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# 金融代写|随机微积分代写STOCHASTIC CALCULUS代考|MA547001 Martingales and Stopping Times

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## 金融代写|随机微积分代写STOCHASTIC CALCULUS代考|Martingales and Stopping Times

Let $M$ be a process defined on a probability space $(\Omega, \mathcal{F}, \mathrm{P})$ and $(\mathcal{F} .)$ be a filtration.
Definition 2.19 $M$ is said to be $\left(\mathcal{F}{.}\right)$-martingale if $M$ is $(\mathcal{F} .)$ adapted, $M{t}$ is integrable for all $t$ and for $0 \leq s<t$ one has
$$\mathrm{E}{\mathrm{P}}\left[M{t} \mid \mathcal{F}{s}\right]=M{s} .$$
Definition $2.20 M$ is said to be $\left(\mathcal{F}{\text {. }}\right)$-submartingale if $M$ is $\left(\mathcal{F}\right.$. ) adapted, $M{t}$ is integrable for all $t$ and for $0 \leq s<t$ one has
$$\mathrm{E}{\mathrm{P}}\left[M{t} \mid \mathcal{F}{s}\right] \geq M{s} .$$
Remark 2.21 Likewise $M$ is said to be a supermartingale if $N$ defined by $N_{t}=$ $-M_{t}$ is a submartingale.

When there is only one filtration under consideration, we will drop reference to it and call $M$ to be a martingale (or a submartingale). If $M$ is a martingale and $\phi$ is a convex function on $\mathbb{R}$, then Jensen’s inequality implies that $\phi(M)$ is a submartingale provided each $\phi\left(M_{t}\right)$ is integrable. In particular, if $M$ is a martingale with $\mathrm{E}\left[M_{t}^{2}\right]<\infty$ for all $t$ then $M^{2}$ is a submartingale. We are going to be dealing with martingales that have r.c.l.l. paths. The next result shows that under minimal conditions on the underlying filtration one can assume that every martingale has r.c.l.l. paths. Theorem 2.22 Suppose that the filtration $\left(\mathcal{F}{.}\right.$) satisfies $$\begin{gathered} N{0} \subseteq N, N \in \mathcal{F}, \mathrm{P}(N)=0 \Rightarrow N_{0} \in \mathcal{F}{0} \ \cap{t>s} \mathcal{F}{t}=\mathcal{F}{s} \quad \forall s \geq 0 \end{gathered}$$
Then every martingale $M$ admits an r.c.l.l.version $\tilde{M}$, i.e. there exists an r.c.l.l.process $\tilde{M}$ such that
$$\mathrm{P}\left(M_{t}=\tilde{M}_{t}\right)=1 \quad \forall t \geq 0 .$$

## 金融代写|随机微积分代写STOCHASTIC CALCULUS代考|A Version of Monotone Class Theorem

The following functional version of the usual monotone class theorem is very useful in dealing with integrals and their extension. Its proof is on the lines of the standard version of monotone class theorem for sets. We will include a proof; because of the central role, this result will play in our development of stochastic integrals.

Definition 2.62 A subset $\mathbb{A} \subseteq \mathbb{B}(\Omega, \mathcal{F})$ is said to be closed under uniform and monotone convergence if $f, g, h \in \mathbb{B}(\Omega, \mathcal{F}), f^{n}, g^{n}, h^{n} \in \mathbb{A}$ for $n \geq 1$ are such that
(i) $f^{n} \leq f^{n+1}$ for all $n \geq 1$ and $f^{n}$ converges to $f$ pointwise
(ii) $g^{n} \geq g^{n+1}$ for all $n \geq 1$ and $g^{n}$ converges to $g$ pointwise
(iii) $h^{n}$ converges to $h$ uniformly
then $f, g, h \in \mathbb{A}$.
Here is a functional version of the monotone class theorem:
Theorem 2.63 Let $\mathbb{A} \subseteq \mathbb{B}(\Omega, \mathcal{F})$ be closed under uniform and monotone convergence. Suppose $\mathbb{G} \subseteq \mathbb{\mathbb { }}$ is an algebra such that
(i) $\sigma(\mathbb{G})=\mathcal{F}$
(ii) $\exists f^{n} \in \mathbb{G}$ such that $f^{n} \leq f^{n+1}$ and $f^{n}$ converges to 1 pointwise.
Then $\mathbb{A}=\mathbb{B}(\Omega, \mathcal{F})$.
Proof Let $\mathbb{K} \subseteq \mathbb{B}(\Omega, \mathcal{F})$ be the smallest class that contains $\mathbb{G}$ and is closed under uniform and monotone convergence. Clearly, $\mathbb{K}$ contains constants and $\mathbb{K} \subseteq \mathbb{A}$. Using arguments similar to the usual (version for sets) monotone class theorem we will first prove that $\mathbb{K}$ itself is an algebra. First we show that $\mathbb{K}$ is a vector space. For $f \in \mathbb{B}(\Omega, \mathcal{F})$, let
$$\mathbb{K}_{0}(f)={g \in \mathbb{K}: \alpha f+\beta g \in \mathbb{K}, \quad \forall \alpha, \beta \in \mathbb{R}}$$

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