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## 金融代写|金融微积分代写Finance Calculus代考|One-Dimensional Diffusion Process

1. Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space and let $\left{W_{t}: t \geq 0\right}$ be a standard Wiener process. Find the $\mathrm{SDE}$ for the random process $X_{t}=W_{t}^{n}, n \in \mathbb{Z}^{+}$.
Show that
$$\mathrm{E}\left(W_{t}^{n}\right)=\frac{1}{2} n(n-1) \int_{0}^{t} \mathrm{E}\left(W_{s}^{(n-2)}\right) d s$$
and using mathematical induction prove that
$$\mathbb{E}\left(W_{t}^{n}\right)= \begin{cases}\frac{n ! t^{\frac{n}{2}}}{2^{\frac{n}{2}}\left(\frac{n}{2}\right) !} & n=2,4,6, \ldots \ 0 & n=1,3,5, \ldots\end{cases}$$
Solution: By expanding $d X_{t}$ using Taylor’s formula and applying Itō’s formula,
\begin{aligned} d X_{t} &=\frac{\partial X_{i}}{\partial t} d t+\frac{\partial X_{i}}{\partial W_{t}} d W_{t}+\frac{1}{2} \frac{\partial^{2} X_{t}}{\partial W_{t}^{2}} d W_{t}^{2}+\ldots \ d W_{t}^{n} &=n W_{t}^{(n-1)} d W_{t}+\frac{1}{2} n(n-1) W_{t}^{(n-2)} d t . \end{aligned}
Taking integrals,
\begin{aligned} \int_{0}^{t} d W_{s}^{n} &=\int_{0}^{t} n W_{s}^{(n-1)} d W_{s}+\frac{1}{2} n(n-1) \int_{0}^{t} W_{s}^{(n-2)} d s \ W_{t}^{n} &=\int_{0}^{t} n W_{s}^{(n-1)} d W_{s}+\frac{1}{2} n(n-1) \int_{0}^{t} W_{s}^{(n-2)} d s \end{aligned}
Finally, by taking expectations.
$$\mathrm{E}\left(W_{t}^{n}\right)=\frac{1}{2} n(n-1) \int_{0}^{t} \mathbb{E}\left(W_{s}^{(n-2)}\right) d s$$

## 金融代写|金融微积分代写Finance Calculus代考|Multi-Dimensional Diffusion Process

1. Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space and consider $n$ assets with prices $S_{t}^{(i)}, i=1,2, \ldots, n$ satisfying the SDEs
\begin{aligned} d S_{t}^{(i)} &=\mu^{(i)} S_{t}^{(i)} d t+\sigma^{(i)} S_{t}^{(i)} d W_{t}^{(i)} \ \left(d W_{t}^{(i)}\right)\left(d W_{t}^{(i)}\right) &=\rho^{(i)} d t \end{aligned}
where $\left{W_{t}^{(i)}: t \geq 0\right}, i=1,2, \ldots, n$ are standard Wiener processes, $\rho^{(i)} \in(-1,1), i \neq j$ and $\rho^{(i i)}=1$.
By considering the function $f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)$, show using Itō’s formula that
\begin{aligned} d f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)=& \sum_{i=1}^{n} \mu^{(i)} S_{t}^{(i)} \frac{\partial f}{\partial S_{t}^{(i)}} d t+\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \rho^{(i)} \sigma^{(i)} \sigma^{(i)} S_{t}^{(i)} S_{t}^{(j)} \frac{\partial^{2} f}{\partial S_{t}^{(i)} \partial S_{t}^{(j)}} d t \ &+\sum_{i=1}^{n} \sigma^{(i)} S_{t}^{(i)} \frac{\partial f}{\partial S_{t}^{(i)}} d W_{t}^{(i)} . \end{aligned}
Solution: By expanding $d f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)$ using Taylor’s formula,
$$d f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)=\sum_{i=1}^{n} \frac{\partial f}{\partial S_{t}^{(i)}} d S_{t}^{(i)}+\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{\partial^{2} f}{\partial S_{t}^{(i)} S_{t}^{(j)}} d S_{t}^{(i)} d S_{t}^{(i)}+\ldots$$

