## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|AEM4060

statistics-lab™ 为您的留学生涯保驾护航 在代写蒙特卡洛树搜索Monte Carlo tree search方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写蒙特卡洛树搜索Monte Carlo tree search代写方面经验极为丰富，各种代写蒙特卡洛树搜索Monte Carlo tree search相关的作业也就用不着说。

• 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

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|MCTS for Go

The game of Go has proved to be one of the biggest challenges for AI due to its complexity and the effect that moves have on long term success of the game [6]. The application to more sophisticated MCTS algorithms like UCT to games like Go have greatly improved the rankings of Go systems to the top of the amateur level [6].

Recently in 2015, Google’s AlphaGo program was the first program to defeat a 9-dan (i.e. top-ranked) professional in a game without handicaps. That version of AlphaGo, named AlphaGo Fan, was fairly complicated and used multiple neural networks trained on expert play to guide decision making. Afterwards, AlphaGo Lee was created that used a similar design as that of AlphaGo Fan, but depended more on data generated from self-play to learn the value functions of states. AlphaGo Lee was the most famous iteration of AlphaGo since it was the one that beat the European champion Lee Sedol in 2016. In 2017, Google released an AI on an online Go server that was able to beat top human players 60-0 named AlphaGo Master. Google published their final design AlphaGo Zero which shared many similarities to AlphaGo Master but removed some complications relating to evaluating states. This final version, AlphaGo Zero, learned the strategies of the game of Go exclusively from games of self-play and using a single neural network to do so.

This section will have no theoretical results but serve rather as an exposition on how to specialize MCTS to be successful in a specific application.

AlphaGo Zero is a specific instantiation of the general MCTS algorithm that uses a neural network learned using reinforcement learning to help guide the decision making process. One of the most important things about this system is that it is trained only by playing against itself and improving from there; it learned the strategy of how to play Go from no prior background. To understand how AlphaGo Zero works, we first specify how its decision process fits in the general MCTS framework and then we will describe how it is trained. This entire section uses [7] as a reference so further citations will be omitted.

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|AlphaGo Zero MCTS

MCTS for AlphaGo works very similarly to how it works in UCT and the terminology is referenced heavily from there. While the computational budget hasn’t been exceeded, episodes are played out to iteratively builds up a game tree from a start state $s$ in order to estimate which action is optimal from from s. During each episode, a game is played out by using a Tree Policy to decide a path down the known tree, a leaf node is expanded and it’s value estimated using a trained netural network, and that information is percolated back up the tree.

Because the game alternates between two players on each turn, each level of the tree models a different player just like a minimax tree. The algorithm stays the same at each level of the tree but just makes decision to make the current player win. This makes sense for games like Go because it’s a perfect information game and you can use the same estimation technique to approximate how the opponent will play optimally with an approximation that improves with computation time.
Phase One: Play-out iterations to build tree The Tree Policy that determines the path down the tree on each episode that the algorithm is rum is:
$$a_t=\underset{a \in \mathcal{A}}{\arg \max } Q\left(s_t, a\right)+U\left(s_t, a\right)$$
This is very similar to UCT because $Q\left(s_t, a\right)$ represents an empirical average of the value of taking action $a$ from state $s_t . U\left(s_t, a\right)$ is a an additional term to encourage exploration when the value estimates lack high-confidence and favors actions that have a higher prior-probability. Each episode $i$ ends the tree policy at a leaf node $s_L^i$. After $n$ episodes in total, we define the terms above as
\begin{aligned} & N(s, a)=\sum_{i=1}^n \mathbb{1}(s, a, i) \ & Q(s, a)=\frac{1}{N(s, a)} \sum_{i=1}^n \mathbb{1}(s, a, i) V\left(s_L^i\right) \ & U(s, a)=c P(s, a) \frac{\sqrt{\sum_{b \in \mathcal{A}} N(s, b)}}{1+N(s, a)} \end{aligned}
where $\mathbb{1}(s, a, i)$ is an indicator that action $a$ was chosen from state $s$ on iteration $i$, $V\left(s_L^i\right)$ is an estimate of the value of a state (in the case of games it is an indication of how likely it is that the current player will win), and $P(s, a)$ is a prior probability of selecting action $a$ from state $s$. These $Q$ and $V$ functions are exactly estimates of the Q-values and V-values of the game.

