Posted on Categories:Bayesian Analysis, 统计代写, 统计代考, 贝叶斯分析

# 统计代写|贝叶斯分析代考Bayesian Analysis代写|Alternative solution

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## 统计代写|贝叶斯分析代考Bayesian Analysis代写|Alternative solution

The above results can also be obtained by working through in the style of the solutions to previous exercises, as follows. Before the data is observed, the Bayesian model may be written:
\begin{aligned} & f(s \mid y, \theta)=\left(\begin{array}{l} N \ n \end{array}\right)^{-1}=\left(\begin{array}{l} 4 \ 2 \end{array}\right)^{-1}=\frac{1}{6}, \ & s=(1,2),(1,3),(1,4),(2,3),(2,4),(3,4) \ & f(y \mid \theta)=\frac{1}{4}, \quad y=(\theta, \theta, \theta, \theta),(\theta, \theta, \theta, 1-\theta), \ & (\theta, \theta, 1-\theta, 1-\theta),(\theta, 1-\theta, 1-\theta, 1-\theta) \ & f(\theta)=1 / 2, \theta=0,1 \quad \text { (the prior density of the parameter). } \end{aligned}
(the prior density of the parameter).
The observed data is $D=\left(s, y_s\right)=((2,3),(1,1))$. At this particular value of the data:
$f(s \mid y, \theta)=\frac{1}{6}, s=(2,3) \quad$ (the value of $s$ actually observed)
$f(y \mid \theta)=\frac{1}{4}, \quad y=(0,1,1,1)$ and $\theta=0$,
$y \in{(1,1,1,1),(1,1,1,0)}$ and $\theta=1$ (where we need only consider values of $y$ consistent with the data)
$f(\theta)=1 / 2, \theta=0,1 \quad$ (since both values of $\theta$ are still possible, i.e. consistent with the observed data).
With the quantities $s=(2,3), y_s=\left(y_2, y_3\right)=(1,1)$ and $y_r=\left(y_1, y_4\right)$ all fixed at these values, the joint density of all quantities in the model may be written
\begin{aligned} & f(\theta, s, y)=f\left(\theta, s, y_s, y_r\right)=f(\theta) f\left(y_s, y_r \mid \theta\right) f\left(s \mid y_s, y_r, \theta\right) \ & =\frac{I(\theta \in{0,1})}{2} \times \frac{I(y=(0,1,1,1), \theta=0)+I(y \in{(1,1,1,1),(1,1,1,0)}, \theta=1)}{4} \times \frac{1}{6} \ & \theta, y_r \ & \propto I\left(y_r=(0,1), \theta=0\right)+I\left(y_r \in{(1,1),(1,0)}, \theta=1\right) \text {. } \ & \end{aligned}

## 统计代写|贝叶斯分析代考Bayesian Analysis代写|The case of SRSWR

In the case of SRSWR, the sampling density
$$f(I \mid y, \lambda)=\frac{n !}{\prod_{i=1}^N I_i} \prod_{i=1}^N\left(\frac{y_i}{y_T}\right)^{I_i}$$
changes to
$$f(I \mid y, \lambda)=\frac{n !}{\prod_{i=1}^N I_i} \prod_{i=1}^N\left(\frac{1}{N}\right)^{I_i}=\frac{n !}{N^n \prod_{i=1}^N I_i},$$
which we note does not depend on $\lambda$ or $y_{r T}$ and so can be ‘ignored’.
The result is then almost the same as before, the only difference being that the term
$$\prod_{i=1}^N y_T^{L_i}=y_T^n=\left(y_{s T}+y_{r T}\right)^n$$
in
$$k\left(y_{r T}\right)=\left(\frac{N-d}{N+\tau}\right)^{y_{r T}} \frac{\Gamma\left(\eta+y_{s T}+y_{r T}\right)}{y_{r T} !\left(y_{s T}+y_{r T}\right)^n}$$
is replaced by 1 .
Thus under SRSWR we find that
$$f\left(y_{r T} \mid D\right)=\frac{K\left(y_{r T}\right)}{C}, y_{r T}=0,1,2, \ldots,$$
where
$$K\left(y_{r T}\right)=\left(\frac{N-d}{N+\tau}\right)^{y_{r T}} \frac{\Gamma\left(\eta+y_{s T}+y_{r T}\right)}{y_{r T} ! \times 1}$$
and
$$C=\sum_{y_{r r}=0}^{\infty} K\left(y_{r T}\right) .$$
As regards the posterior distribution of $\lambda$ under SRSWR, this need no longer be expressed as an infinite mixture of gamma distributions but simply as
$$(\lambda \mid D) \sim G\left(\eta+y_{s T}, \tau+d\right)$$

