### 金融代写|金融数学代写Financial Mathematics代考|MATHS 1009

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## 金融代写|金融数学代写Financial Mathematics代考|The Gambler’s ruin

To conclude this chapter and as an introduction to problems in financial mathematics, let us look at a problem of ruin. A gambler has an initial capital $x$.

They toss for heads or tails using a balanced coin and win 1 if they obtain tails and loses 1 if they obtain heads. The gambler has a fixed objective of a fortune $a \geq x$ and a baseline for losing, $b \leq x$. They play until their fortune reaches $a$ or $b$.

The gambler’s fortune is modeled using a random walk $\left(S_{n}\right){n \in \mathbb{N}}$, with $S{0}=x$ and $S_{n}=S_{0}+\sum_{k=1}^{n} X_{k}$, where $X_{k}$ represents the winnings of the $k$-th toss.

It is known that arriving at $a$ or $b$, starting from $x$ is equivalent to arriving at $a-x$ or $b-x$ starting from 0 . The game then stops almost surely at the end of a finite time.
We use $p_{x}$ to denote the probability of ruin, starting from an initial capital of $x$, and $R_{x}$ to denote the event ruined after starting from $x$, that is,
$$p_{x}=\mathbb{P}\left(T_{b}<T_{a}\right)=\mathbb{P}\left(R_{x}\right)$$
It can be directly noted that $p_{a}=0$ and $p_{b}=1$ because in both these situations the game does not start; ruin is impossible in the first case and is certain in the second.
We will now obtain a recurrence over $p_{x}$. If $b<x<a$, then using the formula for total probabilities, we have
\begin{aligned} p_{x} &=\mathbb{P}\left(R_{x}\right) \ &=\mathbb{P}\left(R_{x}, \mid S_{1}=x+1\right) \mathbb{P}\left(S_{1}=x+1\right)+\mathbb{P}\left(R_{x} \mid S_{1}=x-1\right) \mathbb{P}\left(S_{1}=x-1\right) \ &=\mathbb{P}\left(R_{x+1}\right) \frac{1}{2}+\mathbb{P}\left(R_{x-1}\right) \frac{1}{2} \end{aligned}
by using stationarity. We thus obtain that $\left(p_{x}\right){a{x}=\alpha+\beta x . \alpha$ and $\beta$ are identified with the extremal conditions $p_{a}=0$ and $p_{b}=1$
$$\left{\begin{array} { l } { p _ { a } = 0 = \alpha + \beta a } \ { p _ { b } = 1 = \alpha + \beta b } \end{array} \Leftrightarrow \left{\begin{array}{l} \alpha=\frac{a}{a-b} \ \beta=-\frac{1}{a-b} \end{array}\right.\right.$$

Thus, the probability of ruin, starting from $x$, is
$$p_{x}=\frac{a-x}{a-b} .$$
Using similar reasoning, it can be shown that the duration of the game $d_{x}$ starting from $x$ is the solution to the recurrent linear equation with the right-hand side ${ }^{2}$
$$d_{x}=\frac{1}{2}\left(d_{x+1}+d_{x-1}\right)+1$$
with $d_{a}=d_{b}=0$. Its unique solution is $d_{x}=-a b+x(a+b)-x^{2}$. In particular, if $x=0$ and $b$ tend towards $-\infty$, we obtain that $\mathbb{E}\left[T_{a}\right]=+\infty$. It is known that the walk starting from 0 will reach $a$ almost surely, in finite time. However, on average, this time is infinite.

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

In all that follows, we will work on a filtered probability space $\left(\Omega, \mathcal{F}, \mathbb{P},\left(\mathcal{F}{n}\right){n \in \mathbb{N}}\right)$.
DEFINITION 4.1.-A sequence of random variables $\left(X_{n}\right){n \in \mathbb{N}}$ is an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-martingale [respectively a submartingale, supermartingale] if 1) for any $n \in \mathbb{N}, X{n}$ is integrable;
2) the sequence $\left(X_{n}\right){n \in \mathbb{N}}$ is $\left(\mathcal{F}{n}\right){n \in \mathbb{N} \text {-adapted; }}$ 3) $\forall n \in \mathbb{N}, \mathbb{E}\left[X{n+1} \mid \mathcal{F}{n}\right]=X{n}$ [respectively $\mathbb{E}\left[X_{n+1} \mid \mathcal{F}{n}\right] \geq X{n}$. $\left.\mathbb{E}\left[X_{n+1} \mid \mathcal{F}{n}\right] \leq X{n}\right]$

ReMARK 4.1.- To justify that a sequence of random variables $\left(X_{n}\right){n \in \mathbb{N}}$ is an Indeed, if for an $n \in \mathbb{N}$, we have $\mathbb{E}\left[X{n+1} \mid \mathcal{F}{n}\right]=X{n}$, then $X_{n}$ is the $\mathcal{F}{n}$-measurable. Consequently, the equality in statement 3 implies statement 2 . However, in practice, it is often useful to first demonstrate that $X{n}$ is $\mathcal{F}{n}$-measurable in order to calculate the conditional expectation $\mathbb{E}\left[X{n+1} \mid \mathcal{F}_{n}\right]$.

