# 经济代写|随机微积分代写Stochastic calculus代考|GRA6550

#### Doug I. Jones

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## 经济代写|随机微积分代写Stochastic calculus代考|Multidimensional Ito’s Integral

Let $W=\left(W^1, W^2, \ldots, W^d\right)$ be $d$-dimensional Brownian motion, where $W^j$ is the $j$ th component. In other words, each $W^j$ is a real-valued Brownian motion and $W^1, W^2, \ldots, W^d$ are independent. Suppose further that $(\mathcal{F}$. ) is a filtration such that $\left(W_t, \mathcal{F}t\right){{t \geq 0}}$ is a Wiener martingale. Thus for each $s,\left{W_t-W_s: t \geq s\right}$ is independent of $\mathcal{F}s$. Denoting $\theta=\left(\theta^1, \ldots, \theta^d\right) \in \mathbb{R}^d$ and defining \begin{aligned} &X_t^\theta=\sum{j=1}^d \theta^j W_t^j \ &M_t^\theta=\left(X_t^\theta\right)^2-t, \end{aligned}
we have
$$\forall \theta \in \mathbb{R}^d \text { with }|\theta|=1 ; \quad X^\theta, M^\theta \text { are }(\mathcal{F} \text {.)-martingales . }$$
The argument given in the proof of next lemma is interesting. Throughout this section, the filtration $(\mathcal{F}$.) will remain fixed.
Lemma $3.22$ For $j \neq k, W^j W^k$ is also a martingale.
Proof Let $X_t=\frac{1}{\sqrt{2}}\left(W_t^j+W_t^k\right)$. Then, as seen above, $X$ is a Brownian motion and hence $X_t^2-t$ is a martingale. Note that
\begin{aligned} X_t^2-t &=\frac{1}{2}\left[\left(W_t^j\right)^2+\left(W_t^k\right)^2+2 W_t^j W_t^k\right]-t \ &=\frac{1}{2}\left[\left(W_t^j\right)^2-t\right]+\frac{1}{2}\left[\left(W_t^k\right)^2-t\right]+W_t^j W_t^k \end{aligned}
Since the left-hand side above as well as the first two terms of right-hand side above are martingales, it follows that so is the third term.

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

We are going to consider stochastic differential equations (SDE) of the type
$$d X_t=\sigma\left(t, X_t\right) d W_t+b\left(t, X_t\right) d t .$$
Equation (3.5.1) is to be interpreted as an integral equation:
$$X_t=X_0+\int_0^t \sigma\left(s, X_s\right) d W_s+\int_0^t b\left(s, X_s\right) d s .$$
Here $W$ is an $\mathbb{R}^d$-valued Brownian motion, $X_0$ is an $\mathbb{R}^d$-valued $\mathcal{F}_0$ measurable random variable, $\sigma:[0, \infty) \times \mathbb{R}^m \mapsto \mathrm{L}(m, d)$ and $b:[0, \infty) \times \mathbb{R}^m \mapsto \mathbb{R}^m$ are given functions, and one is seeking a process $X$ such that $(3.5 .2)$ is true. The solution $X$ to the SDE (3.5.1), when it exists, is called a diffusion process with diffusion coefficient $\sigma \sigma^*$ and drift coefficient $b$.
We shall impose the following conditions on $\sigma, b$ :
$$\begin{gathered} \sigma:[0, \infty) \times \mathbb{R}^m \mapsto \mathrm{L}(m, d) \text { is a continuous function } \ b:[0, \infty) \times \mathbb{R}^m \mapsto \mathbb{R}^m \text { is a continuous function } \ \forall T<\infty \exists C_T<\infty \text { such that forr all } t \in[0, T], x^1, x^2 \in \mathbb{R}^d \ \left|\sigma\left(t, x^1\right)-\sigma\left(t, x^2\right)\right| \leq C_T\left|x^1-x^2\right| \ \left|b\left(t, x^1\right)-b\left(t, x^2\right)\right| \leq C_T\left|x^1-x^2\right| \end{gathered}$$

Since $t \mapsto \sigma(t, 0)$ and $t \mapsto b(t, 0)$ are continuous and hence bounded on $[0, T]$ for every $T<\infty$, using the Lipschitz conditions (3.5.4), we can conclude that for each $T<\infty, \exists K_T<\infty$ such that
$$\begin{gathered} |\sigma(t, x)| \leq K_T(1+|x|), \ |b(t, x)| \leq K_T(1+|x|) . \end{gathered}$$
We will need the following lemma, known as Gronwall’s lemma, for proving uniqueness of solution to (3.5.2) under the Lipschitz conditions.

# 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|多维伊藤积分

$$\forall \theta \in \mathbb{R}^d \text { with }|\theta|=1 ; \quad X^\theta, M^\theta \text { are }(\mathcal{F} \text {.)-martingales . }$$

\begin{aligned} X_t^2-t &=\frac{1}{2}\left[\left(W_t^j\right)^2+\left(W_t^k\right)^2+2 W_t^j W_t^k\right]-t \ &=\frac{1}{2}\left[\left(W_t^j\right)^2-t\right]+\frac{1}{2}\left[\left(W_t^k\right)^2-t\right]+W_t^j W_t^k \end{aligned}

## 经济代写|随机微积分代写Stochastic calculus代考|随机微分方程

$$X_t=X_0+\int_0^t \sigma\left(s, X_s\right) d W_s+\int_0^t b\left(s, X_s\right) d s .$$

$$\begin{gathered} \sigma:[0, \infty) \times \mathbb{R}^m \mapsto \mathrm{L}(m, d) \text { is a continuous function } \ b:[0, \infty) \times \mathbb{R}^m \mapsto \mathbb{R}^m \text { is a continuous function } \ \forall T<\infty \exists C_T<\infty \text { such that forr all } t \in[0, T], x^1, x^2 \in \mathbb{R}^d \ \left|\sigma\left(t, x^1\right)-\sigma\left(t, x^2\right)\right| \leq C_T\left|x^1-x^2\right| \ \left|b\left(t, x^1\right)-b\left(t, x^2\right)\right| \leq C_T\left|x^1-x^2\right| \end{gathered}$$

$$\begin{gathered} |\sigma(t, x)| \leq K_T(1+|x|), \ |b(t, x)| \leq K_T(1+|x|) . \end{gathered}$$
。我们将需要以下引理，称为Gronwall引理，来证明在Lipschitz条件下(3.5.2)解的唯一性

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