So... I'll be honest, I don't know anything about anything Bayesian, this problem being no exception (from the Society of Actuaries' Exam C sample questions):
You are given:
(i) The annual number of claims for a policyholder has a binomial distribution with probability function: $$p\left(x \mid q\right) = {2\choose x}q^{x}(1-q)^{2-x}\text{, }\quad x = 0, 1, 2$$
(ii) The prior distribution is $$\pi(q) = 4q^{3}\text{, }\quad 0 < q < 1$$
The policyholder had one claim in each of Years 1 and 2.
Determine the Bayesian estimate of the number of claims in Year 3.
I have decided not to look at the solution, as I am attempting to study for this exam through reverse engineering of questions (i.e., grab a question and learn about the underlying concepts using a text). Unlike the previous questions for which this method has been successful, I hit a brick wall with this one.
So from what I understand, unlike the frequentist mindset, with Bayesian statistics, you assign a "prior" distribution to some variable. The only other thing I know about Bayesian methods is what a posterior distribution is (by definition) and how it can be motivated through Bayes' theorem.
Guidance with this problem is appreciated.
You have a prior distribution $\pi(q)$, a likelihood function $p(x|q)$ and two observations $x_1=1$ and $x_2=1$. So your posterior distribution for $q$ is $$\pi(q|x_1,x_2)=\frac{\pi(q)p(x_1|q)p(x_2|q)}{\displaystyle \int_0^1 \pi(r)p(x_1|r)p(x_2|r)\; dr}$$ in effect taking the product of the prior and the likelihoods and then turning this into a probability density by dividing by its integral over all possible $q$ (represented here by $r$ as $q$ is also a free variable), though knowing $0 \le q \le 1$.
If you want the expected number of accidents in year $3$, then your next step would be to calculate $$\int_0^1 (2 \times p(2|q) + 1 \times p(1|q) + 0 \times p(0|q)) \pi(q|x_1,x_2)\; dq$$
I will leave the calculus to you.