# 统计代写|概率论代写Probability theory代考|STAT4061

## 统计代写|概率论代写Probability theory代考|Martingales

Everyone who does not own a casino would agree without hesitation that the successive payment of gains $Y_1, Y_2, \ldots$, such that $Y_1, Y_2, \ldots$ are i.i.d. with $\mathbf{E}\left[Y_1\right]=$ 0 , could be considered a fair game consisting of consecutive rounds. In this case, the process $X$ of partial sums $X_n=Y_1+\ldots+Y_n$ is integrable and $\mathbf{E}\left[X_n \mid \mathcal{F}_m\right]=X_m$ if $m<n$ (where $\mathbb{F}=o(X)$ ). We want to use this equation for the conditional expectations as the defining equation for a fair game that in the following will be called a martingale. Note that, in particular, this definition does not require that the individual payments be independent or identically distributed. This makes the notion quite a bit more flexible. The momentousness of the following concept will become manifest only gradually.

Definition $9.24$ Let $(\Omega, \mathcal{F}, \mathbf{P})$ be a probability space, $I \subset \mathbb{R}$, and let $\mathbb{F}$ be a filtration. Let $X=\left(X_t\right)_{t \in I}$ be a real-valued, adapted stochastic process with $\mathbf{E}\left[\left|X_t\right|\right]<\infty$ for all $t \in I . X$ is called (with respect to $\mathbb{F}$ ) a martingale if $\mathbf{E}\left[X_t \mid \mathcal{F}_s\right]=X_s$ for all $s, t \in I$ with $t>s$,
submartingale if $\mathbf{E}\left[X_t \mid \mathcal{F}_s\right] \geq X_s$ for all $s, t \in I$ with $t>s$,
supermartingale if $\mathbf{E}\left[X_t \mid \mathcal{F}_s\right] \leq X_s$ for all $s, t \in I$ with $t>s$.

Remark $9.25$ Clearly, for a martingale, the map $t \mapsto \mathbf{E}\left[X_t\right]$ is constant, for submartingales it is monotone increasing and for supermartingales it is monotone decreasing. $\diamond$

Remark 9.26 The etymology of the term martingale has not been resolved completely. The French la martingale (originally Provençal martegalo, named after the town Martiques) in equitation means “a piece of rein used in jumping and cross country riding”. Sometimes the ramified shape, in particular of the running martingale (French la martingale à anneaux), is considered as emblematic for the doubling strategy in the Petersburg game.

This doubling strategy itself is the second meaning of la martingale. Starting here, a shift in the meaning towards the mathematical notion seems plausible. A different derivation, in contrast to the appearance, is based on the function of the rein, which is to “check the upward movement of the horse’s head”. Thus the notion of a martingale might first have been used for general gambling strategies (checking the movements of chance) and later for the doubling strategy in particular. $\diamond$

Remark 9.27 If $I=\mathbb{N}, I=\mathbb{N}0$ or $I=\mathbb{Z}$, then it is enough to consider at each instant $s$ only $t=s+1$. In fact, by the tower property of the conditional expectation (Theorem 8.14(iv)), we get $$\mathbf{E}\left[X{s+2} \mid \mathcal{F}s\right]=\mathbf{E}\left[\mathbf{E}\left[X{s+2} \mid \mathcal{F}_{s+1}\right] \mid \mathcal{F}_s\right]$$

## 统计代写|概率论代写Probability theory代考|Discrete Stochastic Integral

So far we have encountered a martingale as the process of partial sums of gains of a fair game. This game can also be the price of a stock that is traded at discrete times on a stock exchange. With this interpretation, it is particularly evident that it is natural to construct new stochastic processes by considering investment strategies changes as the stock price changes. It is the price multiplied by the number of stocks in the portfolio. In order to describe such processes formally, we introduce the following notion.

Definition 9.37 (Discrete stochastic integral) Let $\left(X_n\right){n \in \mathbb{N}_0}$ be an $\mathbb{F}$-adapted real process and let $\left(H_n\right){n \in \mathbb{N}}$ be a real-valued and $\mathbb{F}$-predictable process. The discrete stochastic integral of $H$ with respect to $X$ is the stochastic process $H \cdot X$ defined by
$$(H \cdot X)n:=\sum{m=1}^n H_m\left(X_m-X_{m-1}\right) \quad \text { for } n \in \mathbb{N}0 .$$ If $X$ is a martingale, then $H \cdot X$ is also called the martingale transform of $X$. Remark $9.38$ Clearly, $H \cdot X$ is adapted to $\mathbb{F} . \diamond$ Let $X$ be a (possibly unfair) game where $X_n-X{n-1}$ is the gain per euro in the $n$th round. We interpret $H_n$ as the number of euros we bet in the $n$th game. $H$ is then a gambling strategy. Clearly, the value of $H_n$ has to be decided at time $n-1$; that is, before the result of $X_n$ is known. In other words, $H$ must be predictable.

Now assume that $X$ is a fair game (that is, a martingale) and $H$ is locally bounded (that is, each $H_n$ is bounded). Then (since $\mathbf{E}\left[X_{n+1}-X_n \mid \mathcal{F}n\right]=0$ ) \begin{aligned} \mathbf{E}\left[(H \cdot X){n+1} \mid \mathcal{F}n\right] &=\mathbf{E}\left[(H \cdot X)_n+H{n+1}\left(X_{n+1}-X_n\right) \mid \mathcal{F}n\right] \ &=(H \cdot X)_n+H{n+1} \mathbf{E}\left[X_{n+1}-X_n \mid \mathcal{F}_n\right] \ &=(H \cdot X)_n \end{aligned}
Thus $H \cdot X$ is a martingale. The following theorem says that the converse also holds; that is, $X$ is a martingale if, for sufficiently many predictable processes, the stochastic integral is a martingale.

# 概率论代考

## 统计代写|概率论代写概率论代考|鞅

submartingale if $\mathbf{E}\left[X_t \mid \mathcal{F}_s\right] \geq X_s$ 为所有人 $s, t \in I$ 用 $t>s$，
supermartingale if $\mathbf{E}\left[X_t \mid \mathcal{F}_s\right] \leq X_s$ 为所有人 $s, t \in I$ 用 $t>s$.

(H \cdot X)n:=\sum{m=1}^n H_m\left(X_m-X{m-1}\right) \quad \text { for } n \in \mathbb{N}0 . $$如果X是一个鞅，那么H \cdot X也被称为X的鞅变换。备注9.38显然，H \cdot X改编为\mathbb{F} . \diamond让X是一个(可能不公平的)游戏，其中X_n-X{n-1}是n第一轮每欧元的收益。我们将H_n解释为我们在n第一次游戏中下注的欧元数。H是一种赌博策略。显然，H_n的值必须在n-1时刻决定;也就是说，在知道X_n的结果之前。换句话说，H必须是可预测的 现在假设X是一个公平的博弈(即鞅)，H是局部有界的(即每个H_n都是有界的)。那么(因为\mathbf{E}\left[X_{n+1}-X_n \mid \mathcal{F}n\right]=0)$$ \begin{aligned} \mathbf{E}\left[(H \cdot X){n+1} \mid \mathcal{F}n\right] &=\mathbf{E}\left[(H \cdot X)n+H{n+1}\left(X{n+1}-X_n\right) \mid \mathcal{F}n\right] \ &=(H \cdot X)n+H{n+1} \mathbf{E}\left[X{n+1}-X_n \mid \mathcal{F}_n\right] \
&=(H \cdot X)_n
\end{aligned}


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