# 经济代写|随机微积分代写Stochastic calculus代考|MTH5500

## 经济代写|随机微积分代写Stochastic calculus代考|Quadratic Variation of Stochastic Integrators

We now show that stochastic integrators also, like Brownian motion, have finite quadratic variation.

Theorem 4.64 Let $X$ be a stochastic integrator. Then there exists an adapted increasing process $A$, written as $[X, X]$, such that
$$X_t^2=X_0^2+2 \int_{11}^t X^{-} d X+[X, X]t, \quad \forall t .$$ Further, let $\delta_m \downarrow 0$ and for $m \geq 1$ let $\left{\tau_n^m: n \geq 0\right}$ be a $\delta_m$-partition for $X$. Then one has $$\sum{n=0}^{\infty}\left(X_{\tau_{n+1}^{\prime \prime} \wedge t}-X_{\tau_n^{m \prime \prime}}\right)^2 \stackrel{\text { ucp }}{\longrightarrow}[X, X]t .$$ Proof For $a, b \in \mathbb{R}$, $$b^2-a^2=2 a(b-a)+(b-a)^2 .$$ Using this with $b=X{\tau_{\mu_{n+1}^m}^m \wedge t}$ and $a=X_{\tau_{n^m} \wedge t}$ and summing with respect to $n$, we get
$$X_t^2=X_0^2+2 V_t^m+Q_t^m$$
where
$$V_t^m=\sum_{n=0}^{\infty} X_{\tau_n^m \wedge t}\left(X_{\tau_{\tau_{n+1}}^m \wedge t}-X_{\tau_n^m \wedge t}\right)$$
and
$$Q_t^m=\sum_{n=0}^{\infty}\left(X_{\tau_{n+1}^m \wedge t}-X_{\tau_n^m \wedge t}\right)^2 .$$
Note that after some $n$ that may depend upon $\omega \in \Omega, \tau_n^m>t$ and hence $X_{\tau_{n+1}^m \wedge t}=X_t$. In view of this, the two sums above have only finitely many nonzero terms.

## 经济代写|随机微积分代写Stochastic calculus代考|Quadratic Variation of Stochastic Integrals

In this section, we will relate the quadratic variation of $Y=\int f d X$ with the quadratic variation of $X$. We will show that for stochastic integrators $X, Z$ and $f \in \mathbb{L}(X)$, $g \in \mathbb{L}(Z)$
$$\left[\int f d X, \int g d Z\right]=\int f g d[X, Z]$$
We begin with a simple result.
Lemma 4.82 Let $X$ be a stochastic integrator; $f \in \mathbb{L}(X), 0 \leq u<\infty$, b be a $\mathcal{F}u$ measurable bounded random variable. Then $$h=b 1{(u, \infty)} f$$
is predictable, $h \in \mathbb{L}(X)$ and
$$\int_0^t b 1_{(u, \infty)} f d X=b \int_0^t 1_{(u, \infty)} f d X=b\left(\int_0^t f d X-\int_0^{u \wedge t} f d X\right)$$
Proof When $f \in \mathbb{S}$, validity of $(4.7 .2)$ can be verified directly as then $h$ is also simple predictable. Then, the class of $f$ such that $(4.7 .2)$ is true can be seen to be closed innder $b p$-convergence and hence hy Theorem $2.66$, (4.7.2) is valid for all hounded predictable processes. Finally, since $b$ is bounded, say by $c,|h| \leq c|f|$ and hence $h \in \mathbb{L}(X)$. Now (4.7.2) can be shown to be true for all $f \in \mathbb{L}(X)$ by approximating $f$ by $f^n=f 1_{{|f| \leq n}}$ and using Dominated Convergence Theorem-Theorem 4.29.

# 随机微积分代考

## 经济代写|随机微积分代写随机微积分代考|随机积分器的二次变分

$$X_t^2=X_0^2+2 \int_{11}^t X^{-} d X+[X, X]t, \quad \forall t .$$ 进一步，让 $\delta_m \downarrow 0$ 而对于 $m \geq 1$ 让 $\left{\tau_n^m: n \geq 0\right}$ 做一个 $\delta_m$-partition for $X$。然后是 $$\sum{n=0}^{\infty}\left(X_{\tau_{n+1}^{\prime \prime} \wedge t}-X_{\tau_n^{m \prime \prime}}\right)^2 \stackrel{\text { ucp }}{\longrightarrow}[X, X]t .$$ 证明 $a, b \in \mathbb{R}$， $$b^2-a^2=2 a(b-a)+(b-a)^2 .$$ 使用这个 $b=X{\tau_{\mu_{n+1}^m}^m \wedge t}$ 和 $a=X_{\tau_{n^m} \wedge t}$ 然后对求和 $n$，我们得到
$$X_t^2=X_0^2+2 V_t^m+Q_t^m$$
where
$$V_t^m=\sum_{n=0}^{\infty} X_{\tau_n^m \wedge t}\left(X_{\tau_{\tau_{n+1}}^m \wedge t}-X_{\tau_n^m \wedge t}\right)$$

$$Q_t^m=\sum_{n=0}^{\infty}\left(X_{\tau_{n+1}^m \wedge t}-X_{\tau_n^m \wedge t}\right)^2 .$$

## 经济代写|随机微积分代写随机积分代考|随机积分的二次变分

$$\left[\int f d X, \int g d Z\right]=\int f g d[X, Z]$$

$$\int_0^t b 1_{(u, \infty)} f d X=b \int_0^t 1_{(u, \infty)} f d X=b\left(\int_0^t f d X-\int_0^{u \wedge t} f d X\right)$$

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