# 贝叶斯分析代写

## 统计代写|贝叶斯分析代考Bayesian Analysis代写|Alternative solution

\begin{aligned} & f(s \mid y, \theta)=\left(\begin{array}{l} N \ n \end{array}\right)^{-1}=\left(\begin{array}{l} 4 \ 2 \end{array}\right)^{-1}=\frac{1}{6}, \ & s=(1,2),(1,3),(1,4),(2,3),(2,4),(3,4) \ & f(y \mid \theta)=\frac{1}{4}, \quad y=(\theta, \theta, \theta, \theta),(\theta, \theta, \theta, 1-\theta), \ & (\theta, \theta, 1-\theta, 1-\theta),(\theta, 1-\theta, 1-\theta, 1-\theta) \ & f(\theta)=1 / 2, \theta=0,1 \quad \text { (the prior density of the parameter). } \end{aligned}
(参数的先验密度)。

$f(s \mid y, \theta)=\frac{1}{6}, s=(2,3) \quad$(实际观察到的$s$的值)
$f(y \mid \theta)=\frac{1}{4}, \quad y=(0,1,1,1)$和$\theta=0$，
$y \in{(1,1,1,1),(1,1,1,0)}$和$\theta=1$(我们只需要考虑$y$与数据一致的值)
$f(\theta)=1 / 2, \theta=0,1 \quad$(因为$\theta$的两个值仍然是可能的，即与观测数据一致)。

\begin{aligned} & f(\theta, s, y)=f\left(\theta, s, y_s, y_r\right)=f(\theta) f\left(y_s, y_r \mid \theta\right) f\left(s \mid y_s, y_r, \theta\right) \ & =\frac{I(\theta \in{0,1})}{2} \times \frac{I(y=(0,1,1,1), \theta=0)+I(y \in{(1,1,1,1),(1,1,1,0)}, \theta=1)}{4} \times \frac{1}{6} \ & \theta, y_r \ & \propto I\left(y_r=(0,1), \theta=0\right)+I\left(y_r \in{(1,1),(1,0)}, \theta=1\right) \text {. } \ & \end{aligned}

## 统计代写|贝叶斯分析代考Bayesian Analysis代写|The case of SRSWR

$$f(I \mid y, \lambda)=\frac{n !}{\prod_{i=1}^N I_i} \prod_{i=1}^N\left(\frac{y_i}{y_T}\right)^{I_i}$$

$$f(I \mid y, \lambda)=\frac{n !}{\prod_{i=1}^N I_i} \prod_{i=1}^N\left(\frac{1}{N}\right)^{I_i}=\frac{n !}{N^n \prod_{i=1}^N I_i},$$

$$\prod_{i=1}^N y_T^{L_i}=y_T^n=\left(y_{s T}+y_{r T}\right)^n$$

$$k\left(y_{r T}\right)=\left(\frac{N-d}{N+\tau}\right)^{y_{r T}} \frac{\Gamma\left(\eta+y_{s T}+y_{r T}\right)}{y_{r T} !\left(y_{s T}+y_{r T}\right)^n}$$

$$f\left(y_{r T} \mid D\right)=\frac{K\left(y_{r T}\right)}{C}, y_{r T}=0,1,2, \ldots,$$

$$K\left(y_{r T}\right)=\left(\frac{N-d}{N+\tau}\right)^{y_{r T}} \frac{\Gamma\left(\eta+y_{s T}+y_{r T}\right)}{y_{r T} ! \times 1}$$

$$C=\sum_{y_{r r}=0}^{\infty} K\left(y_{r T}\right) .$$

$$(\lambda \mid D) \sim G\left(\eta+y_{s T}, \tau+d\right)$$

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## MATLAB代写

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