1. Whether you codes follow your work on the conditional densities;
2. In computing the MH acceptance probabilities and the conditional densities, try using the normalization trick (take log first).
3. Check whether you did something like C/0, where C is some constant.

• Mutivariate normal PDF
  \[
  \int_{\mathbb{R}^p} \exp\left\{ -\frac{1}{2}(\beta - \beta_0)^{\intercal}\Sigma_0^{-1}(\beta - \beta_0)\right\} \, \text{d}\beta = \sqrt{\text{det}(2\pi \Sigma_0)}.
  \]
• Inverse gamma PDF
  \[
  \int_{\mathbb{R}^+} \theta^{-\alpha-1} \exp\left( -\frac{\beta}{\theta}\right) \, \text{d} \theta = \frac{\Gamma(\alpha)}{\beta^{\alpha}}.
  \]
  

mcmc_label = function(env, iUse, switch=TRUE){
  z = env$z[iUse,]
  if (switch) {
    o = t(apply(env$theta[iUse,], 1, order))
    H = env$H
    n = length(iUse)
    xi = vector("list", H)
    for (i in 1:n) {
      if (!identical(o[i,], 1:H)) { #-------------①This step switches the labels.
        for (h in 1:H)	#-------------------------The first loop finds whose label will be swiched to h
          xi[[o[i, h]]] = which(z[i,]==h)
        for (h in 1:H)  #-------------------------②The second for loop performs the swiching
          z[i, xi[[h]]] = h
      }
    }
  }
  apply(z, 2, function(x){
    ux = unique(x)
    ux[which.max(tabulate(match(x, ux)))] #-------③Yes, combining all iterations, 
                                          			the final prediction is the most frequent label
  })
}

z[i,] = o[i,z[i,]]

### Part 2b ###
z[which(z==1)] = 4 #⓹We choose z==1 as year-4 students because the frequency of year-4 students is max
z[which(z==3)] = 5
z[which(z==2)] = 3

1. We cannot directly use $\mu_j,j=1,2$ because $\hat{\theta}^{\pi}_j$ follows a $\href{https://en.wikipedia.org/wiki/Truncated_normal_distribution}{\text{truncated normal distribution}}$ instead of a normal distribution.
2. It is the property of truncated normal distribution. Compared to a normal distribution, an upper bound and a lower bound are imposed to the support of the truncated normal distribution. And $u_j,\ell_j$ are understood as the z-scores of its upper bound and lower bound respectively.

1. Denote the standard normal density as $\phi(\cdot)$. The density of $X$ is given by (exercise) $$f_X(x) = \frac{\phi((x-\mu)/\sigma)}{\{\texttt{pnorm((b-$\mu$)/$\sigma$)}-\texttt{pnorm((a-$\mu$)/$\sigma$)}\}\sigma},\quad x\in[a,b].$$
2. Let $Z:=(X-\mu)/\sigma.$ Then $Z\sim\mathrm{TN}(0,1;\alpha,\beta)$ where $\alpha = (a-\mu)/\sigma, \beta =(b-\mu)/\sigma$. And $$f_Z(z) = \frac{\phi(z)}{\texttt{pnorm($\beta$)}-\texttt{pnorm($\alpha$)}},\quad z\in[\alpha,\beta].$$
3. The expectation of $Z$ is $$E(Z) = \int_{\alpha}^{\beta}\frac{z\phi(z)}{\texttt{pnorm($\beta$)}-\texttt{pnorm($\alpha$)}}\mathrm{d}z=\frac{\int_{\alpha}^{\beta}\frac{z\exp(-z^2/2)}{\sqrt{2\pi}}\mathrm{d}z}{\texttt{pnorm($\beta$)}-\texttt{pnorm($\alpha$)}}=\frac{\int_{\alpha^2/2}^{\beta^2/2}\frac{\exp(-u)}{\sqrt{2\pi}}\mathrm{d}u}{\texttt{pnorm($\beta$)}-\texttt{pnorm($\alpha$)}}=\frac{\texttt{dnorm($\alpha$)}-\texttt{dnorm($\beta$)}}{\texttt{pnorm($\beta$)}-\texttt{pnorm($\alpha$)}}.$$

1. We write in that form because the densities may not exist. If the joint density, marginal densities as well as the conditional densities exist, we can exchange the order. We consider the posterior first in the inner integral first for computational convenience. Note that under L2 loss, $\hat{\theta}_\pi = E[\theta \mid x_{1:n}]$. Therefore, the inner integral equals to $Var(\theta \mid x_{1:n})$. By our assumption, it is free of $x_{1:n}$. Therefore, we can take it out of the outer integral and finish all the calculation easier.
2. You are correct. Therefore is a typo in the equation. But the conclusion is the same. We will update the tutorial note shortly. Thanks for pointing out.
3. Here, $a$ cannot be $0$ since $a < 0$ is considered in this scenario. The subsequent argument is thus not valid.

