# Proof of Skorohod embedding

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 $$S_n=\sum_{i=1}^n T_i$$.

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?

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.