\begin{aligned} =& \sum_{i=1}^{n} \frac{\partial f}{\partial S_{t}^{(i)}}\left(\mu^{(i)} S_{t}^{(i)} d t+\sigma^{(i)} S_{t}^{(i)} d W_{t}^{(i)}\right) \ &+\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{\partial^{2} f}{\partial S_{t}^{(i)} S_{t}^{(j)}}\left(\mu^{(i)} S_{t}^{(i)} d t+\sigma^{(i)} S_{t}^{(i)} d W_{t}^{(i)}\right)\left(\mu^{(j)} S_{t}^{(j)} d t+\sigma^{(j)} S_{t}^{(j)} d W_{t}^{(j)}\right) \ &+\ldots \end{aligned}
By setting $(d t)^{2}=0,\left(d W_{t}^{(i)}\right)^{2}=d t,\left(d W_{t}^{(i)}\right)\left(d W_{t}^{(j)}\right)=\rho^{(i j)} d t, d W_{t}^{(i)} d t=0, i, j=1,2, \ldots, n$ we have
\begin{aligned} d f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)=& \sum_{i=1}^{n} \mu^{(i)} S_{t}^{(i)} \frac{\partial f}{\partial S_{t}^{(i)}} d t+\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \rho^{(i j)} \sigma^{(i)} \sigma^{(j)} S_{t}^{(i)} S_{t}^{(j)} \frac{\partial^{2} f}{\partial S_{t}^{(i)} \partial S_{t}^{(j)}} d t \ &+\sum_{i=1}^{n} \sigma^{(i)} S_{t}^{(i)} \frac{\partial f}{\partial S_{t}^{(i)}} d W_{t}^{(i)} \end{aligned}

# 金融微积分代考

## 金融代写|金融微积分代写Finance Calculus代考|One-Dimensional Diffusion Process

1. 让 $(\Omega, \mathscr{F}, \mathbb{P})$ 是一个概率空间，让 left 的分隔符缺失或无法识别
2. 是一个标准的维纳过程。找出SDE对于随机过程 $X_{t}=W_{t}^{n}, n \in \mathbb{Z}^{+}$.
3. 显示
4. $$5. \mathrm{E}\left(W_{t}^{n}\right)=\frac{1}{2} n(n-1) \int_{0}^{t} \mathrm{E}\left(W_{s}^{(n-2)}\right) d s 6.$$
7. 并使用数学归纳法证明
8. $$9. \mathbb{E}\left(W_{t}^{n}\right)=\left{\frac{n ! t^{\frac{n}{2}}}{2^{\frac{1}{2}}\left(\frac{n}{2}\right) !} \quad n=2,4,6, \ldots 0 \quad n=1,3,5, \ldots\right. 10.$$
11. 解决方案: 通过扩展 $d X_{t}$ 使用泰勒公式并应用伊藤公式，
12. $$13. d X_{t}=\frac{\partial X_{i}}{\partial t} d t+\frac{\partial X_{i}}{\partial W_{t}} d W_{t}+\frac{1}{2} \frac{\partial^{2} X_{t}}{\partial W_{t}^{2}} d W_{t}^{2}+\ldots d W_{t}^{n} \quad=n W_{t}^{(n-1)} d W_{t}+\frac{1}{2} n(n-1) W_{t}^{(n-2)} d t . 14.$$
15. 取积分，
16. $$17. \int_{0}^{t} d W_{s}^{n}=\int_{0}^{t} n W_{s}^{(n-1)} d W_{s}+\frac{1}{2} n(n-1) \int_{0}^{t} W_{s}^{(n-2)} d s W_{t}^{n}=\int_{0}^{t} n W_{s}^{(n-1)} d W_{s}+\frac{1}{2} n(n-1) \int_{0}^{t} W_{s}^{(n-2)} d s 18.$$
19. 最后，通过期待。
20. $$21. \mathrm{E}\left(W_{t}^{n}\right)=\frac{1}{2} n(n-1) \int_{0}^{t} \mathbb{E}\left(W_{s}^{(n-2)}\right) d s 22.$$