EXAMPLE 4.1.- The simple symmetric random walk is a martingale with respect to its natural filtration. Indeed, it must be recalled that $\left(X_{n}\right){n \geq 1}$ is a sequence of random independent variables and with the same Rademacher distribution with the parameter $1 / 2$ $$\mathbb{P}\left(X{1}=1\right)=\mathbb{P}\left(X_{1}=-1\right)=1 / 2 .$$
Let $S_{0}=0$ and let $S_{n}=\sum_{k=1}^{n} X_{k}$ for $n \geq 1$ be the simple symmetric random walk starting from 0 . Let $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ be the natural filtration of $S=\left(S_{n}\right){n \in \mathbb{N}}$ and $n \geq 0$. Then, for any $n,-n \leq S{n} \leq n$, therefore $S_{n}$ is integrable. By construction, $\left(S_{n}\right)$ is adapted with respect to its natural filtration and
$$\mathbb{E}\left[S_{n+1} \mid \mathcal{F}{n}\right]=\mathbb{E}\left[S{n}+X_{n+1} \mid \mathcal{F}{n}\right]=S{n}+\mathbb{E}\left[X_{n+1}\right]=S_{n},$$
because $S_{n}$ is the $\mathcal{F}{n}$-measurable, $X{n+1}$ is independent of $\mathcal{F}{n}$ and $\mathbb{E}\left[X{n+1}\right]=0$. We thus have a martingale. A martingale corresponds to an equitable game.

EXAMPLE 4.2.- Let us now study the simple random asymmetric walk. We thus have $\left(X_{n}\right){n \geq 1}$, a sequence of independent random variables and with the same Rademacher distribution with parameter $p \neq 1 / 2$ $$\mathbb{P}\left(X{1}=1\right)=1-\mathbb{P}\left(X_{1}=-1\right)=p .$$
Let $S_{0}=0$ and $S_{n}=\sum_{k=1}^{n} X_{k}$ for $n \geq 1$. Let $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ be the natural filtration of $S=\left(S_{n}\right){n \in \mathbb{N}}$ and $n \geq 0$. Then, for any $n,-n \leq S{n} \leq n$; therefore $S_{n}$ is integrable. By construction, $\left(S_{n}\right)$ is adapted with respect to its natural filtration and
\begin{aligned} \mathbb{E}\left[S_{n+1} \mid \mathcal{F}{n}\right] &=\mathbb{E}\left[S{n}+X_{n+1} \mid \mathcal{F}{n}\right] \ &=S{n}+\mathbb{E}\left[X_{n+1}\right] \ &=S_{n}+p-(1-p) \ &=S_{n}+2 p-1 \end{aligned}
Therefore, $S=\left(S_{n}\right)_{n \in \mathbb{N}}$ is a supermartingale if $p \leq 1 / 2$, or a submartingale, if $p>1 / 2$. A supermartingale corresponds to a losing game, while a submartingale corresponds to a winning game.

## 金融代写|金融数学代写Financial Mathematics代考|Martingale transform

THEOREM 4.1.-Let $\left(K_{n}\right){n \in \mathbb{N}}$ be a positive and $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ be a predictable process. Consider $\left(X{n}\right){n \in \mathbb{N}}$ an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-martingale [respectively submartingale, supermartingale]. If the process $\left(K{n}\right){n \in \mathbb{N}}$ is bounded, then the process $\left(K \cdot X{n}\right){n \in \mathbb{N}}$, defined by $K \cdot X{0}=X_{0}$ and for any $n \geq 1$,
$$K \cdot X_{n}:=\sum_{k=1}^{n} K_{k}\left(X_{k}-X_{k-1}\right),$$
is an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-martingale [respectively submartingale, supermartingale]. The process $\left(K, X_{n}\right){n \in \mathbb{N}}$ is called a martingale transform of $\left(X{n}\right)$. $K \cdot X_{n}:=\sum_{k=1}^{n} K_{k}\left(X_{k}-X_{k-1}\right)$, is clearly $\mathcal{F}{n}$-measurable. On the contrary, because $\left(K{n}\right)$ is bounded, there exists $M>0$ such that
$$\mathbb{E}\left[\left|K \cdot X_{n}\right|\right] \leq M \sum_{k=1}^{n} \mathbb{E}\left[\left|X_{k}-X_{k-1}\right|\right]<\infty$$
because $\left(X_{n}\right)$ is an integrable.

