统计代写|DATA5711 Bayesian Analysis

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这是悉尼大学贝叶斯分析课程的代写成功案例。

DATA5711课程简介

Increased computing power has meant that many Bayesian methods can now be easily implemented and provide solutions to problems that have previously been intractable. Bayesian methods allow researchers to incorporate prior knowledge into their statistical models. This unit is made up of three distinct modules, each focusing on a different niche in the application of Bayesian statistical methods to complex data in, for example, geophysics, ecology and hydrology. These include (but are not restricted to) Bayesian methods and models; statistical inversion; approximate Bayesian inference for semiparametric regression. Across all modules you will develop expertise in Bayesian computational statistics. On completion of this unit you will be able to apply appropriate Bayesian methods to a variety of applications in science, and other data-heavy disciplines to develop a better understanding of the information inherent in complex datasets.

Prerequisites 

At the completion of this unit, you should be able to:

• LO1. Demonstrate a coherent and advanced understanding of key concepts in computational statistics.
• LO2. Apply fundamental principles and results in statistics to solve given problems.
• LO3. Distinguish and compare the properties of different types of statistical models and statistical methods applicable to them.
• LO4. Identify assumptions required for various statistical methods to be valid and devise methods for testing these assumptions.
• LO5. Devise statistical solutions to complex problems.
• LO6. Adapt various computational techniques to build software for solving particular statistical problems.
• LO7. Communicate coherent statistical arguments appropriately to student and expert audiences, both orally and through written work.

DATA5711 Bayesian Analysis HELP（EXAM HELP， ONLINE TUTOR）

Let $A$ be the event that the person has the disease, and let $B$ be the event that they test positive for the disease. Then:
$P(A)=0.01 \quad$ (the prior probability of the person having the disease)
$P(B \mid A)=0.9$ (the true positive rate, also called the sensitivity of the test)
$P(\bar{B} \mid \bar{A})=0.9$ (the true negative rate, also called the specificity of the test).

So: $P(A B)=P(A) P(B \mid A)=0.01 \times 0.9=0.009$
$$
P(\bar{A} B)=P(\bar{A}) P(B \mid \bar{A})=0.99 \times 0.1=0.099 .
$$
So the unconditional (or prior) probability of the person testing positive is $P(B)=P(A B)+P(\bar{A} B)=0.009+0.099=0.108$.

So the required posterior probability of the person having the disease is
$$
P(A \mid B)=\frac{P(A B)}{P(B)}=\frac{0.009}{0.108}=\frac{1}{12}=0.08333 .
$$

Let: $A=$ ‘The person has Type 1 blood’
$B=$ ‘The person has Type 2 blood’
$C=$ ‘The person has Type 3 blood’
$D=$ ‘The person has the disease’.
Then:
$$
\begin{array}{ll}
P(A)=0.1, & P(D \mid A)=0.02 \
P(B)=0.3, & P(D \mid B)=0.04 \
P(C)=0.6, & P(D \mid C)=0.03
\end{array}
$$
So:
$$
\begin{aligned}
P(D)= & P(A D)+P(B D)+P(C D) \
& =P(A) P(D \mid A)+P(B) P(D \mid B)+P(C) P(D \mid C) \
& =0.1 \times 0.02+0.3 \times 0.04+0.6 \times 0.03 \
& =0.002+0.012+0.018=0.032 .
\end{aligned}
$$
Hence: $P(C \mid D)=\frac{P(C D)}{P(D)}=\frac{0.018}{0.032}=\frac{9}{16}=56.25 \%$.

Recall in Exercise 1.1, where $A=$ ‘Person has disease’ and $B=$ ‘Person tests positive’, the relevant probabilities are $P(A)=0.01, P(B \mid A)=0.9$ and $P(\bar{B} \mid \bar{A})=0.9$, from which can be deduced that $P(A \mid B)=1 / 12$.
We now wish to test $H_0: A$ vs $H_1: \bar{A}$. So we calculate:
$$
\begin{aligned}
& \pi_0=P(A)=0.01, \pi_1=P(\bar{A})=0.99, P R O=\pi_0 / \pi_1=1 / 99 \
& p_0=P(A \mid B)=1 / 12, p_1=P(\bar{A} \mid B)=11 / 12, P O O=p_0 / p_1=1 / 11
\end{aligned}
$$
Hence the required Bayes factor is $B F=P O O / P R O=(1 / 11) /(1 / 99)=9$.
This means the positive test result has multiplied the odds of the person having the disease relative to not having it by a factor of 9 or $900 \%$. Another way to say this is that those odds have increased by $800 \%$.

Let $y$ be the number of times that 6 comes up on the two rolls of the chosen die, and let $\theta$ be the probability of 6 coming up on a single roll of that die. Then the Bayesian model is:

$$
\begin{aligned}
& (y \mid \theta) \sim \operatorname{Bin}(2, \theta) \
& f(\theta)= \begin{cases}1 / 6, & \theta=0.1 \
2 / 6, & \theta=0.2 \
3 / 6, & \theta=0.3 .\end{cases}
\end{aligned}
$$
In this case $y=2$ and so
$$
f(y \mid \theta)=\left(\begin{array}{l}
2 \
y
\end{array}\right) \theta^y(1-\theta)^{2-y}=\left(\begin{array}{l}
2 \
2
\end{array}\right) \theta^2(1-\theta)^{2-2}=\theta^2 .
$$
So $f(y)=\sum_\theta f(\theta) f(y \mid \theta)=\frac{1}{6}(0.1)^2+\frac{2}{6}(0.2)^2+\frac{3}{6}(0.3)^2=0.06$.
So $f(\theta \mid y)=\frac{f(\theta) f(y \mid \theta)}{f(y)}=\left{\begin{array}{c}(1 / 6) 0.1^2 / 0.06=0.02778, \quad \theta=0.1 \ (2 / 6) 0.2^2 / 0.06=0.22222, \quad \theta=0.2 \ (3 / 6) 0.3^2 / 0.06=0.75, \quad \theta=0.3 .\end{array}\right.$

MY-ASSIGNMENTEXPERT™可以为您提供SYDNEY DATA5711 BAYESIAN ANALYSIS贝叶斯分析课程的代写代考和辅导服务！