Stochastic Analysis 随机分析

Quadratic Variation

Definition: Let $E_n = \{0 = t_1^{(n)} < \ldots < t_{l_n}^{(n)}\}$ be a sequence of partitions with $s(E_n) := \sup |t_{i+1}^{(n)} - t_i^{(n)}| \to 0$. Then, the continuous function $X_t$ has continuous quadratic variation along the sequence $E_n$ if

$\langle X \rangle_t = \lim_{n \to \infty} \sum_{\substack{t_j \in E_n \\ t_{j+1} \leq t}} \left( X_{t_{j+1}} - X_{t_j} \right)^2$

exists $P$-a.s. and is a continuous function of $t$.

Remark: For any continuous function $X_t$, its quadratic variation $t \mapsto \langle X \rangle_t$ is monotone increasing (non-decreasing).

其实不难发现，这个quadratic variation与普通的variance非常像，有很多类似的性质。

Lemma: Let $X$ be a continuous function, for $a,b \in \mathbb{R}$ we have

$\langle aX+b \rangle = a^2 \langle X \rangle$

Proof:

By definition, for all $t$ we have:

$\begin{align*} \langle aX+b \rangle_t &= \lim_{n \to \infty} \sum_{\substack{t_j \in E_n \\ t_{j+1} \leq t}} \left( aX_{t_{j+1}}+b - aX_{t_j} -b \right)^2 \\ &= a^2\lim_{n \to \infty} \sum_{\substack{t_j \in E_n \\ t_{j+1} \leq t}} \left( X_{t_{j+1}} - X_{t_j} \right)^2 \\ &= a^2 \langle X \rangle \end{align*}$

$\square$

Cross Variation

在这章里，我们假设$X,Y$都是continuous function，并且拥有continuous quadratic variations $\langle X \rangle, \langle Y \rangle$.

Definition: The cross-variation $\langle X,Y \rangle$ is defined as

$\langle X,Y \rangle_t = \underset{n \to \infty}{\text{lim}} \sum_{\substack{t_i \in E_n \\ t_{i+1} \leq t}} (X_{t_{j+1}} - X_{t_j})(Y_{t_{j+1}} - Y_{t_j})$

(if the limit exists). Clearly,

$\langle X,X \rangle = \langle X \rangle$

Lemma: Let $X,Y$ be continuous functions with continuous quadratic variations $\langle X \rangle, \langle Y \rangle$, for $a,b,c,d \in \mathbb{R}$ we have

$\langle aX+b,cY+d \rangle = ac\langle X,Y \rangle$

Proof:

By definition, for all $t$ we have:

$\begin{align*} \langle aX+b,cY+d \rangle_t &= \underset{n \to \infty}{\text{lim}} \sum_{\substack{t_i \in E_n \\ t_{i+1} \leq t}} (aX_{t_{j+1}}+b - aX_{t_j}-b)(cY_{t_{j+1}}+d - cY_{t_j}-d) \\ &= ac \cdot \underset{n \to \infty}{\text{lim}} \sum_{\substack{t_i \in E_n \\ t_{i+1} \leq t}} (X_{t_{j+1}} - X_{t_j})(Y_{t_{j+1}} - Y_{t_j}) \\ &= ac\langle X,Y \rangle \end{align*}$

$\square$

Lemma: Let $X,Y$ be continuous functions with continuous quadratic variations $\langle X \rangle, \langle Y \rangle$, we have

$\langle X+Y,X+Y \rangle = \langle X \rangle + 2\langle X,Y \rangle + \langle Y \rangle$

Proof:

By definition, for all $t$ we have:

$\begin{align*} \langle X+Y \rangle_t &= \lim_{n \to \infty} \sum_{\substack{t_j \in E_n \\ t_{j+1} \leq t}} \left( X_{t_{j+1}}+ Y_{t_{j+1}} - X_{t_j} - Y_{t_{j}}\right)^2 \\ &= \lim_{n \to \infty} \sum_{\substack{t_j \in E_n \\ t_{j+1} \leq t}} \left( (X_{t_{j+1}}- X_{t_j})+ (Y_{t_{j+1}} - Y_{t_{j}})\right)^2\\ &= \lim_{n \to \infty} \sum_{\substack{t_j \in E_n \\ t_{j+1} \leq t}} \Bigl(\left( X_{t_{j+1}}- X_{t_j}\right)^2 +2\left( X_{t_{j+1}}- X_{t_j}\right)\left(Y_{t_{j+1}} - Y_{t_{j}}\right) +\left(Y_{t_{j+1}} - Y_{t_{j}}\right)^2 \Bigl) \\ &= \langle X \rangle + 2\langle X,Y \rangle + \langle Y \rangle \end{align*}$

$\square$

Lemma: Let $X,Y$ be continuous functions with continuous quadratic variations $\langle X \rangle, \langle Y \rangle$ and $\langle X \rangle \equiv 0$, then $\langle X+Y \rangle =0$.