# 数学代写|随机微积分代写Stochastic calculus代考|MATH530A

## 数学代写|随机微积分代写Stochastic calculus代考|Tanaka’s formula

The preceding application of (3.3.1) is mundane by comparison to the one made by $\mathrm{H}$. Tanaka to represent what Lévy called local time for an $\mathbb{R}$-valued Brownian motion $\left(B(t), \mathcal{F}t, \mathbb{P}\right)$. To describe local time, define the occupation time measures $L(t, \cdot)$ for $t \geq 0$ by $$L(t, \Gamma)=\int_0^t \mathbf{1}{\Gamma}(B(\tau)) d \tau, \quad \Gamma \in \mathcal{B}{\mathbb{R}} .$$ One of Lévy’s many remarkable discoveries is that there is a map $\omega \in \Omega \longmapsto$ $\ell(\cdot, \cdot \cdot)(\omega) \in C([0, \infty) \times \mathbb{R} ;[0, \infty))$ such that $\ell(\cdot, y)$ a is progressively measurable, non-decreasing function with the property that $$\mathbb{P}\left(L(t, \Gamma)=\int{\Gamma} \ell(t, y) d y \text { for all } t \geq 0 \& \Gamma \in \mathcal{B}{\mathbb{R}}\right)=1$$ In other words, with probability $1, L(t, \cdot)$ is absolutely continuous with respect to Lebesgue measure $\lambda{\mathbb{R}}$ for all $t \geq 0$, and
$$\frac{L(t, d y)}{\lambda_{\mathbb{R}}(d y)}=\ell(t, y) .$$
Notice that this result is another manifestation of the non-differentiability of Brownian paths. Indeed, suppose that $p:[0, \infty) \longrightarrow \mathbb{R}$ is a continuously differentiable path, and let $t \rightsquigarrow \mu(t, \cdot)$ be its occupation time measures. If $\dot{p}=0$ on an interval $[a, b]$, then it is clear that, for $t>a, \mu(t,{p(a)}) \geq$ $(t \wedge b-a)$, and therefore $\mu_t$ can’t be absolutely continuous with respect to $\lambda_{\mathbb{R}}$. On the other hand, if $\dot{p}>0$ on $[a, b]$, then, for $t \in(a, b]$,
$$\frac{\mu(t, d y)-\mu(a, d y)}{\lambda_{\mathbb{R}}(d y)}=\frac{\mathbf{1}_{[a, t]}(y)}{\dot{p} \circ\left(p\lceil[p(a), p(t)])^{-1}(y)\right.},$$
and so $\mu(t, \cdot)-\mu(a, \cdot)$ is absolutely continuous but its Radon-Nikodym derivative cannot be continuous. It is only because a Brownian path dithers as it leaves points that its occupation time measure can admit a continuous density.

