这是一次UCL伦敦大学学院决策与风险 Decision and Risk STAT0011课程的代考成功案例，学生和我们配合得比较好提前几周给了老师复习的资料，最后取得了满意的成绩。
这门课程是一门本科生的决策与风险的入门课程。
STAT0011旨在从贝叶斯和多次重复实验的角度介绍风险计算的基本思想，以及理性、一致决策的结构。用通过学习课程希望学生能够理解和计算金融风险度量；了解决策理论的概念；能够为风险事件找到适当的概率模型，并检查基本假设的有效性；熟悉检测风险水平随时间变化的方法。课程的主要内容包括贝叶斯推理简介：条件概率、贝叶斯定理、主观概率的激发。统计决策理论：预期损失、随机和非随机决策规则、决策原则、决策规则比较。风险度量： 风险价值、预期损失。依赖模型：共轭理论及其应用。极值理论：极端事件发生的概率模型、块状最大值、超过阈值的峰值。建立波动性模型和检测/模拟风险随时间变化的时间序列方法： ARMA 模型、ARCH/GARCH 模型。
Suppose that a parameter $\theta$ can take only three values, $\theta=0, \theta=1$ or $\theta=2$. Your prior on $\theta$ before observing any data is $p(\theta=0)=0.3, p(\theta=1)=0.5$ and $p(\theta=2)=0.2$. The distribution of $Y$ is as follows:
$$
\begin{aligned}
& p(Y \mid \theta=0)=\operatorname{Gamma}(2,3) \
& p(Y \mid \theta=1)=\operatorname{Gamma}(1,2) \
& p(Y \mid \theta=2)=\operatorname{Gamma}(3,3)
\end{aligned}
$$
Your task is to estimate the value of $\theta$. Let action $a_0$ corresponds to claiming that $\theta=0, a_1$ corresponds to claiming that $\theta=1$, and $a_2$ corresponds to claiming that $\theta=2$. The losses corresponding to each action and values of $\theta$ are represented by the following loss matrix:
\begin{tabular}{c|ccc}
& $\theta=0$ & $\theta=1$ & $\theta=2$ \
\hline$a_0$ & 0 & 1 & 5 \
$a_1$ & 1 & 0 & 4 \
$a_2$ & 8 & 3 & 0
\end{tabular}
For example, the loss associated with action $a_0$ when the true value is $\theta=1$ is $L(\theta=$ $\left.1, a_0\right)=1$. You obtain one observation $Y=2$.
(a) Compute the posterior distribution for $\theta$.
$[7]$
(b) Compute the Bayesian expected loss associated with all three actions given this observation, and decide which action to take.
[7]
Solution
Using Bayes’ Theorem:
$$
p(\theta=0 \mid y)=\frac{0.013}{0.058}=0.23
$$
\begin{aligned}
&p(\theta=1 \mid y)=\frac{0.018}{0.058}=0.31\
&p(\theta=2 \mid y)=\frac{0.027}{0.058}=0.46
\end{aligned}
After seeing the data $Y=2$, the Bayesian expected loss is:
$$
\begin{aligned}
& \rho\left(\pi^, a_0 \mid y\right)=2.602 \ & \rho\left(\pi^, a_1 \mid y\right)=2.060 \
& \rho\left(\pi^*, a_2 \mid y\right)=2.771
\end{aligned}
$$
The Bayesian expected loss is minimised by taking action $a_1$, hence we choose action $a_1$.
Consider the following historical record for the daily log-returns of a financial stock:
$$
Y_1=0.4, Y_2=-0.61, Y_3=-0.53, Y_4=0.22, Y_5=-0.018, Y_6=-0.39
$$
The log-returns are assumed to be independent and identically distributed draws from a Normal distribution with known mean 0 and unknown variance $\sigma^2$.
(a) Using the Inverse-Gamma $(\alpha, \beta)$ distribution as a conjugate prior for $\sigma^2$, derive its posterior distribution given the data. You do not need to evaluate any integrals or normalising constants.
[5]
(b) Let $\tilde{Y}$ denote the log-return on a particular day in the future. Show that the posterior predictive distribution is of the following form:
$$
p\left(\tilde{y} \mid y_1, \ldots, y_6\right)=\frac{1}{\sqrt{2 \pi}} \frac{\tilde{\beta}^{\tilde{\alpha}}}{\Gamma(\tilde{\alpha})} \frac{\Gamma\left(\tilde{\alpha}+\frac{1}{2}\right)}{\left(\tilde{\beta}+\frac{\bar{y}^2}{2}\right)^{\tilde{\alpha}+\frac{1}{2}}}
$$
for some values of $\tilde{\alpha}$ and $\tilde{\beta}$. Using the numeric values of $Y_1, \ldots, Y_6$ provided above, give expressions for $\tilde{\alpha}$ and $\tilde{\beta}$ in terms of $\alpha$ and $\beta$. State the range of possible values that $\tilde{Y}$ can take.
$[7]$
Solution
$$
\sigma^2 \mid \boldsymbol{y} \sim I G\left(\alpha+\frac{n}{2}, \beta+\frac{\sum y_i^2}{2}\right)
$$
where $n=6$ and $\frac{\sum y_i^2}{2}=0.507$
Let $\tilde{\alpha}$ and $\tilde{\beta}$ be the parameters of the posterior $p\left(\sigma^2 \mid y\right)$ from part (a). Then:
$$
\begin{aligned}
p(\tilde{y} \mid \boldsymbol{y}) & =\int p\left(\tilde{y} \mid \sigma^2\right) p\left(\sigma^2 \mid \boldsymbol{y}\right) d \sigma^2 \
& =\frac{1}{\sqrt{2 \pi}} \frac{\tilde{\beta}^{\tilde{\alpha}}}{\Gamma(\tilde{\alpha})} \frac{\Gamma\left(\tilde{\alpha}+\frac{1}{2}\right)}{\left(\tilde{\beta}+\frac{\tilde{y}^2}{2}\right)^{\tilde{\alpha}+\frac{1}{2}}}
\end{aligned}
$$
where $\tilde{\alpha}=\alpha+\frac{n}{2}=\alpha+3$ and $\tilde{\beta}=\beta+\frac{\sum y_i^2}{2}=\beta+0.507$
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微分几何代写
离散数学代写
Partial Differential Equations代写可以参考一份偏微分方程midterm答案解析