## 金融代写|金融微积分代写Finance Calculus代考|Multi-Dimensional Diffusion Process

1. 让 $(\Omega, \mathscr{F}, \mathbb{P})$ 是一个概率空间并考虑 $n$ 有价格的资产 $S_{t}^{(i)}, i=1,2, \ldots, n$ 满足 SDE
$$d S_{t}^{(i)}=\mu^{(i)} S_{t}^{(i)} d t+\sigma^{(i)} S_{t}^{(i)} d W_{t}^{(i)}\left(d W_{t}^{(i)}\right)\left(d W_{t}^{(i)}\right) \quad=\rho^{(i)} d t$$
在哪里 \left 的分隔符缺失或无法识别 $\quad$ 是标准的维纳过程， $\rho^{(i)} \in(-1,1), i \neq j$ 和 $\rho^{(i i)}=1$.
通过考虑函数 $f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)$ ，用伊藤公式证明
$$d f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)=\sum_{i=1}^{n} \mu^{(i)} S_{t}^{(i)} \frac{\partial f}{\partial S_{t}^{(i)}} d t+\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \rho^{(i)} \sigma^{(i)} \sigma^{(i)} S_{t}^{(i)} S_{t}^{(j)} \frac{\partial^{2} f}{\partial S_{t}^{(i)} \partial S_{t}^{(j)}} d t \quad+\sum_{i=1}^{n} \sigma^{(i)} S_{t}^{(i)} \frac{\partial f}{\partial S_{t}^{(i)}} d W_{t}^{(i)} .$$
解决方案: 通过扩展 $d f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)$ 使用泰勒公式，
$$\begin{gathered} d f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)=\sum_{i=1}^{n} \frac{\partial f}{\partial S_{t}^{(i)}} d S_{t}^{(i)}+\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{\partial^{2} f}{\partial S_{t}^{(i)} S_{t}^{(j)}} d S_{t}^{(i)} d S_{t}^{(i)}+\ldots \ =\sum_{i=1}^{n} \frac{\partial f}{\partial S_{t}^{(i)}}\left(\mu^{(i)} S_{t}^{(i)} d t+\sigma^{(i)} S_{t}^{(i)} d W_{t}^{(i)}\right)+\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{\partial^{2} f}{\partial S_{t}^{(i)} S_{t}^{(j)}}\left(\mu^{(i)} S_{t}^{(i)} d t+\sigma^{(i)} S_{t}^{(i)} d W_{t}^{(i)}\right)\left(\mu^{(j)} S_{t}^{(j)} d t+\sigma^{(j)} S_{t}^{(j)} d W_{t}^{(j)}\right)+\ldots \end{gathered}$$
通过设置 $(d t)^{2}=0,\left(d W_{t}^{(i)}\right)^{2}=d t,\left(d W_{t}^{(i)}\right)\left(d W_{t}^{(j)}\right)=\rho^{(i j)} d t, d W_{t}^{(i)} d t=0, i, j=1,2, \ldots, n$ 我们有
$$d f\left(S_{t}^{(1)}, S_{t}^{(2)}, \ldots, S_{t}^{(n)}\right)=\sum_{i=1}^{n} \mu^{(i)} S_{t}^{(i)} \frac{\partial f}{\partial S_{t}^{(i)}} d t+\frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \rho^{(i j)} \sigma^{(i)} \sigma^{(j)} S_{t}^{(i)} S_{t}^{(j)} \frac{\partial^{2} f}{\partial S_{t}^{(i)} \partial S_{t}^{(j)}} d t \quad+\sum_{i=1}^{n} \sigma^{(i)} S_{t}^{(i)} \frac{\partial f}{\partial S_{t}^{(i)}} d W_{t}^{(i)}$$

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