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统计代写|统计推断代考Statistical Inference代写|Monty Hall Problem

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统计代写|统计推断代考Statistical Inference代写|Monty Hall Problem

This problem was introduced in Section 2.6.
Example 5.1 Is it better to switch doors? – Monty Hall Problem revisited
You may recall that we were presented with a choice of 3 doors where a car is behind one and goats behind the others. Having picked one, the host opens up a door with a goat, and offers you the opportunity to change your answer. In order to assess the probabilities, we must remember that
1 the host never opens your door
2 the host always opens a door with a goat
We’ll go through a specific example, that of you choosing door 1 and the host opening door 2. The analysis proceeds in identical ways for the other possibilities. We apply the Bayes’ Recipe, where the models under consideration are

“car behind door 1 “

“car behind door 2 “

“car behind door 3 “
The Bayes’ Recipe proceeds as follows
1 Specify the prior probabilities for the models being considered
\begin{aligned} & P(\operatorname{car} 1 \mid \text { you } 1)=0.333 \ & P(\operatorname{car} 2 \mid \text { you } 1)=0.333 \ & P(\operatorname{car} 3 \mid \text { you } 1)=0.333 \end{aligned}
where, for example, $P$ (car $1 \mid$ you 1 ) represents the probability that the door contains the car given that you chose door 1 . Since your choice of door doesn’t add any information about the location of the car, all of the probabilities are equal.
2 Write the top of Bayes’ Rule for all models being considered
\begin{aligned} & P(\text { car } 1 \mid \text { you 1, host } 2) \sim P(\text { host } 2 \mid \text { you 1, car 1) } P(\text { car 1 } 1 \text { you 1 }) \ & P(\text { car 2|you 1, host 2) } \sim P(\text { host 2 } \mid \text { you 1, car 2) } P(\text { car 2|you 1 }) \ & P(\text { car } 3 \mid \text { you 1, host 2) }) \sim P(\text { host } 2 \mid \text { you 1, car } 3) P(\text { car } 3 \mid \text { you } 1) \ & \end{aligned}
3 Put in the likelihood and prior values
Due the restrictions on the host above, the host cannot open a door with a car, so $P$ (host $2 \mid$ you 1, car 2 ) $=0$. In the case where you choose door 1 and the car is also behind door, the host has the freedom to choose either door 2 or door 3 , so $P$ (host $2 \mid$ you 1, car 1 ) = 0.5 . Where the information comes in is when the car is behind door 3 and you’ve chosen door 1 . In that case, the host cannot open your door (door 1) or the door with the car (door 3 ) and must open door 2 . Thus, $P\left(\right.$ host $2 \mid$ you $\left.1, \operatorname{car}_3\right)=1$.

统计代写|统计推断代考Statistical Inference代写|Bent Coins

Imagine we have a series of coins bent by various amounts (Figure 6.1). If the coin is bent completely in half, then we could have the coin always flip heads (i.e. $P$ (heads) $=1$ ) or tails (i.e. $P($ tails $)=1$ ) depending on how it is bent. If you don’t bend the coin at all then we’d have a fair coin $(P$ (heads) $=P$ (tails) $=0.5$ ). So, let’s say that we have a collection of bent coins which are bent by different amounts. For convenience we will number them from o to 10 . The Table 6.1 summarizes the probability of each coin flipping heads.

Now I have the following scenario ${ }^1$, with a few questions.
Imagine I have taken a random coin from my collection, flipped it and observed the following data:
T T T H T H T T T T T H (i.e. 9 tails and 3 heads)
I From this data, which coin do I most likely have?

