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# 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|MATH581 Integral with respect to QV

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## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Integral with respect to QV

Integral with respect to QV. Let $a$ be an adpated process satisfying $\int_0^t\left|a_s\right|^2 \mathrm{~d} s<$ $\infty, t \geq 0$ almost surely. Let $\mathcal{P}=\left{0=t_0<\cdots<t_n=t\right}$ be our familiar partition of $[0, t]$, for $t \geq 0$.

The fact that $[W]t=t, t \geq 0$, written heuristically in the form (2.27), suggests that the integral of $a$ with respect to time (which, following (2.27), we also write suggestively below as an integral with respect to QV – this is of course well-defined, as QV induces a finite measure, Lebesgue measure, over any finite time interval), $$J_t:=\int_0^t a_s \mathrm{~d} s=: \int_0^t a_s \mathrm{~d}[W]_s, \quad t \geq 0,$$ should coincide, as the mesh of the partition decreases to zero, with the discrete sum $$\mathcal{J}_t:=\sum{k=0}^{n-1} a_{t_k}\left(W_{t_{k+1}}-W_{t_k}\right)^2,$$
of the squared increments of BM, weighted with values of $a$ at the start of each partition interval.

This is true, and we prove this result below. It will come into play when proving the Itô formula in Section 4.

Lemma $2.37$ (Integral with respect to QV). The integral in (2.28) of the adapted process a with respect to $Q V$ and the discrete sum of weighted squared Brownian increments in (2.29), coincide as the mesh of the partition vanishes:
$$\lim {|\mathcal{P}| \rightarrow 0} \mathcal{J}_t:=\lim {|\mathcal{P}| \rightarrow 0} \sum_{k=0}^{n-1} a_{t_k}\left(W_{t_{k+1}}-W_{t_k}\right)^2=J_t:=\int_0^t a_s \mathrm{~d} s=: \int_0^t a_s \mathrm{~d}[W]_s, \quad \text { a.s. }, \quad t \geq 0 .$$

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Total variation and path length of BM

Total variation and path length of BM. Given a continuous function $f:[0, t] \rightarrow \mathbb{R}$ its total variation over $[0, t]$, over any partition $\mathcal{P}=\left{0=t_0 \leq t_1 \leq \ldots \leq t_n=t\right}$ of $[0, t]$, is
$$\operatorname{TV}(f)t \equiv[f]_t^{(1)}:=\lim {|\mathcal{P}| \rightarrow 0} \sum_{k=0}^{n-1}\left|f\left(t_{k+1}\right)-f\left(t_k\right)\right|, \quad t \geq 0 .$$
This may be infinite, or some finite number, in which case we say that $f$ has bounded variation. Consider an element of arc length $\Delta s_k$ along $f(\cdot)$ in the interval $\left[t_k, t_{k+1}\right]$. If this interval is small, we have $\left(\Delta s_k\right)^2 \approx\left(\Delta t_k\right)^2+\left(\Delta f_k\right)^2$, where we have written $\Delta t_k=t_{k+1}-t_k$ and $\Delta f_k=f\left(t_{k+1}\right)-f\left(t_k\right)$. By the triangle inequality we have
$$\left|\Delta f_k\right| \leq\left|\Delta s_k\right| \leq\left|\Delta f_k\right|+\left|\Delta t_k\right| .$$

Denoting the total arc length (or path length) of $f$ over $[0, t]$ by $s(f)t$ we therefore have, in the limit $|\mathcal{P}| \rightarrow 0$, Therefore, $$\operatorname{TV}(f)_t \leq s(f)_t \leq \operatorname{TV}(f)_t+t, \quad t \geq 0 . .$$ finite path length $\Longleftrightarrow \operatorname{TV}(f)<\infty$. If, on the other hand, we examine the quadratic variation of $f(\cdot)$ over $[0, t]$, we have \begin{aligned} {[f]_t } & =\lim {|\mathcal{P}| \rightarrow 0} \sum_{k=0}^{n-1}\left|\Delta f_k | \Delta f_k\right| \ & \leq \lim {|\mathcal{P}| \rightarrow 0}\left(\max {j=0, \ldots, n-1}\left|\Delta f_j\right|\right) \lim {|\mathcal{P}| \rightarrow 0} \sum{k=0}^{n-1}\left|\Delta f_k\right| \ & =\lim {|\mathcal{P}| \rightarrow 0}\left(\max {j=0, \ldots, n-1}\left|\Delta f_j\right|\right) \mathrm{TV}(f) . \end{aligned}

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Integral with respect to $\mathrm{Q}$

$$J_t:=\int_0^t a_s \mathrm{~d} s=: \int_0^t a_s \mathrm{~d}[W]s, \quad t \geq 0,$$ 应该重合，因为分区的网格减少到零，与离散总和 $$\mathcal{J}_t:=\sum k=0^{n-1} a{t_k}\left(W_{t_{k+1}}-W_{t_k}\right)^2,$$
$\mathrm{BM}$ 的平方增量，加权值 $a$ 在每个分区间隔的开始。

$$\lim |\mathcal{P}| \rightarrow 0 \mathcal{J}t:=\lim |\mathcal{P}| \rightarrow 0 \sum{k=0}^{n-1} a_{t_k}\left(W_{t_{k+1}}-W_{t_k}\right)^2=J_t:=\int_0^t a_s \mathrm{~d} s=: \int_0^t a_s \mathrm{~d}[W]s, \quad \text { a.s. }, \quad t \geq 0$$ path length of BM $B M$ 的总变化和路径长度。给定一个连紏函数 $f:[0, t] \rightarrow \mathbb{R}$ 它的总变化超过 $[0, t]$, 在任何分区 \left 缺少或无法识别的分隔符 $\quad$ 的 $[0, t]$ ，是 $$\operatorname{TV}(f) t \equiv[f]_t^{(1)}:=\lim |\mathcal{P}| \rightarrow 0 \sum{k=0}^{n-1}\left|f\left(t_{k+1}\right)-f\left(t_k\right)\right|, \quad t \geq 0$$

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|Total variation and path length of BM

$$\left|\Delta f_k\right| \leq\left|\Delta s_k\right| \leq\left|\Delta f_k\right|+\left|\Delta t_k\right| .$$

$$\operatorname{TV}(f)t \leq s(f)_t \leq \operatorname{TV}(f)_t+t, \quad t \geq 0 .$$ 有限路径长度 $\Longleftrightarrow \mathrm{TV}(f)<\infty$. 另一方面，如果我们检萛的二次方差 $f(\cdot)$ 超过 $[0, t]$, 我们有 $$[f]_t=\lim |\mathcal{P}| \rightarrow 0 \sum{k=0}^{n-1}\left|\Delta f_k\right| \Delta f_k|\quad \leq \lim | \mathcal{P}\left|\rightarrow 0\left(\max j=0, \ldots, n-1\left|\Delta f_j\right|\right) \lim \right| \mathcal{P}\left|\rightarrow 0 \sum k=0^{n-1}\right| \Delta f_k|=\lim | \mathcal{P} \mid \rightarrow 0(\max j=0, \ldots, n-1$$

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