The following is a homework question so I do not need a full answer, but I do need help understanding the question.
The Question: A coin with an unknown probability of heads ($\theta$) is flipped once, and you observe the result. On the second flip, you can choose the following actions:
- $a_1$: predict the same side
- $a_2$: predict the opposite side
- $a_3$: do not play
If you are correct, you get \$2. If you are incorrect, you lose \$1. If you don't play you net $0.
Assuming a Uniform Prior on $\theta$, what is the Bayes Rule?
My Attempt: The Bayes Rule in the context of our class is typically an estimator ($\hat\theta_{Bayes}$) that minimizes the posterior expected loss of an action: $$\hat\theta_{Bayes}={\arg\min}\int_{\Theta}\ell(\theta,a)p(\theta|x)d\theta $$
First, we find the posterior. The likelihood follows $p(x|\theta)\sim Binomial(2, \theta)$, and the prior can be expressed as $p(\theta)\sim Beta(1,1)$. Using the Beta-Binomial model, our posterior is thus: $p(\theta|x)\sim Beta(x+1, 1+2-x)$
The loss, however, is dependent on the choice of action (negative loss represents a gain): $$\begin{align*} l(\theta,a_1)&=-\$2p(x=0,2|\theta)+\$1p(x=1|\theta)\\ l(\theta,a_2)&=-\$2p(x=1|\theta)+\$1p(x=0,2|\theta)\\ l(\theta,a_3)&=\$0 \end{align*}$$
What I do not understand is: I am used to seeing one loss function (for instance, MSE or absolute loss).
In the case above, is the Bayes Rule/Bayes Estimator unique to each action, or is there one Bayes Rule that satisfies this problem for all actions? Stated differently, how do I incorporate the fact that there are different loss functions into the problem (or have I incorrectly described the loss function)?