0
$\begingroup$

This statement and proof is taken from a book of Ludger Rüschendorf.

Skorohod embedding: $B$ standard brownian motion, $X_1,X_2,\ldots$ iid random variables with $E[X_1]=0$ and $E[X_1^2]<\infty$. Then there exist iid stopping times $T_1, T_2, \ldots$ with $E[T_1]=E[X_1^2]$ and $S_1, S_2,\ldots$ is a increasing sequence of stopping times with respect to the filtration $( \mathcal{F}_t)_{t\in[0,\infty)}$, whereas $ \mathcal{F}_t:=\sigma(B_t:0\le s\le t)$ and $S_n=\sum_{i=1}^n T_i$.

Additionally it holds

  1. $P^{B_{S_{n}}-B_{S_{n-1}}}=P^ {X_1}$
  2. $(B_{S_{n}}-B_{S_{n-1}})_{n\ge 1}\overset{law}{=}(X_n)_{n\ge 1}$
  3. $(B_{S_n})_{n\ge 1}\overset{law}{=}(\sum_{i=1}^n X_i)_{n\ge 1}$

Proof:

Using the Skorohod construction we find a stopping time $T_1$ with $$E[T_1]=E[X_1^2]\quad \text{and}\quad B_{T_1}\overset{law}{=}X_1$$

$(B_{t+T_1}-B_{T_1})_{t\ge 0}$ is again a brownian motion by the strong markov property of brownian motion. Again we can find a stopping time $T_2$ with

$$E[T_2]=E[X_2^2]\quad\text{and}\quad B_{T_2+T_1}-B_{T_2}\overset{law}{=}X_2$$

Since the increments of a brownian motion are independent, $\color{blue}{\text{it follows that } T_2\text{ is independent of } \mathcal{F}_{T_1}\text{ and therefore }T_2\text{ is independent of }T_1.}$ ($\mathcal{F}_{T_1}$ denotes the $\sigma$-algebra of $T_1$-past). By induction we find a sequence of stopping times $S_1,S_2,\ldots$ with

$$S_n=S_{n-1}+T_n,\quad B_{S_n}-B_{S_{n-1}}\overset{law}{=}X_n\quad\text{and}\quad E[T_n]=E[X_n^2], $$ $\color{blue}{\text{such that } T_n \text{ is independent of }T_1,\ldots, T_{n-1}\text{ and } B_{S_n}-B_{S_{n-1}}\text{ is independent of }\mathcal{F}_{T_{n-1}}.}$ From that we find $$(B_{S_{n}}-B_{S_{n-1}})_{n\ge 1}\overset{law}{=}(X_n)_{n\ge 1}$$ $\color{blue}{\text{Therefore}}$

$$\color{blue}{(B_{S_n})_{n\ge 1}\overset{law}{=}(\sum_{i=1}^n X_i)_{n\ge 1}}\quad\text{and}\quad E[S_n]=\sum_{i=1}^nE[X_i^2]$$

First blue sentence: How can I show exactly that $T_2$ and $T_1$ are independent? I know that from the strong markov property it also follows, that $(B_{t+T_1}-B_{T_1})_{t\ge 0}$ is independent of $\mathcal{F}_{T_1}$. Then I was thinking that $B_{T_2}\overset{law}{=}B_{T_2+T_1}-B_{T_1}$ must be independent of $\mathcal{F}_{T_1}$, too. But now I neither understand how I can use this for proving that $T_2$ is independent of $\mathcal{F}_{T_1}$ nor how $T_2$ is independent of $T_1$.

Second blue sentence: I see that again the strong markov property of brownian motion is used, such that

$$B_{T_n+(T_{n-1}+\ldots+ T_1)}-B_{T_{n-1}+\ldots+ T_1}$$ is independent of $\mathcal{F}_{T_{n-1}+\ldots+ T_1}$. Again as in the first blue sentence $T_n$ should be independent of $T_{n-1}+\ldots+ T_1$, but does this already imply, that $T_n$ is independent of each $T_{n-1},\ldots, T_1$?

