## 金融代写|金融微积分代写Finance Calculus代考|Risk-Neutral Measure for Jump Processes

1. Simple Jump Process. Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space and let $\left{N_{t}: 0 \leq t \leq T\right}$ be a Poisson process with intensity $\lambda>0$ relative to the filtration $\mathscr{F}{t}, 0 \leq t \leq T$. Suppose the asset price $S{t}$ follows a simple jump process
$$\frac{d S_{t}}{S_{t^{-}}}=(J-1) d N_{t}$$
where $J$ is a constant jump amplitude if $N$ jumps at time $t$ and
$$d N_{t}= \begin{cases}1 & \text { with probability } \lambda d t \ 0 & \text { with probability } 1-\lambda d t .\end{cases}$$
Let $r$ be the risk-free interest rate.
By considering the Radon-Nikodým derivative process, for $\eta>0$
$$Z_{t}=e^{(\lambda-\eta) t}\left(\frac{\eta}{\lambda}\right)^{N_{t}}, \quad 0 \leq t \leq T$$
show that by changing the measure $\mathbb{P}$ to the risk-neutral measure $\mathbb{Q}$, the above stochastic differential equation is
$$\frac{d S_{t}}{S_{t}}=(J-1) d N_{t}$$
where $N_{t} \sim$ Poisson $(\eta t), \eta=r(J-1)^{-1}>0$ provided $J>1$.
Is the market arbitrage free and complete under the $\mathbb{Q}$ measure?
Solution: Let $\eta>0$ and from the Radon-Nikodým derivative process
$$Z_{t}=e^{(\lambda-\eta) \mathrm{r}}\left(\frac{\eta}{\lambda}\right)^{N_{t}} .$$
By changing the measure $\mathbb{P}$ to the risk-neutral measure $\mathbb{Q}$ on the filtration $\mathscr{F}{s}, 0 \leq s \leq t$ then under $\mathbf{Q}, N{t} \sim$ Poisson $(\eta t)$ and the discounted asset price $e^{-r t} S_{t}$ is a Q-martingale.
By setting $S_{t^{-}}$to denote the value of $S_{t}$ before a jump event and by expanding $d\left(e^{-r} S_{t}\right)$, we have
\begin{aligned} d\left(e^{-r t} S_{t}\right) &=-r e^{-r t} S_{t} d t+e^{-r t} d S_{t} \ &=-r e^{-r t} S_{t} d t+e^{-r t} S_{t}(J-1) d N_{t} \ &=e^{-r t S_{t}}(-r+\eta(J-1)) d t+e^{-r t} S_{t}(J-1)\left(d N_{t}-\eta d t\right) . \end{aligned}

## 金融代写|金融微积分代写Finance Calculus代考|Mathematics Formulae

Indices
$$\begin{gathered} x^{a} x^{b}=x^{a+b}, \quad \frac{x^{a}}{x^{b}}=x^{a-b}, \quad\left(x^{a}\right)^{b}=\left(x^{b}\right)^{a}=x^{a b} \ x^{-a}=\frac{1}{x^{a}}, \quad\left(\frac{x}{y}\right)^{a}=\frac{x^{a}}{y^{a}}, \quad x^{0}=1 . \end{gathered}$$
Surds
$$\begin{gathered} x_{a}^{\frac{1}{a}}=\sqrt[a]{x}, \quad \sqrt[a]{x y}=\sqrt[a]{x} \sqrt[a]{y}, \quad \sqrt[a]{x / y}=\frac{\sqrt[a]{x}}{\sqrt[a]{y}} \ (\sqrt[a]{x})^{a}=x, \quad \sqrt[a]{\sqrt[b]{x}}=\sqrt[a b]{x}, \quad(\sqrt[a]{x})^{b}=x^{\frac{b}{a}} \end{gathered}$$
Exponential and Natural Logarithm
$$\begin{gathered} e^{x} e^{y}=e^{x+y}, \quad\left(e^{x}\right)^{y}=\left(e^{y}\right)^{x}=e^{x y}, \quad e^{0}=1 \ \log (x y)=\log x+\log y, \quad \log \left(\frac{x}{y}\right)=\log x-\log y, \quad \log x^{y}=y \log x \ \log e^{x}=x, \quad e^{\log x}=x, \quad e^{a \log x}=x^{a} \end{gathered}$$
For constants $a, b$ and $c$, the roots of a quadratic equation $a x^{2}+b x+c=0$ are
$$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} .$$
Binomial Formula
\begin{aligned} &\left(\begin{array}{l} n \ k \end{array}\right)=\frac{n !}{k !(n-k) !}, \quad\left(\begin{array}{l} n \ k \end{array}\right)+\left(\begin{array}{c} n \ k+1 \end{array}\right)=\left(\begin{array}{l} n+1 \ k+1 \end{array}\right) \ &(x+y)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l} n \ k \end{array}\right) x^{n-k} y^{k}=\sum_{k=0}^{n} \frac{n !}{k !(n-k) !} x^{n-k} y^{k} . \end{aligned}