# 蒙特卡洛树搜索代考

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|MCTS for Go

AlphaGo Zero 是通用 MCTS 算法的特定实例，它使用通过强化学习学习的神经网络来帮助指导决策过程。这个系统最重要的事情之一是它只能通过与自己对战并从那里改进来训练；它从没有任何先验背景的情况下学习了如何下围棋的策略。为了解 AlphaGo Zero 的工作原理，我们首先说明它的决策过程如何适应通用 MCTS 框架，然后我们将描述它是如何训练的。整个部分使用 [7] 作为参考，因此将省略进一步的引用。

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|AlphaGo Zero MCTS

AlphaGo 的 MCTS 的工作方式与它在 UCT 中的工作方式非常相似，并且大量引用了那里的术语。虽然没 有超过计算预算，但会播放剧集以从开始状态迭代地构建游戏树 $s$ 为了从 $\mathrm{s}$ 估计哪个动作是最优的。在每 一集中，通过使用 Tree Policy 来决定沿着已知树的路径进行游戏，扩展叶节点并使用训练有素的神经网 络估计其值，并将该信息向上渗透到树中。

$$a_t=\underset{a \in \mathcal{A}}{\arg \max } Q\left(s_t, a\right)+U\left(s_t, a\right)$$

$$N(s, a)=\sum_{i=1}^n 1(s, a, i) \quad Q(s, a)=\frac{1}{N(s, a)} \sum_{i=1}^n 1(s, a, i) V\left(s_L^i\right) U(s, a)=c P(s, a) \frac{\sqrt{\sum_b}}{1+}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Math577

statistics-lab™ 为您的留学生涯保驾护航 在代写蒙特卡洛树搜索Monte Carlo tree search方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写蒙特卡洛树搜索Monte Carlo tree search代写方面经验极为丰富，各种代写蒙特卡洛树搜索Monte Carlo tree search相关的作业也就用不着说。

• 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

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Monte Carlo Tree Search

Theorem 2. Under the assumptions of Theorem 1,
$$\left|\mathbb{E}\left[\bar{X}_n\right]-\mu^\right| \leq\left|\delta_n^\right|+O\left(\frac{K\left(C_p^2 \ln n+N_0\right)}{n}\right)$$
where $N_0=N_0(1 / 2)$.
To provide further intuition for the above, recall that $\delta_n^*$ is the distance between the optimal arm’s mean at time $t=n$, and the true optimal mean. Furthermore, the theorem is proven using the value $N_0(\epsilon)$ where $\epsilon=1 / 2$.

Furthermore, using Theorem 1, we can derive a lower bound on the number of times some arm $i$ will be played. This result follows:

Theorem 3. Under the assumptions of Theorem 3, there exists some positive constant $\rho$ such that for all arms $i$ and $n, T_i(n) \geq\lceil\rho \log (n)\rceil$.

We need the above result to prove that the optimal reward converges quickly to its true mean in the drifting multi armed bandit problem. We omit the proof of the result, but it is natural that it would require both the lower bound on $T_i(n)$ and the upper bound on $\mathbb{E}\left[T_i(n)\right]$. This quick convergence is one of the two most important properties for incorporating UCB into MCTS. If the drifting means converge slowly then we cannot hope to quickly find an optimal arm. The bound follows:
Theorem 4. Fix an arbitrary $\delta>0$ and let $\Delta_n=9 \sqrt{2 n \ln 2 / \delta}$. Let $n_0$ be such that
$$\sqrt{n_0} \geq O\left(K\left(C_p^2 \ln n_0+N_0(1 / 2)\right)\right)$$
Then for any $n \geq n_0$, under the assumptions of Theorem 1 the following bounds hold true:
\begin{aligned} & \mathbb{P}\left(n \bar{X}n \geq n \mathbb{E}\left[\bar{X}_n\right]+\Delta_n\right) \leq \delta \ & \mathbb{P}\left(n \bar{X}_n \leq n \mathbb{E}\left[\bar{X}_n\right]-\Delta_n\right) \leq \delta \end{aligned} Another extremely important result for using UCB in MCTS is that when the means can drift, UCB still finds the best arm when given infinite time. This result follows: Theorem 5. Under the assumptions of Theorem 1 it holds that $$\lim {t \rightarrow \infty} P\left(I_t \neq i^*\right)=0$$

