### 金融代写|金融数学代写Financial Mathematics代考|MATH3090

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 金融代写|金融数学代写Financial Mathematics代考|The Doob decomposition

It is possible to obtain a martingale starting from any process.
THEOREM $4.2$ (Doob decomposition theorem).-Let $X=\left(X_{n}\right){n \in \mathbb{N}}$ be a stochastic process that is adapted to the filtration $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ and integrable. It can then be uniquely decomposed in the form $$X{n}=X_{0}+M_{n}+A_{n}$$
with $M_{0}=A_{0}=0, M=\left(M_{n}\right){n \in \mathbb{N}}$ is a martingale, and $A=\left(A{n}\right)_{n \in \mathbb{N}}$ is a predictable process, which is called the compensator of $X$.

PROOF.- Existence We write $A_{0}=0$,
$$A_{n+1}=A_{n}+\mathbb{E}\left[X_{n+1}-X_{n} \mid \mathcal{F}{n}\right]=\sum{k=1}^{n} \mathbb{E}\left[X_{k+1}-X_{k} \mid \mathcal{F}{k}\right],$$ $M{0}=0$ and $M_{n}=X_{n}-A_{n}$ for $n \geq 1 .$
We then directly have that $\left(A_{n}\right)$ is predictable and $\left(M_{n}\right)$ is adapted. Since the $X_{n}$ are integrable, $A_{n}$ and $M_{n}$ are also adaptable. Furthermore
\begin{aligned} \mathbb{E}\left[M_{n+1} \mid \mathcal{F}{n}\right] &=\mathbb{E}\left[X{n+1}-A_{n+1} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[X{n+1} \mid \mathcal{F}{n}\right]-A{n+1} \ &=\mathbb{E}\left[X_{n+1} \mid \mathcal{F}{n}\right]-A{n}-\mathbb{E}\left[X_{n+1}-X_{n} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[X{n+1} \mid \mathcal{F}{n}\right]-A{n}-\mathbb{E}\left[X_{n+1} \mid \mathcal{F}{n}\right]+X{n} \ &=X_{n}-A_{n} \ &=M_{n} \end{aligned}
thus $M=\left(M_{n}\right){n \in \mathbb{N}}$ is a martingale. Unicity Let us assume that there are two such decompositions: $$X{n}=X_{0}+M_{n}+A_{n}=X_{0}+M_{n}^{\prime}-A_{n}^{\prime} .$$
Then, $A_{0}-A_{0}^{\prime}=0$ and since the processes are predictable
\begin{aligned} A_{n+1}-A_{n+1}^{\prime} &=\mathbb{E}\left[A_{n+1}-A_{n+1}^{\prime} \mid \mathcal{F}{n}\right] \ &=\mathbb{E}\left[M{n+1}-M_{n+1}^{\prime} \mid \mathcal{F}{n}\right] \ &=M{n}-M_{n}^{\prime} \ &=A_{n}-A_{n}^{\prime} \end{aligned}
because we have martingales. Therefore, $A_{n}=A_{n}^{\prime}$ for any $n$, and consequently, $M_{n}=M_{n}^{\prime}$ for any $n$. Thus, we do have unicity.
COROLLARY 4.1.- In the Doob decomposition of $X=\left(X_{n}\right){n \in \mathbb{N}}$, $-\left(A{n}\right)$ is increasing if and only if $X=\left(X_{n}\right){n \in \mathbb{N}}$ is a submartingale; $-\left(A{n}\right)$ is decreasing if and only if $X=\left(X_{n}\right){n \in \mathbb{N}}$ is a supermartingale; $-\left(A{n}\right)$ is null if and only if $X=\left(X_{n}\right){n \in \mathbb{N}}$ is a martingale. ProOF.- The final point is a consequence of the first two. By definition, we have $A{n+1}=A_{n}+\mathbb{E}\left[X_{n+1}-X_{n} \mid \mathcal{F}{n}\right]$ and $\mathbb{E}\left[X{n+1}-X_{n} \mid \mathcal{F}_{n}\right]$ is positive for a submartingale, negative for a supermartingale.

## 金融代写|金融数学代写Financial Mathematics代考|Stopping time

We now introduce the concept of stopping time. Informally, a stopping time corresponds to a random date, thus a date that is unknown in advance, but such that at any instant it is possible to say whether or not the date has passed.