2 Can we be significantly confident that this particular coin will result in more tails than heads in the future?
The way we’ve set up this problem is exactly like the model comparison example with the High and Low Deck (Section 4.1), except in this case we have 11 models (one for each coin). Applying the Bayes’ Recipe we have
1 Specify the prior probabilities for the models being considered. Given no further information, we select a uniform distribution for the prior (i.e. all models are initially equally probable):
\begin{aligned} P\left(M_0\right) & =1 / 11 \ P\left(M_1\right) & =1 / 11 \ & \vdots \ P\left(M_{10}\right) & =1 / 11 . \end{aligned}
where $M_0$ is the model defined by “we’re flipping coin o,” $M_1$ is the model defined by “we’re flipping coin 1 ,” etc…
2 Write the top of Bayes’ Rule for all models being considered:
\begin{aligned} P\left(M_0 \mid \text { data }=9 T, 3 H\right) & \sim P\left(\text { data }=9 T, 3 H \mid M_0\right) P\left(M_0\right) \ P\left(M_1 \mid \text { data }=9 T, 3 H\right) & \sim P\left(\text { data }=9 T, 3 H \mid M_1\right) P\left(M_1\right) \ & \vdots \ P\left(M_{10} \mid \text { data }=9 T, 3 H\right) & \sim P\left(\text { data }=9 T, 3 H \mid M_{10}\right) P\left(M_{10}\right) . \end{aligned}

3 Put in the likelihood and prior values. Here we are drawing from a binomial distribution for the likelihood:
\begin{aligned} P\left(M_0 \mid \text { data }=9 T, 3 H\right) & \sim\left(\begin{array}{c} 12 \ 3 \end{array}\right) 0.0^3 \times(1-0.0)^9 \times 1 / 11 \ P\left(M_1 \mid \text { data }=9 T, 3 H\right) & \sim\left(\begin{array}{c} 12 \ 3 \end{array}\right) 0.1^3 \times(1-0.1)^9 \times 1 / 11 \ & \vdots \ P\left(M_{10} \mid \text { data }=9 T, 3 H\right) & \sim\left(\begin{array}{c} 12 \ 3 \end{array}\right) 1.0^3 \times(1-1.0)^9 \times 1 / 11 . \end{aligned}
4 Add these values for all models: see Table 6.2.
5 Divide each of the values by this sum, $K$, to get the final probabilities: see Table 6.2.

When we are dealing with this many models, it is easier to plot the results, shown in Figure 6.2. We are now in a position to address the questions posed at the beginning of the section.

统计推断代写

统计代写|统计推断代考Statistical Inference代写|Monty Hall Problem

” 1号门后面的车”

” 2号门后面的车”

” 3号门后面的车”

＄＄

& P(\operatorname{car} 1 \mid \text {you} 1)=0.333 \
& P(\operatorname{car} 2 \mid \text {you} 1)=0.333 \
& P(\operatorname{car} 3 \mid \text {you} 1)=0.333

＄＄

＄＄

& P(\text {car} 1 \mid \text {you 1, host} 2) \sim P(\text {host} 2 \mid \text {you 1, car 1)} P(\text {car 1} 1 \text {you 1}) \
& P(\text{车2|你1，主机2)}\sim P(\text{主机2}\mid \text{你1，主机2)}P(\text{车2|你1})\
& P(\text {car} 3 \mid \text {you 1, host 2)}) \sim P(\text {host} 2 \mid \text {you 1, host} 3) P(\text {car} 3 \mid \text {you} 1) \

＄＄

＄＄

P\left(M_0\ mid \text {data}=9 T, 3 H\right) & P\left(\text {data}=9 T, 3 H\ mid M_0\right) P\left(M_0\right) \
P\left(M_1\ mid \text {data}=9 T, 3 H\right) & P\left(\text {data}=9 T, 3 H\ mid M_1\right) P\left(M_1\right) \
& \vdots \
P\left(M_{10}\ mid \text {data}=9 T, 3 H\right) & \sim P\left(\text {data}=9 T, 3 H\ mid M_{10}\right) P\left(M_{10}\right)

＄＄

＄＄

P\left(M_0 \mid \text {data}=9 T, 3 H\right) & \sim\left(\begin{array}{c})
12 \
3.
\end{array}\right) 0.0^3 \times(1-0.0)^9 \times 1 / 11 \
P\left(M_1 \mid \text {data}=9 T, 3 H\right) & \sim\left(\begin{a . 1}

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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。