For more detail on any of the theorems discussed in this section, please refer to [5]. In order to understand UCT, we must first relax the restraints on $K$-armed bandits to allow each arm’s mean $\mu_i$ to change over time t. While we can no longer rely on the assumption that the mean of each arm is fixed from $t=1$ onward, we can assume that the expected value of the empirical averages converge. We let $\bar{X}{i, n}=\frac{1}{n} \sum{t=1}^n X_{i, t}$ be the empirical average of arm $i$ at time $n$, and $\mu_{i, n}=\mathbb{E}\left[\tilde{X}{i, n}\right]$ be its expected value. Therefore, we now have a sequence of expected means for each arm $i$, namely $\mu{i, n}$. We assume that these expected means eventually converge to one final mean $\mu_i=\lim {n \rightarrow \infty} \mu{i, n}$. We further define a sequence of offsets for each arm as $\delta_{i, n}=\mu_{i, n}-\mu_i$. We also make the following assumptions about the reward sequence:

Assumption 1. Fix $1 \leq i \leq K$. Let $\left{\mathcal{F}{i, t}\right}_t$ be a fultration such that $\left{X{i, t}\right}_t$ is $\left{\mathcal{F}{i, t}\right}$-adapted and $X{i, t}$ is conditionally independent of $\mathcal{F}{i, t+1}, \mathcal{F}{i, t+2}, \ldots$ given $\mathcal{F}{i, t-1}{ }^1$. Then $0 \leq X{i, t} \leq 1$ and the limit of $\mu_{i, n}=\mathbb{E}\left[\bar{X}{i n}\right]$ exists. Further, we assume that there exist a constant $C_p>0$ and an integer $N_p$ such that for $n \geq N_p$. for any $\delta>0, \Delta_n(\delta)=C_p \sqrt{n \ln (1 / \delta)}$, the following bounds hold: \begin{aligned} & \mathbb{P}\left(n \bar{X}{i, n} \geq n \mathbb{E}\left[\bar{X}{i, n}\right]+\Delta_n(\delta)\right) \leq \delta \ & \mathbb{P}\left(n \bar{X}{i, n} \leq n \mathbb{E}\left[\bar{X}{i, n}\right]-\Delta_n(\delta)\right) \leq \delta \end{aligned} This assumption allows us to define a bias sequence $c{t, s}$ for time $t$ and sample size $s$ which satisfies Equation 3 and Equation 4. This sequence is as follows:
$$c_{t, s}=2 C_p \sqrt{\frac{\ln t}{s}}$$
We define $\Delta_i=\mu^-\mu_i$ to be the loss of arm $i$. Recall that since the expected mean of each arm converges, the mean offset $\delta_{i, t}$ converges to zero. Therefore, there exists a time $N_0(\epsilon)$ at which the uncertainty of the true mean rewards are guaranteed to be within a factor $\epsilon$ of their distance from the optimal mean, and the uncertainty of the optimal mean is guaranteed to be within the same factor $\epsilon$ of its distance to to closest suboptimal mean. Therefore, even though we still have some uncertainty as to what the true means really are, we have enough information to know which is probably the best, as $\mu_{N_0(c)}$ is closer to $\mu^$ than it is to any $\mu_{i, N_0(\epsilon)}$. More formally, $N_0(\epsilon): \mathbb{R} \rightarrow \mathbb{N}$ is a function which returns the minimum $t$ for which $2\left|\delta_{i, t}\right| \leq \epsilon \Delta_i$ for all arms $i$, and $2\left|\delta_{j^*, t}\right| \leq \epsilon \min _i \Delta_i$. Under Assumption 1, and using the preceding definitions, we can upper bound the expected number of times that a suboptimal arm will be played by UCB1 when the means are allowed to drift.