DEFINITION 4.2.- Let $T: \Omega \longrightarrow \mathbb{N} \cup{+\infty}$ be a random variable. It is said that $T$
EXAMPLE 4.7.- Let $T$ be a constant positive random variable. Then, $T$ is an $n$, we have
$$(T \leq n)=\left{\begin{array}{l} \emptyset \text { if } T>n \ \Omega \text { if } T \leq n \end{array}\right.$$
As any $\sigma$-algebra contains the empty set and $\Omega$, we have $(T \leq n) \in \mathcal{F}{n}$. EXAMPLE 4.8.-Let $\left(X{n}\right){n \in \mathbb{N}}$ be a sequence of real random variables. Let $A$ be a Borel set in $\mathbb{R}$ and consider the random variable $$T=\inf \left{k \in \mathbb{N} ; X{k} \in A\right},$$

using the convention $\inf \emptyset=+\infty$. Thus, $T$ is a stopping time with respect to the natural filtration of $\left(X_{n}\right){n \in \mathbb{N}}$, called the hitting time of the set $A$. Indeed, for any $n \in \mathbb{N}$, we have $$(T \leq n)=\left(X{0} \in A\right) \cup \ldots \cup\left(X_{n-1} \in A\right) \cup\left(X_{n} \in A\right) \in \mathcal{F}{n}$$ because $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ is the natural filtration of $\left(X{n}\right){n \in \mathbb{N}}$. Proposition 4.3.-A random variable $T: \Omega \longrightarrow \mathbb{N} \cup{+\infty}$ is an PROOF.- Let us assume that $T$ is an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}^{-s t o p p i n g}}$ time. Then, for any $n \in \mathbb{N}^{*}$, we have: $(T \leq n) \in \mathcal{F}{\mathrm{n}}$ and $(T \leq n-1) \in \mathcal{F}{\mathrm{n}-1} \subset \mathcal{F}{\mathrm{n}}$, and consequently,
$$(T=n)=(T \leq n) \cap(T \leq n-1)^{c} \in \mathcal{F}{n},$$ because a $\sigma$-algebra is closed under finite intersection and complements. It is now assumed that for any $k \in \mathbb{N},(T=k) \in \mathcal{F}{k}$. Therefore, for any fixed $n$, we have
$$(T \leq n)=\bigcup_{k=0}^{n}(T=k) \in \bigcup_{k=0}^{n} \mathcal{F}{k}=\mathcal{F}{n},$$
because a filtration is an increasing sequence of $\sigma$-algebras. Consequently, $T$ is indeed an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-stopping time
DEFINITION 4.3.- Consider an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}^{-s t o p p}}$-sime, T. The set
$$\mathcal{F}{T}=\left{A \in \mathcal{F} ; \forall n \in \mathbb{N}, A \cap(T \leq n) \in \mathcal{F}{n}\right}$$
is called the $\sigma$-algebra of events observable up until time $T$

## 金融代写|金融数学代写Financial Mathematics代考|Stopped martingales

Let us now see how to combine the concept of a martingale and that of stopping time.

Proposition 4.5.-Let $\left(Z_{n}\right){n \in \mathbb{N}}$ be an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-martingale and let $T$ be an $$Z{n}^{T}=Z_{T \wedge n}=Z_{n} \mathbb{1}{{T>n}}+Z{T} \mathbb{1}{{T \leq n}} ; n \in \mathbb{N} .$$ Then, $\left(Z{n}^{T}\right){n \in \mathbb{N}}$ is an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}-$ martingale. PROOF.- We have $$Z{T \wedge n}=Z_{0}+\sum_{l=1}^{T \wedge n}\left(Z_{l}-Z_{l-1}\right)=Z_{0}+\sum_{l \geq 1} \mathbb{1}{{l \leq T \wedge \wedge}}\left(Z{l}-Z_{l-1}\right) .$$
Since $\mathbb{1}{{l \leq T \wedge n}}=\mathbb{1}{{l \leq T}} \mathbb{1}{{l \leq n}}$ and $\mathbb{1}{{l \leq n}}=1$ if and only if $l=1, \ldots ., n$, and 0 if $l>n$, it follows that
$$Z_{T \wedge n}=Z_{0}+\sum_{l=1}^{n} \mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) .$$
On the contrary, for $k \leq n$, we have:
\begin{aligned} &\mathbb{E}\left[Z_{T \wedge n} \mid \mathcal{F}{k}\right] \ &=\mathbb{E}\left[Z{0} \mid \mathcal{F}{k}\right]+\sum{l=1}^{n} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{k}\right] \ &=\mathbb{E}\left[Z{0} \mid \mathcal{F}{k}\right]+\sum{l=1}^{k} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{k}\right] \ &=\sum{l=k+1}^{n} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}_{k}\right] \end{aligned}

