### 金融代写|金融数学代写Financial Mathematics代考|STAT2032

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• 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|Trajectories of the random walk

Let us begin by defining the simple symmetric random walk.

DEFINITION 3.1.-A simple symmetric random walk on $\mathbb{Z}$ is a sequence $\left(S_{n}\right){n \in \mathbb{N}}$ of random variables such that $-S{0}=x \in \mathbb{Z}$ is deterministic;
$-S_{n+1}=S_{n}+X_{n+1}$ for any $n \in \mathbb{N}$,
where $\left(X_{n}\right){n \geq 1}$ is a sequence of independent random variables with the same Rademache distribution with parameter $1 / 2$ : $$\mathbb{P}\left(X{1}=-1\right)=\mathbb{P}\left(X_{1}-1\right)=\frac{1}{2} .$$
We use the term walk because this process may represent the position of a person who moves in a straight line, with steps of unit length and equal probability of moving forward $(+1)$ or backwards $(-1)$. This walk is said to be simple as we only take steps with amplitude 1 and is said to be symmetric as each increment can be $+1$ or $-1$, with equal probability.

This process may also be used to model the wealth of a gambler who tosses for heads or tails with a balanced coin. Starting from an initial fortune of $S_{0}=x>0$, with each time step, the gambler looses or gains $1 €$ depending on the toss. The variables $X_{n}$ thus represent the gain at the $n$-th game, and $S_{n}$ is the total fortune of the gambler after the $n$-th toss.

Based on this definition, we can easily arrive at the average behavior of the walk.
PROPOSITION 3.1.- For any $n \in \mathbb{N}$, we have
$$\mathbb{E}\left[S_{n}\right]=S_{0}, \quad \operatorname{Var}\left(S_{n}\right)=n .$$
PROOF.- Let us first note that
$$\begin{gathered} \mathbb{E}\left[X_{1}\right]=(-1) \times \frac{1}{2}+1 \times \frac{1}{2}=0 \ \operatorname{Var}\left(X_{1}\right)=\mathbb{E}\left[X_{1}^{2}\right]=(-1)^{2} \times \frac{1}{2}+1^{2} \times \frac{1}{2}=1 \end{gathered}$$
the increments are thus centered and reduced. By the linearity of expectation, we obtain
$$\mathbb{E}\left[S_{n}\right]=\mathbb{E}\left[S_{0}+\sum_{k=1}^{n} X_{k}\right]=S_{0}+\sum_{k=1}^{n} \mathbb{E}\left[X_{k}\right]=S_{0}$$

since $S_{0}$ is deterministic. Finally, using the independence of the $X_{k}$, we obtain
$$\operatorname{Var}\left(S_{n}\right)=\operatorname{Var}\left(S_{0}+\sum_{k=1}^{n} X_{k}\right)=\sum_{k=1}^{n} \operatorname{Var}\left(X_{k}\right)=n$$
hence, the result.
Thus, the walk remains constant on average, but its variance becomes greater over time.

## 金融代写|金融数学代写Financial Mathematics代考|Reflection principle

This function defines a bijection between the two sets of paths. Indeed, if we now take a path $\left(\widetilde{S}{m}, \ldots, \widetilde{S}{n}\right)$ from $(m, a)$ to $(n,-b)$, as it takes steps whose amplitude is 1 , since $a>0$ and $-b<0$, this path must necessarily pass through 0 . Let $p$ be the first instant where the path is at 0 . We then construct a path $\left(S_{m}, \ldots, S_{n}\right)$ from $(m, a)$ to $(n, b)$ passing through 0 writing
$$S_{k}= \begin{cases}\widetilde{S}{k} & \text { if } m \leq k \leq p, \ -\widetilde{S}{k} & \text { if } p \leq k \leq n .\end{cases}$$
There are, thus, the same number of paths of both types.
An initial application of this property is the following result, called the ballot theorem.

THEOREM 3.1.-During an election with two opposing candidates, $A$ and $B$, candidate $A$ has obtained a votes and candidate $B$ b votes, with $a>b .$ Thus, the probability that $A$ has obtained the majority (in the broad sense) throughout the counting is
$$p=1-\frac{b}{a+1} .$$
PROOF.- With all the counts being equiprobable, $p$ is obtained as the ratio of the number of counts with $A$ in the lead all the time to that of the total number of counts. A count can be modeled by a random walk $\left(S_{n}\right)$, where $S_{n}$ is the number of votes by which $A$ is ahead of $B$ after counting the $n$-th ballot.

There are $a+b$ ballots in total and ultimately, $A$ has a lead of $a-b$ votes over $B$; thus, the total number of counts is the number of paths from $(0,0)$ to $(a+b, a-b)$ and has the value $C_{a+b}^{a}$.
The number of counts with $A$ in the lead all long corresponds to

• the number of paths from $(0,0)$ to $(a+b, a-b)$, not taking any strictly negative value; in other words, the number of paths from $(0,0)$ to $(a+b, a-b)$ that do not touch $-1$;
• the number of paths from $(0,1)$ to $(a+b, a-b+1)$ not touching 0 , by shifting the ordinates upwards by one unit;
• the total number of paths from $(0,1)$ to $(a+b, a-b+1)$ minus the number of paths from $(0,1)$ to $(a+b, a-b+1)$ touching 0 ;
• the total number of paths from $(0,1)$ to $(a+b, a-b+1)$ minus the number of paths from $(0,1)$ to $(a+b,-(a-b+1))$ by the reflection principle.