## 数学代写|随机微积分代写Stochastic calculus代考|Spacial continuity

Recall the Euler approximations $X_n(t, \mathbf{x})$ in (3.0.1). It is evident that, for all $w \in \mathbb{W}\left(\mathbb{R}^M\right),(t, \mathbf{x}) \rightarrow X_n(t, \mathbf{x})(w)$ is continuous. Now set $\Delta_n(t, \mathbf{x})=$ $X(t, \mathbf{x})-X_n(t, \mathbf{x})$. Using (3.3.2) and arguing as we did when $p=2$, one can show that for each $p \in[2, \infty)$ and $t>0$ there is a $C_p(t)<\infty$ such that
\begin{aligned} &\mathbb{E}^{\mathcal{W}}\left[\left|\Delta_n(\cdot, \mathbf{x})\right|_{[0, t]}^p\right]^{\frac{1}{p}} \leq C_p(t)(1+|\mathbf{x}|) 2^{\frac{n}{2}} \ &\mathbb{E}^{\mathcal{W}}\left[\left|X_n(\cdot, \mathbf{x})-X_n(s, \mathbf{x})\right|_{[s, t]}^p\right]^{\frac{1}{p}} \leq C_p(t)(1+|\mathbf{x}|)(t-s)^{\frac{1}{2}} \ &\mathbb{E}^{\mathcal{W}}\left[\left|X_n(\cdot, \mathbf{y})-X_n(\cdot, \mathbf{x})\right|_{[0, t]}^p\right]^{\frac{1}{p}} \leq C_p(t)|\mathbf{y}-\mathbf{x}| \end{aligned}
for $n \geq 0,0 \leq s<t$, and $\mathbf{x}, \mathbf{y} \in \mathbb{R}^N$. From these it was clear that, for each $\mathbf{x}$, $\left|\Delta_n(\cdot, \mathbf{x})\right|_{[0, t]} \longrightarrow 0$ both (a.s., $\mathcal{W}$ ) and in $L^p\left(\mathcal{W} ; \mathbb{R}^N\right.$ ), but we will now use Kolmogorov’s continuity criterion to show that this convergence is uniform with respect to $\mathbf{x}$ in compact subsets. To this end, note that, for $p \in[2, \infty)$,
\begin{aligned} &\left|\Delta_n(t, \mathbf{y})-\Delta_n(t, \mathbf{x})\right|^p \ &\quad=\left|(X(t, \mathbf{y})-X(t, \mathbf{x}))-\left(X_n(t, \mathbf{y})-X_n(t, \mathbf{x})\right)\right|^{\frac{p}{2}}\left|\Delta_n(t, \mathbf{y})-\Delta_n(t, \mathbf{x})\right|^{\frac{p}{2}}, \end{aligned} apply Schwarz’s inequality to get
\begin{aligned} &\mathbb{E}^{\mathcal{W}}\left[\left|\Delta_n(\cdot, \mathbf{y})-\Delta_n(\cdot, \mathbf{x})\right|_{[0, t]}^p\right] \ &\quad \leq \mathbb{E}^{\mathcal{W}}\left[\left|(X(\cdot, \mathbf{y})-X(\cdot, \mathbf{x}))-\left(X_n(\cdot, \mathbf{y})-X_n(\cdot, \mathbf{x})\right)\right|_{[0, t]}^p\right]^{\frac{1}{2}} \ &\quad \times \mathbb{E}^{\mathcal{W}}\left[\left|\Delta_n(\cdot, \mathbf{y})-\Delta_n(\cdot, \mathbf{x})\right|_{[0, t]}^p\right]^{\frac{1}{2}}, \end{aligned}
and, after combining these with the first and third estimates in (3.4.1), conclude that, for each $R>0$, there is a $K_p(t, R)<\infty$ such that $\mathbb{E}^{\mathcal{W}}\left[\left|\Delta_n(\cdot, \mathbf{y})-\Delta_n(\cdot, \mathbf{x})\right|_{[0, t]}^p\right]^{\frac{1}{p}} \leq K_p(t, R) 2^{-\frac{n}{4}}|\mathbf{y}-\mathbf{x}|^{\frac{1}{2}}$ for $\mathbf{x}, \mathbf{y} \in[-R, R]^N$. Hence, by taking $p>2 N$, we can apply Theorem $2.1 .2$ to see that
$$\sup {n \geq 0} 2^{\frac{n}{4}} \mathbb{E}^{\mathcal{W}}\left[\left|\Delta_n(\cdot, \cdots)\right|{[0, t] \times[-R, R]^N}^p\right]^{\frac{1}{p}}<\infty$$
and therefore that $(t, \mathbf{x}) \rightsquigarrow X(t, \mathbf{x})$ can be chosen so that it is continuous and, $\mathcal{W}$-almost surely, $\left|X_n(\cdot, \mathbf{x})-X(\cdot, \mathbf{x})\right|_{[0, t]} \longrightarrow 0$ uniformly for $\mathbf{x}$ in compact subsets of $\mathbb{R}^N$.

# 随机微积分代考

## 数学代写|随机微积分代写Stochastic calculus代考|Tanaka’s formula

(3.3.1) 的上述应用与由 $H$. Tanaka 代表 Lévy 所称的当地时间 $\mathbb{R}$ 值布朗运动 $(B(t), \mathcal{F} t, \mathbb{P})$. 为了描述当地时 间，定义占用时间度量 $L(t, \cdot)$ 为了 $t \geq 0$ 经过
$$L(t, \Gamma)=\int_0^t 1 \Gamma(B(\tau)) d \tau, \quad \Gamma \in \mathcal{B} \mathbb{R}$$

$$\mathbb{P}\left(L(t, \Gamma)=\int \Gamma \ell(t, y) d y \text { for all } t \geq 0 \& \Gamma \in \mathcal{B} \mathbb{R}\right)=1$$

$$\frac{L(t, d y)}{\lambda_{\mathbb{R}}(d y)}=\ell(t, y) .$$

$$\frac{\mu(t, d y)-\mu(a, d y)}{\lambda_{\mathbb{R}}(d y)}=\frac{\mathbf{1}_{[a, t]}(y)}{\dot{p} \circ\left(p\lceil[p(a), p(t)])^{-1}(y)\right.},$$

## 数学代写|随机微积分代写Stochastic calculus代考|Spacial continuity

$$\sup n \geq 02^{\frac{n}{4}} \mathbb{E}^{\mathcal{W}}\left[\left|\Delta_n(\cdot, \cdots)\right|[0, t] \times[-R, R]^{N^p}\right]^{\frac{1}{p}}<\infty$$

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