1. write $\left(x_{(n)}-x_{(1)}\right)^{3-n}$ as $\left(x_{(n)}-x_{(1)}\right)/\left(x_{(n)}-x_{(1)}\right)^{n-2}$. And write other terms similarly.
2. Multiply both the numeritor and the denominator by $(x_{(n)}-x_{(1)})^{n-2}$.

 for(i.t1 in which(t1.all>theta0)[1]:length(t1.all)){
	 ...
 }

• The other estimators will always have a larger Bayesian risk because each of them is inadmissible.
• In this problem, we have $R(\pi,\hat{\theta})=\pi(\theta_1)R(\theta_1,\hat{\theta})+\pi(\theta_2)R(\theta_2,\hat{\theta})$.
  • For $\hat{\theta}_2$, it is the minimizer of $R(\theta_1,\hat{\theta})$, so we can design $\pi(\theta_1)=1, \pi(\theta_2)=0$ and $\hat{\theta_2}$ will be the corresponding Bayes estimator.
  • The same idea is applicable to $\hat{\theta}_3$.
  • Noticing that $\sum_{i=1}^2R(\theta_i,\hat{\theta}_2)=\sum_{i=1}^2R(\theta_i,\hat{\theta}_6)<\sum_{i=1}^2R(\theta_i,\hat{\theta}_3)$, we can design $\pi(\theta_1)=½, \pi(\theta_2)=½,$ and $\hat{\theta}_2,\hat{\theta}_6$ are the corresponding BEs.

• (Before taking log) Recall that the posterior is given by
  \[
  f(\theta \mid x_{1:n}) = \color{red}{C} f(\theta) f(x_{1:n}\mid \theta).
  \]
  Hence, we can write
  \begin{align}
  f(\theta \mid x_{1:n}) &\propto \color{red}{C} f(\theta) f(x_{1:n}\mid \theta), \qquad \text{or}\\
  f(\theta \mid x_{1:n}) &\propto \color{red}{20 C} f(\theta) f(x_{1:n}\mid \theta), \qquad \text{or}\\
  f(\theta \mid x_{1:n}) &\propto \color{red}{4010 C} f(\theta) f(x_{1:n}\mid \theta), \qquad \text{or}\\
  &\vdots
  \end{align}
  All of them mean the same thing.
• (After taking log) Now, taking log on both side, we get
  \[
  \log f(\theta \mid x_{1:n}) = \color{red}{\log C} + \log f(\theta) + \log f(x_{1:n}\mid \theta).
  \]
  We observe that, after taking log, the proportionality constant $C$ becomes an additive constant $\log C$. Hence, we can write
  \begin{align}
  \log f(\theta \mid x_{1:n}) &= \color{red}{\log C} + \log f(\theta) + \log f(x_{1:n}\mid \theta), \qquad \text{or}\\
  \log f(\theta \mid x_{1:n}) &= \color{red}{\logC-2023} + \log f(\theta) + \log f(x_{1:n}\mid \theta), \qquad \text{or}\\
  \log f(\theta \mid x_{1:n}) &= \color{red}{C-\max\left\{ 10,20,30,40,50\right\}} + \log f(\theta) + \log f(x_{1:n}\mid \theta), \qquad \text{or}\\
  &\vdots
  \end{align}
  where, on the last line, $\max\left\{ 10,20,30,40,50\right\}$ is just a part of the additive constant. Once again, all of them mean the same thing.
• (Answer) In a nutshell, we don't need to adjust back the term $\max\{\cdots\}$ because it is a part of the “meaningless” additive constant. We include it only for facilitating computation.

• In general, it is NOT necessary to know the proportionality constant $c$ to define the posterior distribution.
• However, our answer would be more complete if we also know the proportionality constantat least in the following three cases:
  1. the posterior is a named distribution, e.g., $\text{Beta}(\alpha, \beta)$.
  2. the posterior can be represented as a more interpretable distribution after knowing the value of $c$, e.g., the mixture probabilities of a distribution; see Example 2.9 and Question 1.1 of mock 2.
  3. the computation of the posterior is facilitatedafter knowing the value of $c$, e.g., part 6 of A3.

• The estimand is $\phi = \mathsf{E}(x_1\mid \theta)/y-1$, which is a function of $\theta$. Note that $\phi$ is a random variable.
• A possible estimator is the Bayes estimator $\widehat{\phi} = \mathsf{E}(\phi \mid x_{1:n})$ when the loss is $L(\theta, \widehat{\phi}) = (\widehat{\phi}-\phi)^2$. In this question, you need to consider a more general loss function, and find its Bayes estiamtor.
• From the Bayesian perspective, a point estimator $\widehat{\phi}$ is simply a one-number summary of the distribution of $\phi$.
• Note that if $\widehat{\phi} = \mathsf{E}(\phi \mid x_{1:n})$ then $\mathsf{E}(\widehat{\phi}) = \mathsf{E}(\phi)$, which is the prior expectation of $\phi$. In our case,
  \[
  \mathsf{E}(\phi) = \sum_{j=1}^J t_j p_j,
  \]
  So, $\mathsf{E}(\phi)$ is a fixed and known constant fully determined by the prior distribution. Hence, there is no need to estimate $\mathsf{E}(\phi)$.