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|UCT

UCT is simply a marriage of MCTS and UCB1. The main idea is to treat each internal node in MCTS as a $K$-armed bandit, where the arms are the actions available at that state. $\mathrm{A}$ separate instance of UCB1 is run on each internal state, modified to accommodate the drifting means present in this problem. The necessity of the drifting means generalization can be seen by realizing that UCT is essentially a tree of separate UCB1 instances. The means of the action rewards in each of these multi-armed bandit problems depend on the instances lower in the tree. As we explore the tree, we gain a clearer understanding of the true means, reducing noisiness introduced by a partial exploration deeper in the tree. Simply put, UCT is MCTS where we use UCB1 to select an action in the tree policy. See Algorithm 2 for pseudo code of the action selection.
Recall Theorems 4 and 5 from the previous section. These state that when means can drift in the multi armed bandit problem, UCB still finds the optimal solution quickly with high probability, and given enough time it always finds the optinal arn. When incorporating UCB into MCTS, we also get these properties in UCT, as stated in Theorem 6 .

Theorem 6. Consider algorithm UCT running on a game tree of depth $D$, branching factor $K$ with stochastic payoffs at the leaves. Assume that the payoffs lie in the interval [0,1]. Then the bias of the estimated expected payoff, $\bar{X}_n$, is $O\left(\left(K D \log (n)+K^D\right) / n\right)$. Further, the failure probability at the root converges to zero as the number of samples grows to infinity.

A detailed proof of the preceding theorem is beyond the scope of these lecture notes (see [5] for such a proof), however we will provide a sketch of the proof. We must induct on $D$ to prove the theorem. In the base case of $D=1$, UCT is reduced to a single instantiation of UCB. Therefore Theorems 4 and 5 lead directly to our desired result. In the inductive case, we assume the result holds for any tree of depth $D$ – 1 , and consider a tree of depth $D$. All of the children of the root node are trees of depth $D-1$ which, by induction, satisfy the theorem. Therefore, we consider only the single UCB instance at the root. Using the theorems from the previous section (particularly Theorems 2,4 , and 5 ), we can then show that the theorem holds at the root as well.

# 蒙特卡洛树搜索代考

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Monte Carlo Tree Search

Veft{\mathca|{F}{i, t}\right } } \text { -改编和 } X i , t \text { 有条件地独立于 } \mathcal { F } i , t + 1 , \mathcal { F } i , t + 2 , \ldots \text { 给予 } \mathcal { F } i , t – 1 ^ { 1 } \text { . 然后 } $0 \leq X i, t \leq 1$ 和极限 $\mu_{i, n}=\mathbb{E}[\bar{X} i n]$ 存在。此外，我们假设存在一个常数 $C_p>0$ 和一个整数 $N_p$ 这样 对于 $n \geq N_p$. 对于任何 $\delta>0, \Delta_n(\delta)=C_p \sqrt{n \ln (1 / \delta)}$ ，以下界限成立:
$$\mathbb{P}\left(n \bar{X} i, n \geq n \mathbb{E}[\bar{X} i, n]+\Delta_n(\delta)\right) \leq \delta \quad \mathbb{P}\left(n \bar{X} i, n \leq n \mathbb{E}[\bar{X} i, n]-\Delta_n(\delta)\right) \leq \delta$$

$$c_{t, s}=2 C_p \sqrt{\frac{\ln t}{s}}$$

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|UCT

UCT 只是 MCTS 和 UCB1 的结合。主要思想是将 MCTS 中的每个内部节点视为一个 $K$-armed bandit， 其中 arms 是该状态下可用的动作。AUCB1 的单独实例在每个内部状态上运行，经过修改以适应此问题 中存在的漂移方法。通过认识到 UCT 本质上是一棵独立的 UCB1 实例树，可以看出漂移均值泛化的必要 性。在这些多臂老虎机问题中，行动奖励的方式取决于树中较低的实例。当我们探索这棵树时，我们对真 实均值有了更清晰的理解，减少了在树的更深处进行部分探索所引入的噪音。简单地说，UCT 是 MCTS， 我们使用 UCB1 在树策略中选择一个动作。有关动作选择的伪代码，请参见算法 2。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|NE591

statistics-lab™ 为您的留学生涯保驾护航 在代写蒙特卡洛树搜索Monte Carlo tree search方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写蒙特卡洛树搜索Monte Carlo tree search代写方面经验极为丰富，各种代写蒙特卡洛树搜索Monte Carlo tree search相关的作业也就用不着说。

