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# 统计代写|贝叶斯分析代考Bayesian Analysis代写|The proportionality formula

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## 统计代写|贝叶斯分析代考Bayesian Analysis代写|The proportionality formula

Observe that $f(y)$ is a constant with respect to $\theta$ in the Bayesian equation
$$f(\theta \mid y)=f(\theta) f(y \mid \theta) / f(y)$$
which means that we may also write the equation as
$$f(\theta \mid y)=\frac{f(\theta) f(y \mid \theta)}{k}$$
or as
$$f(\theta \mid y)=c f(\theta) f(y \mid \theta),$$
where $k=f(y)$ and $c=1 / k$.
We may also write
$$f(\theta \mid y) \propto f(\theta) f(y \mid \theta),$$
where $\propto$ is the proportionality sign.

Equivalently, we may write
$$f(\theta \mid y) \stackrel{\theta}{\infty} f(\theta) f(y \mid \theta)$$
to emphasise that the proportionality is specifically with respect to $\theta$.
Another way to express the last equation is
$$f(\theta \mid y) \propto f(\theta) \times L(\theta \mid y)$$
where $L(\theta \mid y)$ is the likelihood function (defined as the model density $f(y \mid \theta)$ multiplied by any constant with respect to $\theta$, and viewed as a function of $\theta$ rather than of $y$ ).
The last equation may also be stated in words as:
The posterior is proportional to the prior times the likelihood.
These observations indicate a shortcut method for determining the required posterior distribution which obviates the need for calculating $f(y)$ (which may be difficult).

This method is to multiply the prior density (or the kernel of that density) by the likelihood function and try to identify the resulting function of $\theta$ as the density of a well-known or common distribution.
Once the posterior distribution has been identified, $f(y)$ may then be obtained easily as the associated normalising constant.

## 统计代写|贝叶斯分析代考Bayesian Analysis代写|Finite and infinite population inference

In the last example (Exercise 1.8), with the model:
\begin{aligned} & (y \mid \theta) \sim \operatorname{Binomial}(n, \theta) \ & \theta \sim \operatorname{Beta}(\alpha, \beta), \end{aligned}
the quantity of interest $\theta$ is the probability of success on a single Bernoulli trial.

This quantity may be thought of as the average of a hypothetically infinite number of Bernoulli trials. For that reason we may refer to derivation of the posterior distribution,
$$(\theta \mid y) \sim \operatorname{Beta}(\alpha+y, \beta+n-y)$$
as infinite population inference.
In contrast, for the ‘buses’ example further above (Exercise 1.6), which involves the model:
\begin{aligned} & f(y \mid \theta)=1 / \theta, y=1, \ldots, \theta \ & f(\theta)=1 / 5, \theta=1, \ldots, 5, \end{aligned}
the quantity of interest $\theta$ represents the number of buses in a population of buses, which of course is finite.
Therefore derivation of the posterior,
$$f(\theta \mid y)=\left{\begin{array}{l} 20 / 47, \theta=3 \ 15 / 47, \theta=4 \ 12 / 47, \theta=5 \end{array}\right.$$
may be termed finite population inference.
Another example of finite population inference is the ‘balls in a box’ example (Exercise 1.7), where the model is:
\begin{aligned} & (y \mid \theta) \sim \operatorname{Hyp}(N, \theta, n) \ & \theta \sim D U(1, \ldots, N), \end{aligned}
and where the quantity of interest $\theta$ is the number of red balls initially in the selected box $(1,2, \ldots, 8$ or 9$)$.

# 贝叶斯分析代写

## 统计代写|贝叶斯分析代考Bayesian Analysis代写|The proportionality formula

$$f(\theta \mid y)=f(\theta) f(y \mid \theta) / f(y)$$

$$f(\theta \mid y)=\frac{f(\theta) f(y \mid \theta)}{k}$$

$$f(\theta \mid y)=c f(\theta) f(y \mid \theta),$$

$$f(\theta \mid y) \propto f(\theta) f(y \mid \theta),$$
$\propto$是比例符号。

$$f(\theta \mid y) \stackrel{\theta}{\infty} f(\theta) f(y \mid \theta)$$

$$f(\theta \mid y) \propto f(\theta) \times L(\theta \mid y)$$

## 统计代写|贝叶斯分析代考Bayesian Analysis代写|Finite and infinite population inference

\begin{aligned} & (y \mid \theta) \sim \operatorname{Binomial}(n, \theta) \ & \theta \sim \operatorname{Beta}(\alpha, \beta), \end{aligned}

$$(\theta \mid y) \sim \operatorname{Beta}(\alpha+y, \beta+n-y)$$

\begin{aligned} & f(y \mid \theta)=1 / \theta, y=1, \ldots, \theta \ & f(\theta)=1 / 5, \theta=1, \ldots, 5, \end{aligned}

$$f(\theta \mid y)=\left{\begin{array}{l} 20 / 47, \theta=3 \ 15 / 47, \theta=4 \ 12 / 47, \theta=5 \end{array}\right.$$

\begin{aligned} & (y \mid \theta) \sim \operatorname{Hyp}(N, \theta, n) \ & \theta \sim D U(1, \ldots, N), \end{aligned}

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## MATLAB代写

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