We know $\mathcal{F}(t)=\sigma(\{W(s):0\leq s \leq t\})$ is the smallest $\sigma$ algebra for which the Brownian Motion $W(s)$ is measurable. We are given these definitions:

$$ X(w) = \lim_{n\rightarrow \infty} X_n(\omega) \text{ where } X_n=g_n(W(t_0^n),W(t_1^n),...,W(t_{N_n}^n))$$ $$ \text{ and } Y=W(t+\delta)-W(t) $$

Let $\phi(x) \text{ and } \psi(x) $ denote any bounded, continuous functions. Use $$f(W(t_0),W(t_1)-W(t_0),...,W(t_{N})-W(t_{N-1}))=g(W(t_0),W(t_1),...,W(t_{N})) $$ and independence of the increments to show $$ E(\phi(X_n)\psi(Y))=E(\phi(X_n))E(\psi(Y)).$$

I understand how independence and expected value come into play, but I'm confused as to how $\phi$ and $\psi$ play a role in this.

  • $\begingroup$ If $U$ and $V$ are independent random variables, then $\phi(U)$ and $\psi(V)$ are independent... that's what you will need to use. (You don't mention it explicitly but I take it that $t_k^n \leq t$ for all $k$ and $n$.) $\endgroup$ – saz Feb 24 at 7:01
  • $\begingroup$ @saz Yes, $t^n_k \leq t$ for all $k$ and $n.$ We start with $\phi(U)$ and $\psi(V)$ and then $E(\phi(U) \psi(V)) $ and it then tells us that $E(\phi(U)) E(\psi(V)),$ but I'm not sure what it all looks like in between. $\endgroup$ – MelMarieHA Feb 24 at 7:18
  • $\begingroup$ All you need to do is to check that $X_n$ and $Y$ are independent, right? Once you have this you can choose $U:=X_n$ and $V:=Y$ to prove the assertion. To prove the independence of $X_n$ and $Y$ you have to use that $(W(t_0^n),\ldots,W(t_{N_n}^n))$ and $W(t+\delta)-W(t)$ are independent, and the hint (in your question) tells you how to approach this. $\endgroup$ – saz Feb 24 at 8:15
  • $\begingroup$ BTW, we don't seem to need $X$ here. $\endgroup$ – AddSup Feb 24 at 8:20
  • $\begingroup$ Okay. I think I understand. I will mull it over. Thank you! $\endgroup$ – MelMarieHA Feb 24 at 20:58

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