• 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

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Multi-armed Bandits

It has been shown that the best possible policy $\pi(t)$ has a regret bounded by $\Omega(\ln n)$, and any policy that is within a constant factor of this is considered to balance the exploration-exploitation trade-off in the tree policy well. One such algorithm is UCB1. [3]

The high level concept of UCB1 is simple: pick the arm with the highest confidence bound on its mean reward. To facilitate this, UCB1 tracks the empirical average of each arm’s rewards $\bar{X}{i, T_c(t-1)}$ up to time $t$. As a result, when incorporating UCB into MCTS, it becomes necessary to track empirical average reward in MCTS. The policy is outlined in Equation 2, where $c{t, s}$ is a bias sequence over time $t$ and sample size $s$, picked such that both Equation 3 and Equation 4 are satisfied.
\begin{aligned} \mathrm{UCB} 1(t)= & \arg \max \left{\bar{X}{i, T_i(t-1)}+c{t-1, T_i(t-1)}\right} \ & \mathbb{P}\left(\bar{X}{i, s} \geq \mu_i+c{t, s}\right) \leq t^{-4} \ & \mathbb{P}\left(\bar{X}{i, s} \leq \mu_i-c{t, s}\right) \leq t^{-4} \end{aligned}
For standard multi-armed bandits, the bias sequence $c_{t, s}=\sqrt{\frac{2 \ln t}{s}}$ is appropriate. Note that the bias sequence is proportional to time and inversely proportional to sample size. It is designed so that confidence bounds expand slowly over time, but contract quickly as we gather more data.

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Drifting Multi-armed Bandits

For more detail on any of the theorems discussed in this section, please refer to [5]. In order to understand UCT, we must first relax the restraints on $K$-armed bandits to allow each arm’s mean $\mu_i$ to change over time t. While we can no longer rely on the assumption that the mean of each arm is fixed from $t=1$ onward, we can assume that the expected value of the empirical averages converge. We let $\bar{X}{i, n}=\frac{1}{n} \sum{t=1}^n X_{i, t}$ be the empirical average of arm $i$ at time $n$, and $\mu_{i, n}=\mathbb{E}\left[\tilde{X}{i, n}\right]$ be its expected value. Therefore, we now have a sequence of expected means for each arm $i$, namely $\mu{i, n}$. We assume that these expected means eventually converge to one final mean $\mu_i=\lim {n \rightarrow \infty} \mu{i, n}$. We further define a sequence of offsets for each arm as $\delta_{i, n}=\mu_{i, n}-\mu_i$. We also make the following assumptions about the reward sequence:

