Welcome to the tutorial on Exam C. Today we are going to go over Question #11, which is part
of the learning objective "Credibility". Here is the view of the problem. Please choose one of the
answers and submit it. If incorrect hit the grey ‘x’ in the top right to try again.
In this question, X denotes losses on a policy and theta denotes the risk parameter,
which varies over different policies. The distribution of losses depends on the risk
parameter, specifically, conditional on theta, X follows a Pareto distribution.
The prior distribution of theta has a mass at 1 and 3 with equal probabilities. Let pi
denote the mass function of the prior distribution, then we have pi(1)=pi(3)=0.5.
For a randomly selected policy, we observe that loss in the first year, denoted by X1, is 5.
The question is asking for the posterior probability that losses for this policy in Year 2
will exceed 8, more precisely, this is a predictive probability rather than a posterior
probability, and it can be expressed as the conditional probability that X2 exceeds 8
given X1=5.
This is a typical credibility problem with the Bayesian Methodology. The solution
contains three steps. First, for the selected policy, we will get a posterior distribution of
the risk parameter theta based on the observed information X1=5. Second, we will
calculate the conditional probability that losses exceed 8 given theta, which can be
obtained by integrating the conditional density function from 8 to infinity. Finally, with
the results from the first two steps, we will calculate the asked predictive probability.
We work on the first step now. We need to calculate several quantities before we are
able to calculate the posterior distribution of theta. From the given conditional density
function, we can calculate the conditional probability that the loss was 5 in the first year,
given theta=1, by plugging in x=5 and theta=1, the result is 1/36. Similarly, given that
theta=3, the conditional probability that x=5 is 3/64. The marginal probability that x=5 is
another quantity needed. It can be calculated with the total probability formula. Since
theta has a mass at 1 and 3, the formula can be simplified as follows. Plug in numbers
for the conditional probabilities and prior distribution, the marginal probability that X=5
is 0.0373.
Now we are ready to calculate the posterior distribution of theta, which is actually the
conditional distribution of theta given the observed loss, X1=5. According to the Bayes’
theorem, the conditional probability of theta given x, is equal to the conditional
probability of x given theta, times the quotient of the marginal probability of theta and
the marginal probability of x. Note that the marginal probability of theta is also the prior
distribution. Since the prior distribution has a mass at 1 and 3, the posterior
distribution also has a mass at 1 and 3. Plug in numbers for the conditional and
marginal probabilities, the conditional probability that theta=1 given x=5 is 0.3721, the
conditional probability that theta=3 given x=5 is 0.6284. Please note that these two
conditional probabilities give the posterior distribution of theta. You may notice that the
sum of the two posterior probabilities is 1, so we don’t have to calculate the marginal
probability f(5) by the total probability formula in advance, instead, we can treat it as a constant
and calculate it by normalizing the constant. Then we get the constant equals 0.0373 and still
get the same result for the posterior probabilities.
In step 2, we will calculate the probability that losses exceed 8 conditional on theta. This equals
the integral of the conditional density function f(x|theta) from 8 to infinity. The given
conditional density is a polynomial function of x, so it's easy find an anti-derivative for it, which
is -theta/(x+theta). Evaluate the anti-derivative at infinity and 8, then take a difference, we get
theta/(8+theta). Plug in theta=1, then the probability that x>8 given theta=1 is 1/9. Similarly, the
probability that x>8 given theta=3 is 3/11. Finally, we are ready to calculate the predictive
probability that X2 >8 given X1=5. Again, this probability can be calculated with the total
probability formula, that is, multiplying the conditional probability that x>8 given theta with the
posterior distribution of theta, then summing up over the support of theta. Because the support
of theta is 1 and 3, the total probability formula can be simplified as the sum of two terms. Then
recall the results from the first two steps, plug in numbers to the formula, the final result is
0.2126. That’s answer D. That's the end of this tutorial. Thanks for watching!