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 $\{X_k, 0\leq k \leq s\}$ ?

  • $\begingroup$ Yes, they are equivalent. $\endgroup$ – saz May 8 '19 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 $0=t_{0}<t_{1}<t_{2}<\ldots<t_{n}$, $X_{t_{1}}-X_{t_{0}}$, $X_{t_{2}}-X_{t_{1}}$, $\ldots$, $X_{t_{n}}-X_{t_{n-1}}$ 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 $0<t_{1}<t_{2}<\ldots<t_{n}=t$, 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.


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 $t<u_{1}<u_{2}<\ldots<u_{n}$, $\mathcal{F}_{t}$ 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.


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

  • $\begingroup$ The direction (b)=>(a) is easy and can be obtained by induction directly. If I have time, I will include it. $\endgroup$ – Danny Pak-Keung Chan May 8 '19 at 20:12

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