### 金融代写|鞅论及其在金融中的应用代写Martingale theory代考| The Swiss Army Formula

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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|鞅论及其在金融中的应用代写Martingale theory代考|The Swiss Army Formula

Depending on which blade is selected, a Swiss army knife transforms itself into various useful tools. The formula obtained in this subsection is called the Swiss army formula of Palm calculus because it contains the main formulas of this theory, as well as some new ones.

Let $\left{T_{n}\right}$ and $\left{\tau_{n}\right}$ be two simple point processes on $R$ and let $A$ and $D$ be the associated counting measures. The sequence $\left{T_{n}\right}$ is supposed to satisfy the usual conventions. The sequence $\left{\tau_{n}\right}$ is also supposed to satisfy $D\left(\mathbb{R}{+}\right)=D\left(\mathbb{R}-\mathbb{R}{+}\right)=\infty$, but it need not be ordered. However, it is required that for each $n \in \mathbb{Z}$
$$\tau_{n}-T_{n} \stackrel{\text { def }}{=} W_{n} \geq 0 .$$
One could imagine that $T_{n}$ is the arrival time of customer $n$ in a system, and that $\tau_{n}$ is its departure time. Let ${X(t)}$ be a non-negative integer-valued process such that
$$X(b)-X(a)=A((a, b])-D((a, b])$$
The point processes $A$ and $D$ can have common points, and as a matter of fact, we shall consider the situation where $\tau_{n}=T_{n+1}$, that is $W_{n}=T_{n+1}-T_{n}$ and therefore $X(t) \equiv$ constant.

## 金融代写|鞅论及其在金融中的应用代写Martingale theory代考|Renewal Process

Let $\left(N, \theta_{t}, P\right)$ be a stationary point process with finite intensity $\lambda$ and $P_{N}^{0}$ be its Palm probability. Suppose moreover that under $P_{N}^{0}$, the inter-event sequence $\left(S_{n}, n \in \mathbb{Z}\right)$ defined by
$$S_{n}=T_{n+1}-T_{n}$$
is i.i.d. Then $\left(N, P_{N}^{0}\right)$ is called a renewal process and $(N, P)$ a stationary renewal process. The existence of such a mathematical object is granted by the results of $\S$ 1.3.5.

Property 1.4.1. The distribution of the sequence $S^{*}=\left{S_{n}\right}_{n \in Z-{0}}$ is the same under $P$ and $P_{N}^{0}$.

Proof: Let $g:\left(\mathbb{R}^{Z}, \mathcal{B}^{Z}\right) \rightarrow(\mathbb{R}, \mathcal{B})$ be an arbitrary non-negative measurable function. It suffices to show that
$$E\left[g\left(S^{}\right)\right]=E_{N}^{0}\left[g\left(S^{}\right)\right] .$$
By the inversion formula (1.2.25):
$$E\left[g\left(S^{}\right)\right]=\lambda E_{N}^{0}\left[\int_{0}^{T_{1}} g\left(S^{}\left(\theta_{u}\right)\right) d u\right] .$$
But if $u$ is in $\left[0, T_{1}\right)$, then $S^{}\left(\theta_{u}\right)=S^{}$, so that
\begin{aligned} E\left[g\left(S^{}\right)\right] &=\lambda E_{N}^{0}\left[\int_{0}^{T_{1}} g\left(S^{}\right) d u\right] \ &=\lambda E_{N}^{0}\left[T_{1}\right] E_{N}^{0}\left[g\left(S^{}\right)\right]=E_{N}^{0}\left[T_{1} g\left(S^{}\right)\right] \end{aligned}
where we have used the independence of $T_{1}=S_{0}$ and $S^{}$ under $P_{N^{}}^{0}$.

## 金融代写|鞅论及其在金融中的应用代写Martingale theory代考|Superposition of Independent Point Processes

The situation is that described in Example 1.1.2 with the additional assumption

$$0<\lambda_{i}<\infty, \quad(1 \leq i \leq k)$$
where $\lambda_{i}$ is the intensity of $N_{i}$. Recall that for each $1 \leq i \leq k,\left(\Phi_{i}, S_{t}^{(i)}, \mathcal{P}{i}\right)$ is a stationary point process. We shall denote by $\mathcal{P}{i}^{0}$ the associated Palm probability. We now prove the following formula:
$$P_{N}^{0}=\sum_{i=1}^{k} \frac{\lambda_{i}}{\lambda}\left(\bigotimes_{j=1}^{i-1} \mathcal{P}{j}\right) \otimes \mathcal{P}{i}^{0} \otimes\left(\bigotimes_{j=i+1}^{k} \mathcal{P}{j}\right)$$ where $\lambda=\sum{i=1}^{k} \lambda_{i}$ is the intensity of $\left(N, \theta_{t}, P\right)$, the superposition of the independent point processes $\left(\Phi_{i}, S_{t}^{(i)}, \mathcal{P}{i}\right), 1 \leq i \leq k$. Proof: By definition, for $A=\prod{i=1}^{k} A_{i}$, where $A_{i} \in \mathcal{M}{i}$ \begin{aligned} &P{N}^{0}(A)=\frac{1}{\lambda} E\left[\int_{(0,1]}\left(1_{A} \circ \theta_{s}\right) N(d s)\right] \ &=\frac{1}{\lambda} \int_{M_{1}} \ldots \int_{M_{k}} \int_{(0,1]} \sum_{j=1}^{k}\left(\prod_{i=1}^{k} 1_{A_{i}} \circ S_{t}^{(i)}\right) \Phi_{j}(d t) \mathcal{P}{1}\left(d m{1}\right) \ldots \mathcal{P}{k}\left(d m{k}\right) \ &=\sum_{j=1}^{k} \frac{1}{\lambda} \int_{M_{1}} \ldots \int_{M_{k}}\left{\int_{(0,1]}\left(\prod_{i=1}^{k} 1_{A_{i}} \circ S_{t}^{(i)}\right) \Phi_{j}(d t)\right} \mathcal{P}{1}\left(d m{1}\right) \ldots \mathcal{P}{k}\left(d m{k}\right) \end{aligned}
But by Fubini’s theorem and the definition of Palm probability $\mathcal{P}{j}^{0}$ \begin{aligned} \frac{1}{\lambda{j}} \int_{M_{1}} & \cdots \int_{M_{k}}\left{\int_{(0,1]} \prod_{i=1}^{k}\left(1_{A_{i}} \circ S_{t}^{(i)}\right) \Phi_{j}(d t)\right} \mathcal{P}{1}\left(d m{1}\right) \ldots \mathcal{P}{k}\left(d m{k}\right) \ &=\mathcal{P}{j}^{0}\left(A{j}\right) \prod_{i=1, i \neq j}^{k} \mathcal{P}{i}\left(A{i}\right) \end{aligned}
where we have taken into account the $S_{t}^{(i)}$-invariance of $\mathcal{P}{i}$. Therefore $$P{N}^{0}\left(\prod_{i=1}^{k} A_{i}\right)=\sum_{i=1}^{k}\left{\frac{\lambda_{i}}{\lambda} \mathcal{P}{i}^{0}\left(A{i}\right) \prod_{\substack{1 \leq j \leq k \ j \neq i}} \mathcal{P}{j}\left(A{j}\right)\right}$$
which implies (1.4.5).

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