1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Yes, they are equivalent. $\endgroup$ – saz May 8 at 17:00
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 at 20:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.