Finally, for any $n \geq 1$, upon simplification we obtain:
\begin{aligned} \mathbb{E}\left[K \cdot X_{n}-K \cdot X_{n-1} \mid \mathcal{F}{n-1}\right] &=\mathbb{E}\left[K{n}\left(X_{n}-X_{n-1}\right) \mid \mathcal{F}{n-1}\right] \ &=K{n} \mathbb{E}\left[X_{n}-X_{n-1} \mid \mathcal{F}{n-1}\right] \end{aligned} where the final equality is due to the fact that $\left(K{n}\right)$ is $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$-predictable. Now, when $\left(X_{n}\right)$ is an $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ martingale [respectively submartingale, supermartingale], we have
$$\mathbb{E}\left[X_{n}-X_{n-1} \mid \mathcal{F}{n-1}\right]=0[\text { resp. } \geq 0, \leq 0] .$$ The result of the theorem follows from this. EXAMPLE 4.5.- The typical example of a martingale transform is the fortune of a gambler with a predictable betting strategy. Let $\left(X{n}\right)$ be a sequence of independent and identically distributed (i.i.d.) random variables representing the net winnings (positive or negative) of a given game. It is assumed that the game is equitable $\mathbb{E}\left[X_{1}\right]=0$, such that the gambler’s fortune $M_{n}=M_{0}+\sum_{k=1}^{n} X_{k}$ is a martingale. Let $\left(K_{n}\right)$ be its (predictable) betting strategy. For instance, the gambler begins with an initial stake of 1 . Each time they win, they continue with the same stake, each time they lose, they double the stakes for the next attempt.
$$K_{n+1}=K_{n} \mathbb{1}{X{n}>0}+2 K_{n} \mathbb{1}{X{n}<0} .$$
Therefore, the fortune of the gambler following this strategy
$$M_{0}+\sum_{k=1}^{n} K_{n} X_{n}=M_{0}+\sum_{k=1}^{n} K_{n}\left(M_{n}-M_{n-1}\right)$$
is the transform of the martingale $M$ using the betting process $K$.
The fact that a martingale transform remains a martingale means that if we start with an equitable game, then regardless of the (predictable) betting strategy adopted, it is not possible to increase (or to decrease) the expectation of a win and the game remains equitable.

## 金融代写|金融数学代写Financial Mathematics代考|The Gambler’s ruin

pX=磷(吨b<吨一个)=磷(RX)

