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Suppose that $0=t_{0,n}0$ is strictly-positive a cadlag stochastic process and $W(t)$ a Wiener process. What I want to prove is that

$$ \sum_{j=1}^n\sigma\left(\frac{i}{n}\right)\,\left(W\left(\frac{i}{n}\right)-W\left(\frac{i-1}{n}\right)\right)^2\stackrel{p}{\longrightarrow}\int_0^1\sigma(s)\,ds $$

My idea is to approximate the increment $\left(W\left(\frac{i}{n}\right)-W\left(\frac{i-1}{n}\right)\right)^2$ with its expected value, writing something like

$$ \left(W\left(\frac{i}{n}\right)-W\left(\frac{i-1}{n}\right)\right)^2 = \frac{1}{n}+o_p\left(\frac{1}{n}\right) $$

however using for example Markov's inequality I am not able to prove $(1)$, admitted that it is true of course. Any idea?

Attempt to use Markov's inequality:

Let $X=\left(W\left(\frac{i}{n}\right)-W\left(\frac{i-1}{n}\right)\right)^2 $ so that $\mathbb{E}\left[X\right]=\frac{1}{n}$ and $\mathbb{E}\left[X^2\right]=\frac{3}{n^2}$ , hence

$$ \mathbb{P}\left[\frac{\left|X-1/n\right|}{1/n}\geq \varepsilon\right]\leq \frac{n}{\varepsilon}\mathbb{E}\left[\left|X-1/n\right|\right] $$

Nevertheless $\mathbb{E}\left[\left|X-1/n\right|\right]=\frac{2}{n\,\sqrt{\pi}}$ so that the Markov's inequality is inconclusive.

1 Answers 1

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I take it that you assume that $\sigma$ is adapted to the canonical filtration of the Brownian motion; otherwise the reasoning presented in this answer does not work.

Since we know that

$$\frac{1}{n} \sum_{i=1}^n \sigma \left( \frac{i-1}{n} \right) = \sum_{i=1}^n \sigma \left( \frac{i-1}{n} \right) \left( \frac{i}{n}- \frac{i-1}{n} \right) \to \int_0^1 \sigma(s) \, ds$$

it suffices to show

$$Q_n := \sum_{i=1}^n \sigma \left( \frac{i-1}{n} \right) \left[ \left( W(i/n)-W((i-1)/n) \right)^2 - \frac{1}{n} \right] \stackrel{\mathbb{P}}{\to} 0.$$

Fix $\epsilon>0$. Since $\sigma(\cdot,\omega)$ is bounded on $[0,1]$ for each $\omega \in \Omega$, the continuity of the probability measure $\mathbb{P}$ implies

$$\mathbb{P} \left( \sup_{t \in [0,1]} |\sigma(t)| > R \right) \xrightarrow[]{R \to \infty} 0. \tag{1}$$

Applying Markov's inequality, we find

$$\begin{align*} \mathbb{P}(|Q_n| \geq \epsilon) &= \mathbb{P} \left( |Q_n| \geq \epsilon, \sup_{t \in [0,1]} |\sigma(t)| \leq R \right) + \mathbb{P} \left( |Q_n| \geq \epsilon, \sup_{t \in [0,1]} |\sigma(t)| > R \right) \\ &\leq \frac{1}{\epsilon^2} \mathbb{E}(Q_n^2 1_{\{\sup_{t \in [0,1]} |\sigma(t)| \leq R\}}) + \mathbb{P} \left( \sup_{t \in [0,1]} |\sigma(t)| > R \right). \tag{2} \end{align*}$$

The next step is to show that the first term on the right-hand side converges to $0$ as $n \to \infty$ (for fixed $R>0$). To this end, we note that

$$\begin{align*} \mathbb{E}(Q_n^2 1_{\{\sup_{t \in [0,1]} |\sigma(t)| \leq R\}}) \leq \mathbb{E} \left( \sum_{i=1}^n 1_{\{\sup_{t \leq (i-1)/n} |\sigma(t)| \leq R\}} \sigma \left( \frac{i-1}{n} \right)^2 \left[ (W(i/n)-W((i-1)/n))^2 - \frac{1}{n} \right]^2 \right); \end{align*}$$

here we have used that the mixed terms (..when squaring $Q_n$...) vanish since $(W_t^2-t)_{t \geq 0}$ is a martingale and $\sigma$ is adapted to the filtration of $(W_t)_{t \geq 0}$. As $$W(i/n)-W((i-1)/n) \sim W(1/n) \sim 1/\sqrt{n} W_1$$ we get

$$\begin{align*} \mathbb{E}(Q_n^2 1_{\{\sup_{t \in [0,1]} |\sigma(t)| \leq R\}}) &\leq \frac{1}{n^2} R^2 \sum_{i=1}^n \mathbb{E}((W_1^2-1))^2) \\ &= \frac{1}{n} R^2 \mathbb{E}((W_1^2-1)^2). \end{align*}$$

Letting $n \to \infty$, we find $ \mathbb{E}(Q_n^2 1_{\{\sup_{t \in [0,1]} |\sigma(t)| \leq R\}}) \to 0$. Now it follows from $(2)$ that

$$\limsup_{n \to \infty} \mathbb{P}(|Q_n| \leq \epsilon) \leq \mathbb{P} \left( \sup_{t \in [0,1]} |\sigma(t)>R \right)$$

for all $R>0$. Finally, by $(1)$, this implies

$$\lim_{n \to \infty} \mathbb{P}(|Q_n| \geq \epsilon)=0.$$