Third blue sentence: My idea would be using $$f:\mathbb{R}^\mathbb{N}\rightarrow \mathbb{R}^\mathbb{N},\quad f(x_1,x_2,x_3,\ldots)=(x_1,x_2+x_1,x_3+x_2+x_2,\ldots),$$ such that

$$ (B_{S_n})_{n\ge 1}=f\big( (B_{S_{n}}-B_{S_{n-1}})_{n\ge 1}\big)\overset{law}{=}f\big((X_n)_{n\ge 1}\big) =(\sum_{i=1}^n X_i)_{n\ge 1}$$

But is this really possible here? Is $f$ a well defined function here?

$\endgroup$
0
$\begingroup$

By the Skorohod construction, there is a stopping time $T_1$ with respect to $\big(\sigma(B_s:0\le s\le t)\big)_{t\ge 0}$ with $$E[T_1]=E[X_1^2]\quad \text{and}\quad B_{T_1}\overset{law}{=}X_1$$

By the strong markov property, $(B_{t+T_1}-B_{T_1})_{t\ge 0}$ is again a Brownian motion and $$\sigma (B_{s+T_1}-B_{T_1}:s\ge 0)\text{ is independent of }\mathcal{F}_{T_1}$$

By the Skorohod construction, there is a stopping time $T_2$ with respect to $\big(\sigma(B_{s+T_1}-B_{T_1}:0\le s\le t\big))_{t\ge 0}$ with

$$E[T_2]=E[X_2^2]\quad\text{and}\quad B_{T_2+T_1}-B_{T_2}\overset{law}{=}X_2$$

By definition $T_2$ is $\sigma (B_{s+T_1}-B_{T_1}:s\ge 0)$-measurable and therefore independent of $\mathcal{F}_{T_1}$. Since $T_1$ is $\mathcal{F}_{T_1}$-measurable, we find that from the property above, that $T_2$ is independent of $T_1$.

By induction, there are stopping times $T_n$ independent of $\sum_{i=1}^{n-1}T_i$. Since $T_i\ge 0$ for all $i\ge 1$, $\sum_{i=1}^{n-1}T_i$ is in fact a stopping time. Using $\mathcal{F}_{T_1},\ldots, \mathcal{F}_{T_n}\subset\mathcal{F}_{T_1+\ldots+T_n}$, $T_n$ is independent of each $T_i$ with $0\le i\le n-1$.

By that, we have found a sequence of stopping times $S_1,S_2,\ldots$ with

$$S_n=S_{n-1}+T_n,\quad B_{S_n}-B_{S_{n-1}}\overset{law}{=}X_n\quad\text{and}\quad E[T_n]=E[X_n^2], $$ such that $T_n$ is independent of $T_1,\ldots, T_{n-1}$ and $B_{S_n}-B_{S_{n-1}}$ is independent of $\mathcal{F}_{T_{n-1}}$. Note that any $S_n$ is stopping time with respect to $\big(\sigma(B_s:0\le s\le t)\big)_{t\ge 0}$, since $T_1$ is stopping time with respect to $\big(\sigma(B_s:0\le s\le t)\big)_{t\ge 0}$ and by this notation we find that $\{S_n\le t\}\in \mathcal{F}_t$, since all $T_i\ge 0$.

Therefore $$(B_{S_{n}}-B_{S_{n-1}})_{n\ge 1}\overset{law}{=}(X_n)_{n\ge 1}$$ and

$$ (B_{S_n})_{n\ge 1}=f\big( (B_{S_{n}}-B_{S_{n-1}})_{n\ge 1}\big)\overset{law}{=}f\big((X_n)_{n\ge 1}\big) =(\sum_{i=1}^n X_i)_{n\ge 1}$$ for $f:\mathbb{R}^\mathbb{N}\rightarrow \mathbb{R}^\mathbb{N}$, $f(x_1,x_2,x_3,\ldots)=(x_1,x_2+x_1,x_3+x_2+x_2,\ldots)$.

At least $E[S_n]=\sum_{i=1}^nE[X_i^2]$ holds by linearity.

$\endgroup$

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.