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# 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|MAST90059 Markovian diffusions

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## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Markovian diffusions

Markovian diffusions. Consider an Itô process $X:=\left(X_t\right)_{t \geq 0}$ satisfying
$$X_t=X_0+\int_0^t a_s \mathrm{~d} s+\int_0^t b_s \mathrm{~d} W_s, \quad t \geq 0$$

which we usually write in differential form
$$\mathrm{d} X_t=a_t \mathrm{~d} t+b_t \mathrm{~d} W_t,$$
for adapted $a, b$ almost surely satisfying $\int_0\left|a_s\right| \mathrm{d} s<\infty, \int_0^* b_s^2 \mathrm{~d} s<\infty$ (or $\mathbb{E}\left[\int_0^* b_s^2 \mathrm{~d} s\right]<\infty$ if we want the local martingale in (6.6) (or $(6.7)$ ) to be a martingale).
Suppose now that
$$a_t \equiv a\left(X_t\right), \quad b_t \equiv b\left(X_t\right), \quad t \geq 0$$
(or we could also have $a_t \equiv a\left(t, X_t\right), b_t \equiv b\left(t, X_t\right)$ ), for well-behaved functions $a(\cdot), b(\cdot)$ such that the process in (6.6) is well-defined. We call the resulting equation
$$\mathrm{d} X_t=a\left(X_t\right) \mathrm{d} t+b\left(X_t\right) \mathrm{d} W_t,$$
a stochastic differential equation (SDE) for $X$, with initial condition $X_0$. Then $X$ is a Markov process, because its drift and diffuson coefficients depend only on the current value of $X$, and not on its history prior to the current time. (Note that proving this intuitive result rigorously needs quite a bit of preparatory work on the theory of Markov processes, and we will not delve into this theory here.) The process $X$ in (6.8) is also called a (time-homogeneous) diffusion process in the case where the coefficient functions do not have explicit time dependence. (The term diffusion is often also used for the case when the coefficient functions do have explicit time dependence – in this case, the two-dimensional process $\left(t, X_t\right)_{t \geq 0}$ is the (Markov) process, but the “diffusion” $X$ is not time-homogeneous.)
We (of course) interpret the SDE (6.8) as the integral equation
$$X_t=X_0+\int_0^t a\left(X_s\right) \mathrm{d} s+\int_0^t b\left(X_s\right) \mathrm{d} W_s, \quad t \geq 0 .$$

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Feynman-Kac theorem

Feynman-Kac theorem. Because of the Markov property, we have, for any $h(\cdot)$ such that $h\left(X_T\right)$ is integrable, for some $T<\infty$, and with $t \in[0, T]$,
$$\mathbb{E}\left[h\left(X_T\right) \mid \mathcal{F}t\right]=\mathbb{E}\left[h\left(X_T\right) \mid X_t\right]=: v\left(t, X_t\right), \quad t \in[0, T] .$$ For $X_t=x \in \mathbb{R}(6.9)$, defines a function $v:[0, T] \times \mathbb{R} \mapsto \mathbb{R}$. Lemma 6.4. With $v(\cdot, \cdot)$ defined by $(6.9)$, assume that $Y_t:=v\left(t, X_t\right), t \in[0, T]$ is integrable for all $t \in[0, T]$ (so $\mathbb{E}\left[\left|Y_t\right|\right]<\infty, \forall t \in[0, T]$ ). Then, $Y=\left(Y_t\right){t \in[0, T]}=\left(v\left(t, X_t\right)\right)_{t \in[0, T]}$ is a martingale.

Proof. For $0 \leq s \leq t \leq T$, on exploiting the Markov property and the tower property, we have
\begin{aligned} \mathbb{E}\left[Y_t \mid \mathcal{F}s\right]=\mathbb{E}\left[v\left(t, X_t\right) \mid \mathcal{F}_s\right] & =\mathbb{E}\left[\mathbb{E}\left[h\left(X_T\right) \mid X_t\right] \mid \mathcal{F}_s\right] \ & =\mathbb{E}\left[\mathbb{E}\left[h\left(X_T\right) \mid \mathcal{F}_t\right] \mid \mathcal{F}_s\right] \ & =\mathbb{E}\left[h\left(X_T\right) \mid \mathcal{F}_s\right] \ & =\mathbb{E}\left[h\left(X_T\right) \mid X_s\right] \ & =v\left(s, X_s\right)=Y_s \end{aligned} This leads to a first statement of the Feynman-Kac theorem. Theorem 6.5 (Feynman-Kac I). With $$\mathrm{d} X_t=a\left(X_t\right) \mathrm{d} t+b\left(X_t\right) \mathrm{d} W_t$$ a continuous diffusion, define the function $v:[0, T] \times \mathbb{R} \mapsto \mathbb{R}$ by $$v(t, x):=\mathbb{E}\left[h\left(X_T\right) \mid X_t=x\right], t \in[0, T],$$ for some Borel function $h(\cdot)$. Assume that $v\left(t, X_t\right)$ is integrable for all $t \in[0, T]$, and that $v \in C^{1,2}([0, T] \times \mathbb{R})$. Then, $v(\cdot, \cdot)$ solves the $P D E$ $$\frac{\partial v}{\partial t}(t, x)+\mathcal{A} v(t, x), \quad v(T, x)=h(x),$$ where $$\mathcal{A} v(t, x):=a(x) v_x(t, x)+\frac{1}{2} b^2(x) v{x x}(t, x)$$
defines the generator of $X$.

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Markovian diffusions

$$X_t=X_0+\int_0^t a\left(X_s\right) \mathrm{d} s+\int_0^t b\left(X_s\right) \mathrm{d} W_s, \quad t \geq 0 .$$

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Feynman-Kac theorem

$$\mathbb{E}\left[h\left(X_T\right) \mid \mathcal{F} t\right]=\mathbb{E}\left[h\left(X_T\right) \mid X_t\right]=: v\left(t, X_t\right), \quad t \in[0, T] .$$

$$Y=\left(Y_t\right) t \in[0, T]=\left(v\left(t, X_t\right)\right)_{t \in[0, T]} \text { 是一个轫。 }$$

$$\mathbb{E}\left[Y_t \mid \mathcal{F}_s\right]=\mathbb{E}\left[v\left(t, X_t\right) \mid \mathcal{F}_s\right]=\mathbb{E}\left[\mathbb{E}\left[h\left(X_T\right) \mid X_t\right] \mid \mathcal{F}_s\right] \quad=\mathbb{E}\left[\mathbb{E}\left[h\left(X_T\right) \mid \mathcal{F}_t\right] \mid \mathcal{F}_s\right]=\mathbb{E}\left[h\left(X_T\right) \mid \mathcal{F}_s\right] \quad=\mathbb{E}\left[h\left(X_T\right) \mid X_s\right]=v\left(s, X_s\right)=Y_s$$

$$\mathrm{d} X_t=a\left(X_t\right) \mathrm{d} t+b\left(X_t\right) \mathrm{d} W_t$$

$$v(t, x):=\mathbb{E}\left[h\left(X_T\right) \mid X_t=x\right], t \in[0, T],$$

$$\frac{\partial v}{\partial t}(t, x)+\mathcal{A} v(t, x), \quad v(T, x)=h(x),$$

$$\mathcal{A} v(t, x):=a(x) v_x(t, x)+\frac{1}{2} b^2(x) v x x(t, x)$$

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