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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.

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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.

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  • $\begingroup$ Thank you for your response. I have two questions - in your first equation, I noticed that you have $p(x_1 \mid q)p(x_2 \mid q)$ as a product in the numerator; is this because we assume that conditional on $q$, $X_1$ and $X_2$ are independent? Also, is the second equation supposed to have $0 \times p(0\mid q)$ instead of $0 \times p(1\mid q)$? $\endgroup$ – Clarinetist Sep 6 '14 at 19:57
  • $\begingroup$ Yes, and yes ($0$ corrected) $\endgroup$ – Henry Sep 6 '14 at 20:45

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