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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $B=(B_t)_{t\ge 0}$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$
  • $\mathcal P$ be a sequence of countable subsets $$\mathcal P^n:=\left\{\cdots<t_i^{(n)}<t_{i+1}^{(n-1)}<\cdots\right\}$$ of $[0,\infty)$ such that
    • $0\in\mathcal P^n\subseteq\mathcal P^{n+1}$
    • $\sup\mathcal P^n=\infty$
    • $\displaystyle|\mathcal P^n|:=\sup_{t\in\mathcal P^n}\min_{s\in\mathcal P^n:s\ne t}|s-t|\stackrel{n\to\infty}{\to}0$
  • $\mathcal P_T^n:=\mathcal P^n\cap [0,T)$
  • $t':=t_{i+1}^{(n)}\wedge T$, if $t=t_i^{(n)}$

How can we show, that $$\sum_{t\in\mathcal P_T^n}f(B_t)(B_{t'}-B_t)^2\stackrel{n\to\infty}{\to}\int_0^Tf(B_s)\;ds$$ in probability?


I've found a proof (Theorem 7.12), but I absolutely don't understand what they are doing. They state, that "[if $\tau$ is] the first exit time from a compact interval [...] it suffices to prove the statement for [the] Brownian motion stopped at $\tau$".

What do they mean by "first exit from a compact interval $K$? Maybe $$\tau:=\inf\left\{t\ge 0:B_t\not\in K\right\}\;?$$

Why is that sufficient?

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Define $\tau_L = \inf\{t \geq 0, |B_t|>L\}$. Note that $\tau_L \uparrow \infty$ and therefore $\Bbb{P}(\tau_L<n) \xrightarrow[L \to \infty]{} 0$ by the bounded convergence theorem.

That is sufficient because once you prove $$ \lim \sum_{j=1}^n f(B_{t^{(n)}_j})(t^{(n)}_{j+1} \wedge T - t^{(n)}_{j}\wedge T) \xrightarrow[n\to \infty]{\text{prob}}\sum\int_0^{t\wedge T} f(B_s)\, ds$$

You will know that

$$ \Bbb{P}\bigg(\bigg\vert\lim\sum_{j=1}^n f(B_{t^{(n)}_j})(t^{(n)}_{j+1} - t^{(n)}_{j}) -\sum\int_0^{t} f(B_s)\, ds \bigg\vert \geq \epsilon\bigg) \leq \\\Bbb{P}\bigg(\bigg\vert\lim\sum_{j=1}^n f(B_{t^{(n)}_j})(t^{(n)}_{j+1} \wedge \tau_L - t^{(n)}_{j}\wedge \tau_L) -\sum\int_0^{t\wedge \tau_L} f(B_s)\, ds \bigg\vert \geq \epsilon\bigg) + \Bbb{P}(t\geq \tau_L)$$

$\Bbb{P}(t \geq \tau_L) \xrightarrow[L\to \infty]{} 0$

Choose $L $ such that $\Bbb{P}(t \geq \tau_L) <\epsilon/2$ and then choose $n$ such that $$\Bbb{P}\bigg(\bigg\vert\lim\sum_{j=1}^n f(B_{t^{(n)}_j})(t^{(n)}_{j+1} \wedge \tau_L - t^{(n)}_{j}\wedge \tau_L) -\sum\int_0^{t\wedge \tau_L} f(B_s)\, ds \bigg\vert \geq \epsilon\bigg) <\epsilon/2.$$

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  • $\begingroup$ What you've written is clear to me. However, I don't see how this is related to the statement. $\endgroup$ – 0xbadf00d Jul 9 '15 at 11:02
  • $\begingroup$ So, we' ve got $$\sum_{t\in\mathcal P_T^n}f(B_t)(B_{t'}-B_t)^2\stackrel{n\to\infty}{\to}\int_0^Tf(B_s)\;ds\tag{1}$$ in probability. And by the Itô isometry $$\operatorname E\left[\int_0^T F'(B_s)^2\;ds\right]=\operatorname E\left[\left(\int_0^T F'(B_s)\;dB_s\right)^2\right]\;.\tag{2}$$ Clearly, $(1)$ implies the existence of a almost surely convergent subsequence. But can we conclude by $(2)$, that $$\sum_{t\in\mathcal P_T^n}f(B_t)(B_{t'}-B_t)^2\stackrel{n\to\infty}{\to}\int_0^Tf(B_s)\;dB_s$$ almost surely (for a subsequence)? $\endgroup$ – 0xbadf00d Jul 9 '15 at 15:24
  • $\begingroup$ I will come back in a while. $\endgroup$ – Conrado Costa Jul 9 '15 at 15:29
  • $\begingroup$ could you supply more background? Is this a theorem in Peres? $\endgroup$ – Conrado Costa Jul 9 '15 at 15:30
  • $\begingroup$ No, but in the proof of Theorem 7.13 (you can find the link to the textbook below) they state, that by $(1)$ and "the definition of stochastic integral, we can choose a subsequence such that the sum in $(1)$ converges almost surely to $\int_0^T f(B_s)\;dB_s$". I assume that the "definition of stochastic integral "part is referring to the Itô isometry. people.bath.ac.uk/maspm/book.pdf $\endgroup$ – 0xbadf00d Jul 9 '15 at 15:44

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