pX=磷(RX) =磷(RX,∣小号1=X+1)磷(小号1=X+1)+磷(RX∣小号1=X−1)磷(小号1=X−1) =磷(RX+1)12+磷(RX−1)12

\左{

p一个=0=一个+b一个 pb=1=一个+bb\左右箭头 \左{

$$因此，破产的概率，从X， 是 pX=一个−X一个−b. 使用类似的推理，可以证明游戏的持续时间dX从…开始X是具有右手边的递归线性方程的解2 dX=12(dX+1+dX−1)+1 和d一个=db=0. 其独特的解决方案是dX=−一个b+X(一个+b)−X2. 特别是，如果X=0和b倾向于−∞，我们得到和[吨一个]=+∞. 已知从 0 开始的 walk 会到达一个几乎可以肯定，在有限的时间内。但是，平均而言，这个时间是无限的。 ## 金融代写|金融数学代写Financial Mathematics代考|Martingales 在接下来的所有内容中，我们将处理过滤的概率空间(Ω,F,磷,(Fn)n∈ñ). 定义 4.1.-随机变量序列(Xn)n∈ñ是一个(Fn)n∈ñ-martingale [分别是 submartingale, supermartingale] 如果 1) 对任何n∈ñ,Xn是可积的； 2) 序列(Xn)n∈ñ是(Fn)n∈ñ-适应; 3) ∀n∈ñ,和[Xn+1∣Fn]=Xn[分别和[Xn+1∣Fn]≥Xn. 和[Xn+1∣Fn]≤Xn] 备注 4.1.- 证明一系列随机变量的合理性(Xn)n∈ñ是一个确实，如果对于一个n∈ñ， 我们有和[Xn+1∣Fn]=Xn， 然后Xn是个Fn- 可测量的。因此，陈述 3 中的相等意味着陈述 2。然而，在实践中，首先证明Xn是Fn- 可测量以计算条件期望和[Xn+1∣Fn]. 例 4.1.- 简单的对称随机游走就其自然过滤而言是鞅。确实，必须记住的是(Xn)n≥1是随机自变量序列，具有与参数相同的 Rademacher 分布1/2 磷(X1=1)=磷(X1=−1)=1/2. 让小号0=0然后让小号n=∑ķ=1nXķ为了n≥1是从 0 开始的简单对称随机游走。让(Fn)n∈ñ成为自然过滤小号=(小号n)n∈ñ和n≥0. 那么，对于任何n,−n≤小号n≤n， 所以小号n是可积的。通过施工，(小号n)适应于其自然过滤和 和[小号n+1∣Fn]=和[小号n+Xn+1∣Fn]=小号n+和[Xn+1]=小号n, 因为小号n是个Fn- 可测量的，Xn+1独立于Fn和和[Xn+1]=0. 因此，我们有一个鞅。鞅对应于公平游戏。 例 4.2.- 现在让我们研究简单的随机不对称游走。因此我们有(Xn)n≥1, 一系列独立随机变量，具有相同的 Rademacher 分布和参数p≠1/2 磷(X1=1)=1−磷(X1=−1)=p. 让小号0=0和小号n=∑ķ=1nXķ为了n≥1. 让(Fn)n∈ñ成为自然过滤小号=(小号n)n∈ñ和n≥0. 那么，对于任何n,−n≤小号n≤n; 所以小号n是可积的。通过施工，(小号n)适应于其自然过滤和 和[小号n+1∣Fn]=和[小号n+Xn+1∣Fn] =小号n+和[Xn+1] =小号n+p−(1−p) =小号n+2p−1 所以，小号=(小号n)n∈ñ如果是超鞅p≤1/2，或亚鞅，如果p>1/2. 上鞅对应于失败的游戏，而下鞅对应于获胜的游戏。 ## 金融代写|金融数学代写Financial Mathematics代考|Martingale transform 定理 4.1.-让(ķn)n∈ñ做一个积极的和(Fn)n∈ñ是一个可预测的过程。考虑(Xn)n∈ñ一个(Fn)n∈ñ-martingale [分别是submartingale，supermartingale]。如果过程(ķn)n∈ñ是有界的，那么过程(ķ⋅Xn)n∈ñ， 被定义为ķ⋅X0=X0并且对于任何n≥1, ķ⋅Xn:=∑ķ=1nķķ(Xķ−Xķ−1), 是一个(Fn)n∈ñ-martingale [分别是submartingale，supermartingale]。过程(ķ,Xn)n∈ñ称为鞅变换(Xn). ķ⋅Xn:=∑ķ=1nķķ(Xķ−Xķ−1), 很明显Fn- 可测量的。相反，因为(ķn)有界，存在米>0这样 和[|ķ⋅Xn|]≤米∑ķ=1n和[|Xķ−Xķ−1|]<∞ 因为(Xn)是一个可积的。 最后，对于任何n≥1，简化后我们得到： 和[ķ⋅Xn−ķ⋅Xn−1∣Fn−1]=和[ķn(Xn−Xn−1)∣Fn−1] =ķn和[Xn−Xn−1∣Fn−1]最终的相等是由于以下事实(ķn)是(Fn)n∈ñ-可预见。现在，当(Xn)是一个(Fn)n∈ñmartingale [分别为submartingale，supermartingale]，我们有 和[Xn−Xn−1∣Fn−1]=0[ 分别 ≥0,≤0].定理的结果由此而来。示例 4.5.- 鞅变换的典型示例是具有可预测投注策略的赌徒的财富。让(Xn)是一系列独立且同分布 (iid) 的随机变量，表示给定游戏的净赢利（正或负）。假设游戏是公平的和[X1]=0, 这样赌徒的财富米n=米0+∑ķ=1nXķ是鞅。让(ķn)成为其（可预测的）投注策略。例如，赌徒以 1 的初始赌注开始。每次他们获胜时，他们都会继续使用相同的赌注，每次失败时，他们都会将赌注加倍以进行下一次尝试。$$
K_{n+1}=K_{n} \mathbb{1} {X {n}>0}+2 K_{n} \mathbb{1}{X{n}<0} 。

M_{0}+\sum_{k=1}^{n} K_{n} X_{n}=M_{0}+\sum_{k=1}^{n} K_{n}\left(M_{ n}-M_{n-1}\right)


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