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# 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|STAT433 Preliminaries

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

Throughout this chapter, we will be working with one fixed filtration $(\mathcal{F}$.) such that $\mathcal{F}_0$ contains all null sets. We do not assume that the filtration is right continuous. All notions-martingale, local martingale, stopping time, adapted process, predictable process-are with reference to this fixed filtration. Since in this chapter, we need to deal with martingales which may not have r.c.l.l. paths a priori, we will explicitly assume r.c.l.l. paths when it is needed.

For r.c.l.l.semimartingales $X^1, X^2, \ldots, X^d$, we introduce the class of semimartingales that admit integral representation w.r.t. $X^1, X^2, \ldots, X^d$ :
\begin{aligned} & \mathbb{I}\left(X^1, X^2, \ldots, X^d\right) \ & =\left{Y: \exists g^j \in \mathbb{L}\left(X^j\right), 1 \leq j \leq d \text { with } Y_t=Y_0+\sum_{j=1}^d \int_0^t g^j d X^j \forall t\right} \end{aligned}
Let us note that if $Y \in \mathbb{I}\left(X^1, X^2, \ldots, X^d\right)$ then for any stopping time $\tau, \tilde{Y}$ defined by $\tilde{Y}t=Y{t \wedge \tau}$ also belongs to $\mathbb{I}\left(X^1, X^2, \ldots, X^d\right)$. Also, if $Y \in \mathbb{I}\left(X^1, X^2, \ldots, X^d\right)$ then we can always choose $g^j \in \mathbb{L}\left(X^j\right)$ with $g_0^j=0$ for $1 \leq j \leq d$ such that
$$Y_t=Y_0+\sum_{j=1}^d \int_0^t g^j d X^j \forall t<\infty .$$
If $Y \in \mathbb{I}\left(X^1, X^2, \ldots, X^d\right)$, the semimartingale $Y$ is said to have an integral representation w.r.t. semimartingales $X^1, X^2, \ldots, X^d$. Here is an elementary observation on the class $\mathbb{I}\left(X^1, X^2, \ldots, X^d\right)$.

Lemma 10.1 Let $Y$ be a semimartingale such that for a sequence of stopping times $\tau_n \uparrow \infty, Y^n$ defined by $Y_t^n=Y_{t \wedge \tau_n}$ admits an integral representation w.r.t. r.c.l.l. semimartingales $X^1, X^2, \ldots, X^d$ for each $n \geq 1$. Then $Y$ also admits an integral representation w.r.t. $X^1, X^2, \ldots, X^d$.
Proof Let $f^{n, j} \in \mathbb{L}\left(X^j\right), 1 \leq j \leq d, n \geq 1$ be such that for all $n$,
$$Y_t^n=Y_0^n+\sum_{j=1}^d \int_0^t f^{n, j} d X^j$$
Define $f^j$ by
$$f^j=\sum_{n=1}^{\infty} 1_{\left(\tau_{n-1}, \tau_n\right]} f^{n, j}$$
Then it is easy to check (using Theorem 4.43) that $f^j \in \mathbb{L}\left(X^j\right)$ and
$$Y_t=Y_0+\sum_{i=1}^d \int_0^t f^j d X^j$$

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|One-Dimensional Case

In this section, we will fix a local martingale $M$ and explore as to when $\mathbb{I}(M)$ contains all martingales. The next lemma gives an important property of $\mathbb{I}(M)$.

Lemma 10.3 Let $M$ be an r.c.l.l. local martingale and let $N^n \in \mathbb{I}(M)$ be martingales such that $\mathrm{E}\left[\left|N_t^n-N_t\right|\right] \rightarrow 0$ for all $t$. Then $N \in \mathbb{I}(M)$.

Proof The assumptions imply that $N$ is a martingale (see Theorem 2.23). In view of Theorem $5.39$ and the assumptions on $N^n, N$, it follows that
$N^n$ converges to $N$ in Emery topology.
Thus, invoking Theorem $4.111$, we conclude that
$$\left[N^n-N, N^n-N\right]_T \rightarrow 0 \text { in probability as } n \rightarrow \infty$$
Hence, using $\left[N^n-N^m, N^n-N^m\right]_T \leq 2\left(\left[N^n-N, N^n-N\right]_T+\left[N^m-N, N^m-\right.\right.$ $N]_T$ ) (see $\left.(4.6 .13)\right)$, we have
$$\left[N^n-N^m, N^n-N^m\right]_T \rightarrow 0 \text { in probability as } n, m \rightarrow \infty$$
Since $N^n \in \mathbb{I}(M)$, there exists predictable process $g^n \in \mathbb{L}(M)$ such that

$$N_t^n=N_0^n+\int_0^t g^n d M .$$
As a consequence, for all $T<\infty$,
\begin{aligned} {\left[N^n-N^m, N^n-N^m\right]T } & =\int_0^T\left(g_s^n-g_s^m\right)^2 d[M, M]_s . \ & \rightarrow 0 \text { in probability as } n, m \rightarrow \infty . \end{aligned} By taking a subsequence, if necessary and relabelling, we assume that for $1 \leq k \leq n$, $$\mathrm{P}\left(\left(\int_0^k\left(g_s^n-g_s^k\right)^2 d[M, M]_s\right)^{\frac{1}{2}} \geq \frac{1}{2^k}\right) \leq \frac{1}{2^k} .$$ Then by Borel-Cantelli Lemma, we conclude $$\sum{k=1}^{\infty}\left(\int_0^T\left(g_s^{k+1}-g_s^k\right)^2 d[M, M]_s\right)^{\frac{1}{2}}<\infty \text { a.s. }$$

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

$\mathbb{I}\left(X^1, X^2, \ldots, X^d\right)$.

$$Y_t^n=Y_0^n+\sum_{j=1}^d \int_0^t f^{n, j} d X^j$$

$$f^j=\sum_{n=1}^{\infty} 1_{\left(\tau_{n-1}, \tau_n\right]} f^{n, j}$$

$$Y_t=Y_0+\sum_{i=1}^d \int_0^t f^j d X^j$$

## 数学代写|随机微积分代写STOCHASTIC CALCULUS代考|One-Dimensional Case

$$\left[N^n-N, N^n-N\right]_T \rightarrow 0 \text { in probability as } n \rightarrow \infty$$

$$\left[N^n-N^m, N^n-N^m\right]_T \rightarrow 0 \text { in probability as } n, m \rightarrow \infty$$

$$N_t^n=N_0^n+\int_0^t g^n d M .$$

$$\left[N^n-N^m, N^n-N^m\right] T=\int_0^T\left(g_s^n-g_s^m\right)^2 d[M, M]_s . \quad \rightarrow 0 \text { in probability as } n, m \rightarrow \infty .$$

$$\mathrm{P}\left(\left(\int_0^k\left(g_s^n-g_s^k\right)^2 d[M, M]_s\right)^{\frac{1}{2}} \geq \frac{1}{2^k}\right) \leq \frac{1}{2^k} .$$

$$\sum k=1^{\infty}\left(\int_0^T\left(g_s^{k+1}-g_s^k\right)^2 d[M, M]_s\right)^{\frac{1}{2}}<\infty \text { a.s. }$$

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