Assumption 1. Fix $1 \leq i \leq K$. Let $\left{\mathcal{F}{i, t}\right}_t$ be a fultration such that $\left{X{i, t}\right}_t$ is $\left{\mathcal{F}{i, t}\right}$-adapted and $X{i, t}$ is conditionally independent of $\mathcal{F}{i, t+1}, \mathcal{F}{i, t+2}, \ldots$ given $\mathcal{F}{i, t-1}{ }^1$. Then $0 \leq X{i, t} \leq 1$ and the limit of $\mu_{i, n}=\mathbb{E}\left[\bar{X}{i n}\right]$ exists. Further, we assume that there exist a constant $C_p>0$ and an integer $N_p$ such that for $n \geq N_p$. for any $\delta>0, \Delta_n(\delta)=C_p \sqrt{n \ln (1 / \delta)}$, the following bounds hold: \begin{aligned} & \mathbb{P}\left(n \bar{X}{i, n} \geq n \mathbb{E}\left[\bar{X}{i, n}\right]+\Delta_n(\delta)\right) \leq \delta \ & \mathbb{P}\left(n \bar{X}{i, n} \leq n \mathbb{E}\left[\bar{X}{i, n}\right]-\Delta_n(\delta)\right) \leq \delta \end{aligned} This assumption allows us to define a bias sequence $c{t, s}$ for time $t$ and sample size $s$ which satisfies Equation 3 and Equation 4. This sequence is as follows:
$$c_{t, s}=2 C_p \sqrt{\frac{\ln t}{s}}$$
We define $\Delta_i=\mu^-\mu_i$ to be the loss of arm $i$. Recall that since the expected mean of each arm converges, the mean offset $\delta_{i, t}$ converges to zero. Therefore, there exists a time $N_0(\epsilon)$ at which the uncertainty of the true mean rewards are guaranteed to be within a factor $\epsilon$ of their distance from the optimal mean, and the uncertainty of the optimal mean is guaranteed to be within the same factor $\epsilon$ of its distance to to closest suboptimal mean. Therefore, even though we still have some uncertainty as to what the true means really are, we have enough information to know which is probably the best, as $\mu_{N_0(c)}$ is closer to $\mu^$ than it is to any $\mu_{i, N_0(\epsilon)}$. More formally, $N_0(\epsilon): \mathbb{R} \rightarrow \mathbb{N}$ is a function which returns the minimum $t$ for which $2\left|\delta_{i, t}\right| \leq \epsilon \Delta_i$ for all arms $i$, and $2\left|\delta_{j^*, t}\right| \leq \epsilon \min _i \Delta_i$. Under Assumption 1, and using the preceding definitions, we can upper bound the expected number of times that a suboptimal arm will be played by UCB1 when the means are allowed to drift.

# 蒙特卡洛树搜索代考

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Multi-armed Bandits

UCB1 的高级概念很简单: 选择对其平均奖励具有最高置信度的手臂。为了促进这一点， UCB1 跟踪每个 手臂奖励的经验平均值 $\bar{X} i, T_c(t-1)$ 及时 $t$. 因此，将 UCB 纳入 MCTS 时，有必要跟踪 MCTS 中的经验 平均奖励。公式 2 概述了该策略，其中 $c t, s$ 是随时间变化的偏置序列 $t$ 和样本量 $s$ ，被选中使得方程 3 和方 程 4 都得到满足。

Ibegin{aligned $} \backslash \operatorname{mathrm}{U C B} 1(\mathrm{t})=\& \backslash \arg \backslash \max \backslash \mathrm{left}\left{\backslash b a r{X}\left{\mathrm{i}, T_{-} \mathrm{i}(\mathrm{t}-1)\right}+c\left{\mathrm{t}-1, T_{-} \mathrm{i}(\mathrm{t}-1)\right} \backslash r i g h t\right} \backslash \& \backslash \operatorname{mathbb}{\mathrm{P}} \backslash \mathrm{ft}(\mathbf{x}$

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Drifting Multi-armed Bandits

Veft{\mathcal{F}{i, t}\right } } \text { -改编和 } X i , t \text { 有条件地独立于 } \mathcal { F } i , t + 1 , \mathcal { F } i , t + 2 , \ldots \text { 给予 } \mathcal { F } i , t – 1 ^ { 1 } \text { . 然后 } $0 \leq X i, t \leq 1$ 和极限 $\mu_{i, n}=\mathbb{E}[\bar{X} i n]$ 存在。此外，我们假设存在一个常数 $C_p>0$ 和一个整数 $N_p$ 这样 对于 $n \geq N_p$. 对于任何 $\delta>0, \Delta_n(\delta)=C_p \sqrt{n \ln (1 / \delta)}$ ，以下界限成立:
$$\mathbb{P}\left(n \bar{X} i, n \geq n \mathbb{E}[\bar{X} i, n]+\Delta_n(\delta)\right) \leq \delta \quad \mathbb{P}\left(n \bar{X} i, n \leq n \mathbb{E}[\bar{X} i, n]-\Delta_n(\delta)\right) \leq \delta$$

$$c_{t, s}=2 C_p \sqrt{\frac{\ln t}{s}}$$

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## MATLAB代写

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