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Let $T>0$ and $P^{n}:=\lbrace0=t_{0}^{n}<t_{1}^{n}<...<t_{m_{n}}^{n}=T\rbrace$ be the $n$- th division of the interval $[0,T]$ such that $\delta(P^{n})\to0$, as $n\to\infty$, where $\delta(P^{n})=\text{max}\lbrace |t_{k+1}^{n}-t_{k}^{n}|:k=0,1,...,m_{n}-1\rbrace$. Show that $$\sum_{k=0}^{m_{n}-1}(W(t_{k+1}^{n})-W(t_{k}^{n}))^{2}\to T $$ in $L^{2}(\Omega)$, where $L^{2}(\Omega):=\lbrace X:\Omega\to\mathbb{R}: \ \mathbb{E}|X|^{2}<\infty\rbrace$ and $\lbrace W(t)\rbrace$ a Wiener process. Is there some elegant way to prove this?

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1 Answer 1

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Let $$S_2^{P^n}(W) := \sum_{j=1}^{m_n-1} (W(t_{j+1}^n)-W(t_j^n))^2$$ Then $$\mathbb{E}S_2^{P^n}(W) = \mathbb{E} \left(\sum_{j=1}^{m_n-1} (W(t_{j+1}^n)-W(t_j^n))^2\right) = \sum_{j=1}^{m_n-1} (t_{j+1}^n-t_{j}^n)=T$$ hence

$$\begin{align} \mathbb{E}\big((S_2^{P_n}(W)-T)^2 \big) &= \text{var}(S_2^{P_n}) \stackrel{(1)}{=} \sum_{j=1}^{m_n-1} \underbrace{\text{var}((W(t_{j+1}^n)-W(t_j^n)^2)}_{\mathbb{E} \bigg( \big( (W(t_{j+1}^n)-W(t_j^n))^2-(t_{j+1}^n-t_j^n) \big)^2 \bigg)} \\ &\stackrel{(2)}{=} \sum_{j=1}^{m_n-1} \mathbb{E} \bigg( \big( W(t_{j+1}^n-t_j^n)^2-(t_{j+1}^n-t_j^n) \big)^2 \bigg) \\ &\stackrel{W_t \sim \sqrt{t} \cdot W_1}{=} \sum_{j=1}^{m_n-1} (t_{j+1}^n-t_j^n)^2 \cdot \underbrace{\mathbb{E}((W_1^2-1)^2)}_{2} \\ &\leq 2 \cdot \delta(P^n) \cdot T \to 0 \qquad (n \to \infty) \end{align}$$

where we used the independence of the increments in (1) and the stationarity of the increments in (2).

The limit

$$L^2-\lim_{n \to \infty} S_2^{P_n}(W)$$

is called quadratic variation. The calculation above shows that the quadratic variation of a Wiener process on $[0,T]$ is equal to $T$.

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Thank you! That was really helpful :) –  czachur Feb 5 '13 at 7:45

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