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# 数学代写|随机过程Stochastic Porcesses代考|Math466+564 Diffusion Processes

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## 数学代写|随机过程Stochastic Porcesses代考|Diffusion Processes

Let us consider a continuous time and continuous state space stochastic process ${X(t), t \in T}$ with $T=[0, \infty), S=(-\infty,+\infty)$.

More specifically let ${X(t), t \in T}$ be a Markov process on a probability space $(\Omega, \mathscr{D}, P)$ assuming values in a set $S$. Denote by $\mathscr{L}$ a $\sigma$-algebra of subsets of $S$. The measure space $(S, \mathscr{L})$ will be called the “state space” of the stochastic process $X(t)$. The process $X(t)$ is said to be a homogeneous Markov process if its transition probability has the property $P(X(t+s) \in B \mid X(s)=x)=P(x, t, B)$, $B \in \mathscr{L}$. One may imagine a continuous parameter Markov process ${X(t), t \in T}$ that is not a process with independent increments. Suppose that given $X(s)=x$, for small times $t$, the displacement $X(s+t)-X(s)=X(s+t)-x$ has mean and variance approximately $t \mu(x)$ and $t \sigma^2(x)$, respectively. Here $\mu(x)$ and $\sigma^2(x)$ are functions of the state of $x$, and not constants as in the case of Brownian motion $W(t)$. The distinction between ${W(t)}$ and ${X(t)}$ is analogous to that between simple random walk and a birth-death chain. More precisely, suppose
$$\left.\begin{array}{l} E(X(s+t)-X(s) \mid X(s)=x)=t \mu(x)+o(t) \ E\left(\left(X(s+t)-X(s)^2 \mid X(s)=x\right)=t \sigma^2(x)+o(t)\right. \ E\left(|X(s+t)-X(s)|^3 \mid X(s)=x\right)=o(t) \end{array}\right}$$
hold, as $t \downarrow 0$ for every $x \in \mathscr{L}$.
Note that (8.1) holds for Brownian motion.

## 数学代写|随机过程Stochastic Porcesses代考|Kolmogorov Backward and Forward Diffusion Equations

There are various methods for determining transition probability function or transition distribution of a Markov process, ranging from purely analytical to purely probabilistic. The method presented here was developed by Kolmogorov in 1931.

Theorem 8.1 Let ${X(t), t \in T}$ be a diffusion process. Assuming that $\frac{\partial}{\partial y} F(y, s ; x, t)$ and $\frac{\partial^2}{\partial y^2} F(y, s ; x, t)$ exist and are jointly continuous in all the variables, we get
$$\frac{\partial}{\partial s} F(y, s ; x, t)=-a(y, s) \frac{\partial}{\partial y} F(y, s ; x, t)-\frac{1}{2} b(y, s) \frac{\partial^2}{\partial y^2} F(y, s ; x, t),$$
the Generalized Heat equation or the Backward Kolomogorov’s Differential equation.

Assume that $f(y, s ; x, t)=\frac{\partial}{\partial x} F(y, s ; x, t)$ exists and that $f(x, t)=f(y, s ; x, t)$. Also assume that $\frac{\partial f}{\partial t}, \frac{\partial}{\partial x}[a(x, t) f(x, t)]$ and $\frac{\partial^2}{\partial x^2}[b(x, t) f(x, t)]$ exist and are continuous in all the variables. Then we get
$$\frac{\partial f(x, t)}{\partial t}=-\frac{\partial}{\partial x}[a(x, t) f(x, t)]+\frac{1}{2} \frac{\partial^2}{\partial x^2}[b(x, t) f(x, t)],$$
the Kolmogorov’s forward differential equation or the Fokker-Planck equation. Proof of Backward Equation
\begin{aligned} F(y, s-\Delta s ; x, t)-F(y, s ; x, t)= & \int_{-\infty}^{\infty} F(z, s ; x, t) d F_z(y, s-\Delta s ; z, s) \ & -F(y, s ; x, t) \int_{-\infty}^{\infty} d F_z(y, s-\Delta s ; z, s) \end{aligned}
Let $\quad I=\int_{-\infty}^{\infty}[F(z, s ; x, t)-F(y, s ; x, t)] d F_z(y, s-\Delta s ; z, s)$
By assumptions for $|z-y| \leq \delta$ and for some $\delta>0$,
$$\begin{gathered} F(z, s ; x, t)=F(y, s ; x, t)+(z-y) \frac{\partial F(y, s ; x, t)}{\partial y} \ +\frac{1}{2}(z-y)^2 \frac{\partial^2 F(y, s ; x, t)}{\partial y^2}+o\left((z-y)^2\right) \end{gathered}$$

## 数学代写|随机过程Stochastic Porcesses代考|Diffusion Processes

}right 缺少或无法识别的分隔符

## 数学代写|随机过程Stochastic Porcesses代考|Kolmogorov Backward and Forward Diffusion Equations

$$\frac{\partial}{\partial s} F(y, s ; x, t)=-a(y, s) \frac{\partial}{\partial y} F(y, s ; x, t)-\frac{1}{2} b(y, s) \frac{\partial^2}{\partial y^2} F(y, s ; x, t)$$

$$\frac{\partial f(x, t)}{\partial t}=-\frac{\partial}{\partial x}[a(x, t) f(x, t)]+\frac{1}{2} \frac{\partial^2}{\partial x^2}[b(x, t) f(x, t)]$$
Kolmogorov 的正微分方程或 Fokker-Planck 方程。倒向方程的证明
$$F(y, s-\Delta s ; x, t)-F(y, s ; x, t)=\int_{-\infty}^{\infty} F(z, s ; x, t) d F_z(y, s-\Delta s ; z, s) \quad-F(y, s ; x, t) \int_{-\infty}^{\infty} d F_z(y, s-\Delta s ; z, s)$$

$$F(z, s ; x, t)=F(y, s ; x, t)+(z-y) \frac{\partial F(y, s ; x, t)}{\partial y}+\frac{1}{2}(z-y)^2 \frac{\partial^2 F(y, s ; x, t)}{\partial y^2}+o\left((z-y)^2\right)$$

## MATLAB代写

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