### 统计代写|贝叶斯分析代写Bayesian Analysis代考|The posterior distribution

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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

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|The posterior distribution

Bayesian inference requires determination of the posterior probability distribution of $\theta$. This task is equivalent to finding the posterior pdf of $\theta$, which may be done using the equation
$$f(\theta \mid y)=\frac{f(\theta) f(y \mid \theta)}{f(y)} .$$
Here, $f(y)$ is the unconditional (or prior) pdf of $y$, as given by $f(y)=\int f(y \mid \theta) d F(\theta)= \begin{cases}\int_{\theta} f(\theta) f(y \mid \theta) d \theta & \text { if } \theta \text { is continuous } \ \sum_{\theta} f(\theta) f(y \mid \theta) & \text { if } \theta \text { is discrete. }\end{cases}$
Note: Here, $\int f(y \mid \theta) d F(\theta)$ is a Lebesgue-Stieltjes integral, which may need evaluating by breaking the integral into two parts in the case where $\theta$ has a mixed distribution. In the continuous case, think of $d F(\theta)$ as $\frac{d F(\theta)}{d \theta} d \theta=f(\theta) d \theta$

Consider six loaded dice with the following properties. Die A has probability $0.1$ of coming up 6, each of Dice B and C has probability $0.2$ of coming up 6, and each of Dice D, E and F has probability $0.3$ of coming up $6 .$
A die is chosen randomly from the six dice and rolled twice. On both occasions, 6 comes up.

What is the posterior probability distribution of $\theta$, the probability of 6 coming up on the chosen die.

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|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
$$f(\theta \mid y)^{\theta} \propto f(\theta) f(y \mid \theta)$$
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|贝叶斯分析代写Bayesian Analysis代考|Continuous parameters

The examples above have all featured a target parameter which is discrete. The following example illustrates Bayesian inference involving a continuous parameter. This case presents no new problems, except that the prior and posterior densities of the parameter may no longer be interpreted directly as probabilities.

Consider the following Bayesian model:
\begin{aligned} &(y \mid \theta) \sim \operatorname{Binomial}(n, \theta) \ &\theta \sim \operatorname{Beta}(\alpha, \beta) \quad \text { (prior). } \end{aligned}
Find the posterior distribution of $\theta$.
Solution to Exercise $1.8$
The posterior density is
\begin{aligned} f(\theta \mid y) & \propto f(\theta) f(y \mid \theta) \ &=\frac{\theta^{\alpha-1}(1-\theta)^{\beta-1}}{B(\alpha, \beta)} \times\left(\begin{array}{l} n \ y \end{array}\right) \theta^{y}(1-\theta)^{n-y} \ & \propto \theta^{\alpha-1}(1-\theta)^{\beta-1} \times \theta^{y}(1-\theta)^{n-y} \quad \text { (ignoring constants which } \ &=\theta^{(\alpha+y)-1}(1-\theta)^{(\beta+n-y)-1}, 0<\theta<1 \end{aligned}
This is the kernel of the beta density with parameters $\alpha+y$ and $\beta+n-y$. It follows that the posterior distribution of $\theta$ is given by
$$(\theta \mid y) \sim \operatorname{Beta}(\alpha+y, \beta+n-y) \text {, }$$
and the posterior density of $\theta$ is (exactly)
$$f(\theta \mid y)=\frac{\theta^{(\alpha+y)-1}(1-\theta)^{(\beta+n-y)-1}}{B(\alpha+y, \beta+n-y)}, 0<\theta<1 .$$
For example, suppose that $\alpha=\beta=1$, that is, $\theta \sim B e t a(1,1)$.
Then the prior density is $f(\theta)=\frac{\theta^{1-1}(1-\theta)^{1-1}}{B(1,1)}=1,0<\theta<1$.
Thus the prior may also be expressed by writing $\theta \sim U(0,1)$.
Also, suppose that $n=2$. Then there are three possible values of $y$, namely 0,1 and 2 , and these lead to the following three posteriors, respectively:
\begin{aligned} &(\theta \mid y) \sim \operatorname{Beta}(1+0,1+2-0)=\operatorname{Beta}(1,3) \ &(\theta \mid y) \sim \operatorname{Beta}(1+1,1+2-1)=\operatorname{Beta}(2,2) \ &(\theta \mid y) \sim \operatorname{Beta}(1+2,1+2-2)=\operatorname{Beta}(3,1) \end{aligned}
These three posteriors and the prior are illustrated in Figure 1.5.

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|The posterior distribution

F(θ∣是)=F(θ)F(是∣θ)F(是).

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

F(θ∣是)=F(θ)F(是∣θ)/F(是),

F(θ∣是)=F(θ)F(是∣θ)ķ

F(θ∣是)=CF(θ)F(是∣θ),

F(θ∣是)∝F(θ)F(是∣θ),

F(θ∣是)θ∝F(θ)F(是∣θ)

F(θ∣是)∝F(θ)×大号(θ∣是),

## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Continuous parameters

(是∣θ)∼二项式⁡(n,θ) θ∼贝塔⁡(一个,b) （事先的）。

F(θ∣是)∝F(θ)F(是∣θ) =θ一个−1(1−θ)b−1乙(一个,b)×(n 是)θ是(1−θ)n−是 ∝θ一个−1(1−θ)b−1×θ是(1−θ)n−是 （忽略常数  =θ(一个+是)−1(1−θ)(b+n−是)−1,0<θ<1

(θ∣是)∼贝塔⁡(一个+是,b+n−是),

F(θ∣是)=θ(一个+是)−1(1−θ)(b+n−是)−1乙(一个+是,b+n−是),0<θ<1.

(θ∣是)∼贝塔⁡(1+0,1+2−0)=贝塔⁡(1,3) (θ∣是)∼贝塔⁡(1+1,1+2−1)=贝塔⁡(2,2) (θ∣是)∼贝塔⁡(1+2,1+2−2)=贝塔⁡(3,1)

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