(i) In the context of a decision-theoretic approach to statistics, what is a loss function? a decision rule? the risk function of a decision rule? the Bayes risk of a decision rule? the Bayes rule with respect to a given prior distribution?

Show how the Bayes rule with respect to a given prior distribution is computed.

(ii) A sample of $n$ people is to be tested for the presence of a certain condition. A single real-valued observation is made on each one; this observation comes from density $f_{0}$ if the condition is absent, and from density $f_{1}$ if the condition is present. Suppose $\theta_{i}=0$ if the $i^{\text {th }}$person does not have the condition, $\theta_{i}=1$ otherwise, and suppose that the prior distribution for the $\theta_{i}$ is that they are independent with common distribution $P\left(\theta_{i}=1\right)=p \in(0,1)$, where $p$ is known. If $X_{i}$ denotes the observation made on the $i^{\text {th }}$person, what is the posterior distribution of the $\theta_{i}$ ?

Now suppose that the loss function is defined by

$$L_{0}(\theta, a) \equiv \sum_{j=1}^{n}\left(\alpha a_{j}\left(1-\theta_{j}\right)+\beta\left(1-a_{j}\right) \theta_{j}\right)$$

for action $a \in[0,1]^{n}$, where $\alpha, \beta$ are positive constants. If $\pi_{j}$ denotes the posterior probability that $\theta_{j}=1$ given the data, prove that the Bayes rule for this prior and this loss function is to take $a_{j}=1$ if $\pi_{j}$ exceeds the threshold value $\alpha /(\alpha+\beta)$, and otherwise to take $a_{j}=0$.

In an attempt to control the proportion of false positives, it is proposed to use a different loss function, namely,

$$L_{1}(\theta, a) \equiv L_{0}(\theta, a)+\gamma I_{\left\{\sum a_{j}>0\right\}}\left(1-\frac{\sum \theta_{j} a_{j}}{\sum a_{j}}\right)$$

where $\gamma>0$. Prove that the Bayes rule is once again a threshold rule, that is, we take action $a_{j}=1$ if and only if $\pi_{j}>\lambda$, and determine $\lambda$ as fully as you can.