# 金融代写|金融微积分代写Finance Calculus代考|MAT9740

## 金融代写|金融微积分代写Finance Calculus代考|Jump Diffusion Process

1. Pure Jump Process. Let $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space and let $\left{N_{t}: t \geq 0\right}$ be a Poisson process with intensity $\lambda>0$ relative to the filtration $\mathscr{F}{t}, t \geq 0$. Suppose $S{t}$ follows a pure jump process
$$\frac{d S_{t}}{S_{t^{-}}}=\left(J_{t}-1\right) d N_{t}$$
where $J_{t}$ is the jump size variable if $N$ jumps at time $t, J_{t} \Perp N_{t}$ and
$$d N_{t}= \begin{cases}1 & \text { with probability } \lambda d t \ 0 & \text { with probahility } 1-\lambda d t\end{cases}$$
Explain why the term in $d N_{t}$ is $J_{t}-1$ and not $J_{t}$.

Show that the above differential equation can also be written as
$$\frac{d S_{t}}{S_{f^{*}}}=d M_{t}$$
where $M_{t}=\sum_{i=1}^{N_{t}}\left(J_{i}-1\right)$ is a compound Poisson process such that $J_{i}, i=1,2, \ldots$ is a sequence of independent and identically distributed random variables which are also independent of $N_{t}$.

Solution: Let $S_{t^{-}}$be the value of $S_{t}$ just before a jump and assume there occurs an instantaneous jump (i.e., $d N_{t}=1$ ) in which $S_{t}$ changes from $S_{t^{-}}$to $J_{t} S_{t^{-}}$where $J_{t}$ is the jump size. Thus,
$$d S_{t}=J_{t} S_{t^{-}}-S_{t^{-}}=\left(J_{t}-1\right) S_{t^{-}}$$
or
$$\frac{d S_{t}}{S_{t}}=\left(J_{t}-1\right) \text {. }$$
Therefore, we can write
$$\frac{d S_{t}}{S_{t^{-}}}=\left(J_{t}-1\right) d N_{t}$$
where
$$d N_{t}= \begin{cases}1 & \text { with probability } \lambda d t \ 0 & \text { with probability } 1-\lambda d t\end{cases}$$
Given the compound Poisson process
$$M_{t}=\sum_{i=1}^{N_{t}}\left(J_{i}-1\right)$$
we let $M_{t}$ – be the value of $M_{t}$ just before a jump event. If $N$ jumps at time $t$ then
$$d M_{t}=M_{t}-M_{t^{-}}=J_{t}-1 .$$
Thus, in general, we can write
$$d M_{t}=\left(J_{t}-1\right) d N_{t}$$
which implies the pure jump process can also be expressed as $\frac{d S_{t}}{S_{t^{-}}}=d M_{t}$.

## 金融代写|金融微积分代写Finance Calculus代考|Girsanov’s Theorem for Jump Processes

1. Let $\left{N_{t}: 0 \leq t \leq T\right}$ be a Poisson process with intensity $\lambda>0$ defined on the probability space $(\Omega, \mathscr{F}, \mathbb{P})$ with respect to the filtration $\mathscr{F}{t}, 0 \leq t \leq T$. Let $\eta>0$ and consider the Radon-Nikodým derivative process $$Z{t}=e^{(\lambda-\eta) t}\left(\frac{\eta}{\lambda}\right)^{N_{t}} .$$
Show that
$$\frac{d Z_{t}}{Z_{t^{-}}}=(\lambda-\eta) d t-\left(\frac{\lambda-\eta}{\lambda}\right) d N_{t}$$
for $0 \leq t \leq T$.
Solution: From Taylor’s theorem,
$$d Z_{t}=\frac{\partial Z_{t}}{\partial t} d t+\frac{\partial Z_{t}}{\partial N_{t}} d N_{t}+\frac{\partial^{2} Z_{t}}{\partial t \partial N_{t}}\left(d N_{t} d t\right)+\frac{1}{2 !} \frac{\partial^{2} Z_{t}}{\partial N_{t}^{2}}\left(d N_{t}\right)^{2}+\frac{1}{3 !} \frac{\partial^{3} Z_{t}}{\partial N_{t}^{3}}\left(d N_{t}\right)^{3}+\ldots$$

Since $d N_{t} d t=0,\left(d N_{t}\right)^{2}=\left(d N_{t}\right)^{3}=\ldots=d N_{t}$, we have
$$d Z_{t}=\frac{\partial Z_{t}}{\partial t} d t+\left(\frac{\partial Z_{t}}{\partial N_{t}}+\frac{1}{2 !} \frac{\partial^{2} Z_{t}}{\partial N_{t}^{2}}+\frac{1}{3 !} \frac{\partial^{3} Z_{t}}{\partial N_{t}^{3}}+\ldots\right) d N_{t} .$$
From $Z_{t}=e^{(\lambda-\eta) t}\left(\frac{\eta}{\lambda}\right)^{N_{t}}$ we can express
\begin{aligned} \frac{\partial Z_{t}}{\partial t} &=(\lambda-\eta) e^{(\lambda-\eta))}\left(\frac{\eta}{\lambda}\right)^{N_{t}}=(\lambda-\eta) Z_{t} \ \frac{\partial Z_{t}}{\partial N_{t}} &=e^{(\lambda-\eta) t} \log \left(\frac{\eta}{\lambda}\right)\left(\frac{\eta}{\lambda}\right)^{N_{t}}=\log \left(\frac{\eta}{\lambda}\right) Z_{t} \ \frac{\partial^{2} Z_{t}}{\partial N_{t}^{2}} &=\log \left(\frac{\eta}{\lambda}\right) \frac{\partial Z_{t}}{\partial N_{t}}=\left[\log \left(\frac{\eta}{\lambda}\right)\right]^{2} Z_{t} . \end{aligned}
In general, we can deduce that
$$\frac{\partial^{m} Z_{t}}{\partial N_{t}^{m}}=\left[\log \left(\frac{\eta}{\lambda}\right)\right]^{m} Z_{t}, \quad m=1,2, \ldots$$