It is now assumed that the walk starts from $0: S_{0}=0$ and we wish to know whether the walk will return to 0 almost surely. Let us start by observing that if $S_{n}=0$, then $n$ is necessarily even.
PROPOSITION 3.9. – For any $n \in \mathbb{N}$, we have
$$\mathbb{P}\left(S_{2 n}=0\right)=\frac{1}{4^{n}} C_{2 n^{*}}^{n}$$
ProOF.- This probability corresponds to the number of paths from $(0,0)$ to $(2 n, 0)$ divided by the total number of paths with length $2 n$, since all the paths are equiprobable. We thus directly have $\mathbb{P}\left(S_{2 n}=0\right)=\frac{C_{2 n}^{n}}{4^{n}}$.

We now look at the first instant of the return to 0 . It is denoted by $T_{0}$. It is, therefore, the random variable
$$T_{0}=\inf \left{n \geq 1 ; S_{n}=0\right},$$
using the convention that $\inf \emptyset=+\infty$. It is, therefore, a discrete random variable taking values in $\mathbb{N}^{} \cup{+\infty}$. The distribution of $T_{0}$ can be explicitly calculated. PROPOSITION 3.10.- For any $n \in \mathbb{N}^{}$, we have
$$\mathbb{P}\left(T_{0}=2 n\right)=\frac{(2 n-2) !}{2^{2 n-1} n !(n-1) !}$$
ProOF.- The event $\left(T_{0}=2 n\right)$ corresponds to $\left(S_{2} \neq 0, \ldots, S_{2 n-2} \neq 0, S_{2 n}=0\right)$ because then $2 n$ is the first time the walk returns to 0 . In particular, between the instant 0 and the instant $2 n$, the walk does not change in sign and retains the same sign as $S_{1}$. Therefore, we have
\begin{aligned} \mathbb{P}\left(T_{0}=2 n\right)=& \mathbb{P}\left(S_{1}=1, S_{2}>0, \ldots, S_{2 n-2}>0, S_{2 n}=0\right) \ &+\mathbb{P}\left(S_{1}=-1, S_{2}<0, \ldots, S_{2 n-2}<0, S_{2 n}=0\right) \end{aligned} These two probabilities are equal by symmetry. Further, if $S_{2 n-2}>0$ and $S_{2 n}=0$, then we necessarily have $S_{2 n-1}=1$. Therefore, it follows
\begin{aligned} \mathbb{P}\left(T_{0}=2 n\right) \ =& 2 \mathbb{P}\left(S_{1}=1, S_{2}>0, \ldots, S_{2 n-2}>0, S_{2 n-1}=1, S_{2 n}=0\right) \ =& 2 \mathbb{P}\left(S_{2 n}=0 \mid S_{1}=1, S_{2}>0, \ldots, S_{2 n-2}>0, S_{2 n-1}=1\right) \ & \times \mathbb{P}\left(S_{1}=1, S_{2}>0, \ldots, S_{2 n-2}>0, S_{2 n-1}=1\right) \end{aligned}

\begin{aligned} &=2 \mathbb{P}\left(S_{2 n}=0 \mid S_{2 n-1}=1\right) \mathbb{P}\left(S_{1}=1, S_{2}>0, \ldots, S_{2 n-2}>0, S_{2 n-1}=1\right) \ &=2 \mathbb{P}\left(S_{1}=-1 \mid S_{0}=0\right) \mathbb{P}\left(S_{1}=1, S_{2}>0, \ldots, S_{2 n-2}>0, S_{2 n-1}=1\right) \ &=\mathbb{P}\left(S_{1}=1, S_{2}>0, \ldots, S_{2 n-2}>0, S_{2 n-1}=1\right) \end{aligned}
using the Markov property, the stationary property and then the fact that
$$\mathbb{P}\left(S_{1}=-1 \mid S_{0}=0\right)=\mathbb{P}\left(S_{1}=-1\right)=\mathbb{P}\left(X_{1}=-1\right)=\frac{1}{2}$$

## 金融代写|金融数学代写Financial Mathematics代考|Trajectories of the random walk

−小号n+1=小号n+Xn+1对于任何n∈ñ,

## 金融代写|金融数学代写Financial Mathematics代考|Reflection principle

p=1−b一个+1.

• 从路径数(0,0)至(一个+b,一个−b)，不取任何严格的负值；换句话说，从(0,0)至(一个+b,一个−b)不接触的−1;
• 从路径数(0,1)至(一个+b,一个−b+1)不接触 0 ，将纵坐标向上移动一个单位；
• 路径总数(0,1)至(一个+b,一个−b+1)减去从(0,1)至(一个+b,一个−b+1)触摸 0 ;
• 路径总数(0,1)至(一个+b,一个−b+1)减去从(0,1)至(一个+b,−(一个−b+1))通过反射原理。

ProOF.- 这个概率对应于从(0,0)至(2n,0)除以有长度的路径总数2n，因为所有路径都是等概率的。因此我们直接有磷(小号2n=0)=C2nn4n.

T_{0}=\inf \left{n \geq 1 ; S_{n}=0\right},T_{0}=\inf \left{n \geq 1 ; S_{n}=0\right},

=2磷(小号2n=0∣小号2n−1=1)磷(小号1=1,小号2>0,…,小号2n−2>0,小号2n−1=1) =2磷(小号1=−1∣小号0=0)磷(小号1=1,小号2>0,…,小号2n−2>0,小号2n−1=1) =磷(小号1=1,小号2>0,…,小号2n−2>0,小号2n−1=1)

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## MATLAB代写

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