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# 金融代写|随机微积分代写STOCHASTIC CALCULUS代考|MATH581 Locally Bounded Processes

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## 金融代写|随机微积分代写STOCHASTIC CALCULUS代考|Locally Bounded Processes

We will introduce an important class of integrands, namely that of locally bounded predictable processes, that is contained in $\mathbb{L}(X)$ for every stochastic integrator $X$. For a stopping time $\tau,[0, \tau]$ will denote the set ${(t, \omega) \in \widetilde{\Omega}: 0 \leq t \leq \tau(\omega)}$ and thus $g=f 1_{[0, \tau]}$ means the following: $g_{t}(\omega)=f_{t}(\omega)$ if $t \leq \tau(\omega)$ and $g_{t}(\omega)$ is zero if $t>\tau(\omega)$.

The next result gives interplay between stopping times and stochastic integration.
Lemma 4.36 Let $X$ be a stochastic integrator and $f \in \mathbb{L}(X)$. Let $\tau$ be a stopping time. Let $g=f 1_{[0, \tau]}$. Let
$$Y_{t}=\int_{0}^{t} f d X$$
and $V=\int g d X .$ Then $V_{t}=Y_{t \wedge \tau}$, i.e.
$$Y_{t \wedge \tau}=\int_{0}^{t} f 1_{[0, \tau]} d X .$$
Proof When $f \in \mathbb{S}$ is a simple predictable process and $\tau$ is a stopping time taking only finitely many values, then $g \in \mathbb{S}$ and (4.4.1)-(4.4.2) can be checked as in that case, the integrals $\int f d X$ and $\int g d X$ are both defined directly by (4.2.2). Thus fix $f \in \mathbb{S}$. Approximating a bounded stopping time $\tau$ from above by stopping time taking finitely many values (as seen in the proof of Theorem 2.54), it follows that (4.4.2) is true for any bounded stopping time, then any stopping time $\tau$ can be approximated by $\tilde{\tau}^{n}=\tau \wedge n$ and one can check that (4.4.2) continues to be true. Thus we have proven the result for simple integrands.
Now fix a stopping time $\tau$ and let
$$\mathbb{K}={f \in \mathbb{B}(\widetilde{\Omega}, \mathcal{P}):(4.4 .1)-(4.4 .2) \text { is true for all } t \geq 0 .} \text {. }$$

## 金融代写|随机微积分代写STOCHASTIC CALCULUS代考|Approximation by Riemann Sums

The next result shows that for an r.c.l.l. process $Y$ and a stochastic integrator $X$, the stochastic integral $\int Y^{-} d X$ can be approximated by Riemann-like sums. The difference is that the integrand must be evaluated at the lower end point of the interval as opposed to any point in the interval in the Riemann-Stieltjes integral.
Theorem 4.55 Let $Y$ be an r.c.l.l. adapted process and $X$ be a stochastic integrator. Let
$$0=t_{0}^{m}<t_{1}^{m}<\ldots<t_{n}^{m}<\ldots ; \quad t_{n}^{m} \uparrow \infty \text { as } n \uparrow \infty$$
be a sequence of partitions of $[0, \infty)$ such that for all $T<\infty$,
$$\delta_{m}(T)=\left(\sup {\left{n::{n}^{m} \leq T\right}}\left(t_{n+1}^{m}-t_{n}^{m}\right)\right) \rightarrow 0 \quad \text { as } m \uparrow \infty$$
Let
$$Z_{t}^{m}=\sum_{n=0}^{\infty} Y_{t_{n}^{m} \wedge t}\left(X_{t_{n+1}^{m} \wedge t}-X_{t_{n}^{m} \wedge t}\right)$$
and $Z=\int Y^{-} d X$. Note that for each $t, m$, the sum in (4.5.3) is a finite sum since $t_{n}^{m} \wedge t=t$ from some n onwards. Then
$$Z^{m} \stackrel{u c p}{\longrightarrow} Z$$
or in other words
$$\sum_{n=0}^{\infty} Y_{t_{n}^{m} \wedge t}\left(X_{t_{n+1}^{\mathrm{m}} \wedge t}-X_{t_{n}^{\mathrm{m}} \wedge t}\right) \stackrel{u c p}{\longrightarrow} \int_{0}^{t} Y^{-} d X$$
Proof Let $Y^{m}$ be defined by
$$Y_{t}^{m}=\sum_{n=0}^{\infty} Y_{t_{n}^{m} \wedge t} 1_{\left(t_{n}^{m}, t_{n+1}^{m}\right]}(t)$$
We will first prove
$$\int Y^{m} d X=Z^{m}$$

## 金融代写|随机微积分代写STOCHASTIC CALCULUS代考|Locally Bounded Processes

$$Y_{t}=\int_{0}^{t} f d X$$

$$Y_{t \wedge \tau}=\int_{0}^{t} f 1_{[0, \tau]} d X .$$

$$\mathbb{K}=f \in \mathbb{B}(\widetilde{\Omega}, \mathcal{P}):(4.4 .1)-(4.4 .2) \text { is true for all } t \geq 0 .$$

## 金融代写|随机微积分代写STOCHASTIC CALCULUS代考|Approximation by Riemann Sums

$$0=t_{0}^{m}<t_{1}^{m}<\ldots<t_{n}^{m}<\ldots ; \quad t_{n}^{m} \uparrow \infty \text { as } n \uparrow \infty$$

\left 的分隔符缺失或无法识别

$$Z_{t}^{m}=\sum_{n=0}^{\infty} Y_{t_{n} \cap t}\left(X_{t_{m 11}^{m} \wedge t}-X_{t_{m} \cap t}\right)$$

$$Z^{m} \stackrel{u c p}{\longrightarrow} Z$$

$$\sum_{n=0}^{\infty} Y_{t_{n}^{m} \wedge t}\left(X_{t_{n+1}^{\mathrm{m}} \wedge t}-X_{t_{\Pi}^{\mathrm{m}} \wedge t}\right) \stackrel{u c p}{\longrightarrow} \int_{0}^{t} Y^{-} d X$$

$$Y_{t}^{m}=\sum_{n=0}^{\infty} Y_{t_{m} \wedge t} 1_{\left(t_{m}, t_{n+1}^{m}\right]}(t)$$

$$\int Y^{m} d X=Z^{m}$$

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