For the first two terms, because the random variables $Z_{0}$ and $\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right)$ are $\mathcal{F}{k}$-measurable, for any $l \leq k$, we have $$\mathbb{E}\left[Z{0} \mid \mathcal{F}{k}\right]=Z{0} \text { et } \sum_{l=1}^{k} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{k}\right]=\sum{l=1}^{k} \mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right)$$
For the third term, it can be written,
\begin{aligned} \sum_{l=k+1}^{n} \mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{k}\right] &=\sum{l=k+1}^{n} \mathbb{E}\left[\mathbb{E}\left[\mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{l-1}\right] \mid \mathcal{F}{k}\right] \ &=\sum_{l=k+1}^{n} \mathbb{E}\left[\mathbb{1}{{l \leq T}} \mathbb{E}\left[\left(Z{l}-Z_{l-1}\right) \mid \mathcal{F}{l-1}\right] \mid \mathcal{F}{k}\right] \ &=0, \end{aligned}
because, as $\left(Z_{n}\right)$ is an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-martingale, we have $\mathbb{E}\left[\left(Z_{l}-Z_{l-1}\right) \mid \mathcal{F}{l-1}\right]=0$, for any $l \geq 1$. It follows from this that $$\mathbb{E}\left[Z{T \wedge n} \mid \mathcal{F}{k}\right]=Z{0}+\sum_{l=1}^{k} \mathbb{1}{{l \leq T}}\left(Z{l}-Z_{l-1}\right)$$
and consequently, for any $k \leq n$, we have
$$\mathbb{E}\left[Z_{T \wedge n} \mid \mathcal{F}{k}\right]=Z{T \wedge k} .$$

## 金融代写|金融数学代写Financial Mathematics代考|The Doob decomposition

Xn=X0+米n+一个n

Xn=X0+米n+一个n=X0+米n′−一个n′.

## 金融代写|金融数学代写Financial Mathematics代考|Stopping time

$$(T \leq n)=\left{ ∅ 如果 吨>n Ω 如果 吨≤n\正确的。 一个s一个n是σ−一个lG和br一个C○n吨一个一世ns吨H和和米p吨是s和吨一个ndΩ,在和H一个在和(吨≤n)∈Fn.和X一个米磷大号和4.8.−大号和吨(Xn)n∈ñb和一个s和q在和nC和○Fr和一个lr一个nd○米在一个r一世一个bl和s.大号和吨一个b和一个乙○r和ls和吨一世nR一个ndC○ns一世d和r吨H和r一个nd○米在一个r一世一个bl和T=\inf \left{k \in \mathbb{N} ; X{k} \in A\right},$$

(吨≤n)=(X0∈一个)∪…∪(Xn−1∈一个)∪(Xn∈一个)∈Fn因为(Fn)n∈ñ是自然过滤(Xn)n∈ñ. 命题 4.3.-一个随机变量吨:Ω⟶ñ∪+∞是一个证明。-让我们假设吨是一个(Fn)n∈ñ−s吨○pp一世nG时间。那么，对于任何n∈ñ∗， 我们有：(吨≤n)∈Fn和(吨≤n−1)∈Fn−1⊂Fn， 因此，

(吨=n)=(吨≤n)∩(吨≤n−1)C∈Fn,因为一个σ-代数在有限交集和补集下是封闭的。现在假设对于任何ķ∈ñ,(吨=ķ)∈Fķ. 因此，对于任何固定n， 我们有

(吨≤n)=⋃ķ=0n(吨=ķ)∈⋃ķ=0nFķ=Fn,

\mathcal{F}{T}=\left{A \in \mathcal{F} ; \forall n \in \mathbb{N}, A \cap(T \leq n) \in \mathcal{F}{n}\right}\mathcal{F}{T}=\left{A \in \mathcal{F} ; \forall n \in \mathbb{N}, A \cap(T \leq n) \in \mathcal{F}{n}\right}

## 金融代写|金融数学代写Financial Mathematics代考|Stopped martingales

∑l=ķ+1n和[1l≤吨(从l−从l−1)∣Fķ]=∑l=ķ+1n和[和[1l≤吨(从l−从l−1)∣Fl−1]∣Fķ] =∑l=ķ+1n和[1l≤吨和[(从l−从l−1)∣Fl−1]∣Fķ] =0,

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## MATLAB代写

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