## 金融代写|金融微积分代写Finance Calculus代考|Jump Diffusion Process

$$\frac{d S_{t}}{S_{t^{-}}}=\left(J_{t}-1\right) d N_{t}$$

$d N_{t}={1 \quad$ with probability $\lambda d t 0 \quad$ with probahility $1-\lambda d t$

$$\frac{d S_{t}}{S_{f^{*}}}=d M_{t}$$

$$d S_{t}=J_{t} S_{t^{-}}-S_{t^{-}}=\left(J_{t}-1\right) S_{t^{-}}$$

$$\frac{d S_{t}}{S_{t}}=\left(J_{t}-1\right)$$

$$\frac{d S_{t}}{S_{t^{-}}}=\left(J_{t}-1\right) d N_{t}$$

$d N_{t}={1 \quad$ with probability $\lambda d t 0 \quad$ with probability $1-\lambda d t$

$$M_{t}=\sum_{i=1}^{N_{t}}\left(J_{i}-1\right)$$

$$d M_{t}=M_{t}-M_{t^{-}}=J_{t}-1 .$$

$$d M_{t}=\left(J_{t}-1\right) d N_{t}$$

## 金融代写|金融微积分代写Finance Calculus代考|Girsanov’s Theorem for Jump Processes

1. 让《left(NN_{t}: 0 \eq $t$ \eq \right } } \text { 是一个有强度的泊松过程 } \lambda > 0 \text { 在概率空间上定义 } ( \Omega , \mathscr { F } , \mathbb { P } ) \text { 关于过滤 } $\mathscr{F} t, 0 \leq t \leq T$. 让 $\eta>0$ 并考虑 Radon-Nikodým 导数过程
$$Z t=e^{(\lambda-\eta) t}\left(\frac{\eta}{\lambda}\right)^{N_{k}} .$$
显示
$$\frac{d Z_{t}}{Z_{t^{-}}}=(\lambda-\eta) d t-\left(\frac{\lambda-\eta}{\lambda}\right) d N_{t}$$
为了 $0 \leq t \leq T$.
解: 根据泰勒定理，
$$d Z_{t}=\frac{\partial Z_{t}}{\partial t} d t+\frac{\partial Z_{t}}{\partial N_{t}} d N_{t}+\frac{\partial^{2} Z_{t}}{\partial t \partial N_{t}}\left(d N_{t} d t\right)+\frac{1}{2 !} \frac{\partial^{2} Z_{t}}{\partial N_{t}^{2}}\left(d N_{t}\right)^{2}+\frac{1}{3 !} \frac{\partial^{3} Z_{t}}{\partial N_{t}^{3}}\left(d N_{t}\right)^{3}+\ldots$$
自从 $d N_{t} d t=0,\left(d N_{t}\right)^{2}=\left(d N_{t}\right)^{3}=\ldots=d N_{t}$ ，我们有
$$d Z_{t}=\frac{\partial Z_{t}}{\partial t} d t+\left(\frac{\partial Z_{t}}{\partial N_{t}}+\frac{1}{2 !} \frac{\partial^{2} Z_{t}}{\partial N_{t}^{2}}+\frac{1}{3 !} \frac{\partial^{3} Z_{t}}{\partial N_{t}^{3}}+\ldots\right) d N_{t} .$$
从 $Z_{t}=e^{(\lambda-\eta) t}\left(\frac{\eta}{\lambda}\right)^{N_{t}}$ 我们可以表达
$$\frac{\partial Z_{t}}{\partial t}=(\lambda-\eta) e^{(\lambda-\eta))}\left(\frac{\eta}{\lambda}\right)^{N_{t}}=(\lambda-\eta) Z_{t} \frac{\partial Z_{t}}{\partial N_{t}} \quad=e^{(\lambda-\eta) t} \log \left(\frac{\eta}{\lambda}\right)\left(\frac{\eta}{\lambda}\right)^{N_{t}}=\log \left(\frac{\eta}{\lambda}\right) Z_{t} \frac{\partial^{2} Z_{t}}{\partial N_{t}^{2}}=$$
一般来说，我们可以推断出
$$\frac{\partial^{m} Z_{t}}{\partial N_{t}^{m}}=\left[\log \left(\frac{\eta}{\lambda}\right)\right]^{m} Z_{t}, \quad m=1,2, \ldots$$

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