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

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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

## 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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