## 金融代写|金融微积分代写Finance Calculus代考|Risk-Neutral Measure for Jump Processes

1. 简单的跳转过程。让 $(\Omega, \mathscr{F}, \mathbb{P})$ 是一个概率空间，让 \left } { N _ { – } \text { t } } : 0 \backslash \text { leq } t \backslash \text { leq T\right } } \text { 是一个有强度的泊松 } 过程 $\lambda>0$ 相对于过滤 $\mathscr{F} t, 0 \leq t \leq T$. 假设资产价格 $S t$ 遒循一个简单的跳转过程 $$\frac{d S{t}}{S_{t^{-}}}=(J-1) d N_{t} 2.$$
3. 在哪里 $J$ 是一个恒定的跳跃幅度，如果 $N$ 时而跳跃 $t$ 和
4. $d N_{t}={1 \quad$ with probability $\lambda d t 0 \quad$ with probability $1-\lambda d t .$
5. 让 $r$ 为无风险利率。
6. 通过考虑 Radon-Nikodým 导数过程，对于 $\eta>0$
7. $$8. Z_{t}=e^{(\lambda-\eta) t}\left(\frac{\eta}{\lambda}\right)^{N_{t}}, \quad 0 \leq t \leq T 9.$$
10. 表明通过改变措施 $\mathbb{P} 风$ 险中性测度 $\mathbb{Q}$ ，上述随机微分方程为
11. $$12. \frac{d S_{t}}{S_{t}}=(J-1) d N_{t} 13.$$
14. 在哪里 $N_{t} \sim$ 泊松 $(\eta t), \eta=r(J-1)^{-1}>0$ 假如 $J>1$.
15. 解决方案: 让 $\eta>0$ 并来自 Radon-Nikodým 导数过程
16. $$17. Z_{t}=e^{(\lambda-\eta) \mathrm{r}}\left(\frac{\eta}{\lambda}\right)^{N_{t}} . 18.$$
19. 通过改变测量 $\mathbb{P}$ 风险中性测度 $\mathbb{Q}$ 关于过滤 $\mathscr{F} s, 0 \leq s \leq t$ 然后在 $\mathbf{Q}, N t \sim$ 泊松 $(\eta t)$ 和折现的资产价格 $e^{-r t} S_{t}$ 是一个 Q-鞅。
20. 通过设置 $S_{t^{-}}$来表示 $S_{t}$ 在跳跃事件之前并通过扩展 $d\left(e^{-r} S_{t}\right)$ ，我们有
21. $$22. d\left(e^{-r t} S_{t}\right)=-r e^{-r t} S_{t} d t+e^{-r t} d S_{t} \quad=-r e^{-r t} S_{t} d t+e^{-r t} S_{t}(J-1) d N_{t}=e^{-r t S_{t}}(-r+\eta(J-1)) d t 23.$$

## 金融代写|金融微积分代写Finance Calculus代考|Mathematics Formulae

$$x^{a} x^{b}=x^{a+b}, \quad \frac{x^{a}}{x^{b}}=x^{a-b}, \quad\left(x^{a}\right)^{b}=\left(x^{b}\right)^{a}=x^{a b} x^{-a}=\frac{1}{x^{a}}, \quad\left(\frac{x}{y}\right)^{a}=\frac{x^{a}}{y^{a}}, \quad x^{0}=1 .$$

$$x_{a}^{\frac{1}{a}}=\sqrt[a]{x}, \quad \sqrt[a]{x y}=\sqrt[4]{x} \sqrt[a]{y}, \quad \sqrt[a]{x / y}=\frac{\sqrt[x]{x}}{\sqrt[a]{y}}(\sqrt[a]{x})^{a}=x, \quad \sqrt[a]{\sqrt[b]{x}}=\sqrt[a b]{x}, \quad(\sqrt[a]{x})^{b}=x^{\frac{b}{a}}$$

$$e^{x} e^{y}=e^{x+y}, \quad\left(e^{x}\right)^{y}=\left(e^{y}\right)^{x}=e^{x y}, \quad e^{0}=1 \log (x y)=\log x+\log y, \quad \log \left(\frac{x}{y}\right)=\log x-\log y,$$

$$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} .$$

$$(n k)=\frac{n !}{k !(n-k) !}, \quad(n k)+(n k+1)=(n+1 k+1) \quad(x+y)^{n}=\sum_{k=0}^{n}(n k) x^{n-k} y^{k}=\sum_{k=0}^{n} \frac{\bar{k}}{k}$$

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