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# 数学代写|随机分析代写Stochastic Analysis in Finance代考|GRA6550 General Case

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## 数学代写|随机分析代写Stochastic Analysis in Finance代考|General Case

We consider more general model than the Black-Scholes model. Let $(\Omega, \mathcal{F}, P)$ be a complete probability space and $B=\left(B^1, \ldots, B^d\right)$ be a $d$-dimensional Brownian motion. Let $\mathcal{F}t=\sigma\left{\sigma{B(s) ; s \leqq t} \cup \mathcal{N}_0\right}, t \geqq 0$. Then by Proposition 3.7.3 we see that $\left{\mathcal{F}_t\right}{t \in[0, \infty)}$ is a filtration satisfying the standard condition, and $B$ is an $\left{\mathcal{F}t\right}{t \in[0, \infty)^{-}}$ Brownian motion.

Let $S^i, i=0,1, \ldots, N$, be $\left{\mathcal{F}t\right}{t \in[0, \infty)}$-continuous semi-martingales satisfying the following condition.
(S-1) $P\left(\min {t \in[0, T]} S^0(t)>0\right)=1$ for any $T>0$. Then by Corollary $4.3 .1$ we see that $\tilde{S}^i(t)=S^0(t)^{-1} S^i(t), t \geqq 0, i=1, \ldots, N$, are continuous semi-martingales. Moreover, we assume the following. (S-2) There are progressively measurable processes $\sigma_j^i, b^i, i=1, \ldots, N, j=$ $1,2, \ldots, d$, such that \begin{aligned} &P\left(\int_0^T \sigma_j^i(t)^2 d t<\infty\right)=1, \quad T>0, i=1, \ldots, N, \ &P\left(\int_0^T\left|b^i(t)\right| d t<\infty\right)=1, \quad T>0, i=1, \ldots, N \end{aligned} and $$\tilde{S}^i(t)=\sum{j=1}^d \int_0^t \sigma_j^i(s) d B^j(s)+\int_0^t b^i(s) d s, \quad t \geqq 0, i=1, \ldots, N$$

## 数学代写|随机分析代写Stochastic Analysis in Finance代考|American Derivative

We consider the same situation as the previous section, i.e., $(\Omega, \mathcal{F}, P)$ is a complete probability space, $B=\left(B^1, \ldots, B^d\right)$ is a $d$-dimensional Brownian motion, $\mathcal{F}t$ $=\sigma\left{\sigma{B(s) ; s \leqq t} \cup \mathcal{N}_0\right}, t \geqq 0, S^i, i=0,1, \ldots, N$, are $\left{\mathcal{F}_t\right}{t \in[0, \infty)}$-continuous semi-martingales, and Conditions (S-1) and (S-2) are satisfied. We also assume that Condition (S-3) is satisfied. Then we see that $\Gamma$ has a unique element $\rho$.

Let $Z:[0, \infty) \times \Omega \rightarrow \mathbf{R}$ be a continuous progressively measurable process and $T>0$. We assume that the following contract is traded in a market.
(A) If the holder exercises his right at a time $t \in[0, T]$, then he will receive $Z(t)$ (yen) from the writer.
This type of contract is called an American derivative.
Let $Y(t)=Z(t) \vee 0$. When $Z(t)$ is negative, the holder will not exercise the right at time $t$, and he may give up the right. So we consider this contract (A) to be equivalent to the following contract (A’).
(A’) If the holder exercises his right at a time $t \in[0, T]$, then he will receive $Y(t)$ (yen) from the writer.
We consider the price of this contract.
Let $Q_T$ be a probability measure on $(\Omega, \mathcal{F})$ given by
$$Q_T(A)=E[\rho(T), A], \quad A \in \mathcal{F} .$$
Also, let $X$ be a continuous progressively measurable process given by $X(t)=$ $S^0(t \wedge T)^{-1} Y(t \wedge T), t \in[0, \infty)$. Note that $X(t) \geqq 0$. Now we make the following assumption (SA).
(SA) There is a $C_0 \in(0, \infty)$ such that $X(t) \leqq C_0, t \in[0, T]$, with probability 1 . Then we see that
$$E^{Q_T}\left[\sup _{t \in[0, T]} X(t)^2\right]<\infty .$$

## 数学代写|随机分析代写STOCHASTIC ANALYSIS IN FINANCE 代考|GENERAL CASE

$S-1 P\left(\min t \in[0, T] S^0(t)>0\right)=1$ 对于任何 $T>0$. 然后由推论 $4.3 .1$ 我们看到 $\tilde{S}^i(t)=S^0(t)^{-1} S^i(t), t \geqq 0, i=1, \ldots, N$, 是连续的半鞅。此外，我们假设 如下。 $S-2$ 有逐步可衡量的过程 $\sigma_j^i, b^i, i=1, \ldots, N, j=1,2, \ldots, d$, 这样
$$P\left(\int_0^T \sigma_j^i(t)^2 d t<\infty\right)=1, \quad T>0, i=1, \ldots, N, \quad P\left(\int_0^T\left|b^i(t)\right| d t<\infty\right)=1, \quad T>0, i=1, \ldots, N$$

$$\bar{S}^i(t)=\sum j=1^d \int_0^t \sigma_j^i(s) d B^j(s)+\int_0^t b^i(s) d s, \quad t \geqq 0, i=1, \ldots, N$$

## 数学代写|随机分析代写STOCHASTIC ANALYSIS IN FINANCE 代考|AMERICAN DERIVATIVE

$A$ 如果持有人一次行使其权利 $t \in[0, T]$ ，那么他将收到 $Z(t)$ yen 来自作家。

$A^{\prime}$ 如果持有人一次行使其权利 $t \in[0, T]$ ，那么他将收到 $Y(t)$ yen 来目作家。

$$Q_T(A)=E[\rho(T), A], \quad A \in \mathcal{F} .$$

$S A$ 有一个 $C_0 \in(0, \infty)$ 这样 $X(t) \leqq C_0, t \in[0, T]$ ，概率为 1 。然后我们看到
$$E^{Q_T}\left[\sup _{t \in[0, T]} X(t)^2\right]<\infty .$$

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