-1}}$$ and $$\sigma\left(X_{u_{n}}-X_{u_{n-1}}\right)$$ are independent (by Claim 2), we have $$P(AB_{1}B_{2}\ldots B_{n-1}B_{n})=P(AB_{1}B_{2}\ldots B_{n-1})P(B_{n})$$. By the same argument, observe that $$\mathcal{F}_{u_{n-2}}$$ and $$\sigma\left(X_{u_{n-1}}-X_{u_{n-2}}\right)$$ are independent, so $$P(AB_{1}B_{2}\cdots B_{n-1})=P(AB_{1}\cdots B_{n-2})P(B_{n-1})$$. By repeating the argument, we have $$P(AB)=P(A)P(B_{1})P(B_{2})\cdots P(B_{n})=P(A)P(B)$$. Here, observe that $$X_{u_{1}}-X_{u_{0}},\ldots,X_{u_{n}}-X_{u_{n-1}}$$ are independent, so $$P(B)=P(B_{1})\cdots P(B_{n})$$. By Dynkin's Theorem, $$\mathcal{F}_{t}$$ and $$\sigma(\mathcal{C})$$ are independent. Observe that $$\sigma(\mathcal{C})=\sigma\left(X_{u_{1}}-X_{t},X_{u_{2}}-X_{t},\ldots,X_{u_{u}}-X_{t}\right)$$. For, $$X_{u_{2}}-X_{t}=(X_{u_{2}}-X_{u_{1}})+(X_{u_{1}}-X_{u_{0}})$$ which is $$\sigma(\mathcal{C})$$-measurable, $$X_{u_{3}}-X_{t}=(X_{u_{3}}-X_{u_{2}})+(X_{u_{2}}-X_{t})$$ which is $$\sigma(\mathcal{C})$$-measurable, etc... Therefore $$\sigma\left(X_{u_{1}}-X_{t},X_{u_{2}}-X_{t},\ldots,X_{u_{n}}-X_{t}\right)\subseteq\sigma(\mathcal{C})$$. For the reversed inclusion, observe that $$X_{u_{i}}-X_{u_{i-1}}=(X_{u_{i}}-X_{t})+(X_{u_{i-1}}-X_{t})$$ which is $$\sigma\left(X_{u_{1}}-X_{t},X_{u_{2}}-X_{t},\ldots,X_{u_{n}}-X_{t}\right)$$-measurable.

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Claim 4: For any $$t\in(0,\infty)$$, $$\mathcal{F}_{t}$$ and $$\mathcal{H}_{t}$$ are independent.

Proof of Claim 4: Let $$t\in(0,\infty)$$ be fixed. For each non-empty finite set $$I\subseteq(t,\infty)$$, let $$\mathcal{C}_{I}=\sigma\left(\{X_{u}-X_{t}\mid u\in I\}\right)$$. Let $$\mathcal{C}=\cup\{\mathcal{C}_{I}\mid I\subseteq(t,\infty)\mbox{ is a non-empty finite subset.\}}$$. By Claim 3, $$\mathcal{F}_{t}$$ and $$\mathcal{C}$$ are independent. Observe that $$\mathcal{C}$$ is a $$\pi$$-class. By Dynkin's theorem again, it follows that $$\mathcal{F}_{t}$$ and $$\sigma(\mathcal{C})$$ are independent. However, $$\sigma(\mathcal{C})=\mathcal{H}_{t}$$. Q.E.D

• The direction (b)=>(a) is easy and can be obtained by induction directly. If I have time, I will include it. – Danny Pak-Keung Chan May 8 at 20:12