This section of sample problems and solutions is a part of The Actuary’s Free Study Guide for Exam 4 / Exam C, authored by Mr. Stolyarov. This is Section 68 of the Study Guide. See an index of all sections by following the link in this paragraph.
Bühlmann credibility, also known as least squares credibility, differs from limited fluctuation credibility in the manner the credibility factor Z is calculated. Under Bühlmann credibility, Z = N/(N + K), where N is the number of observations, and K is calculated as follows:
K = EPV/VHM.
Let θ be the parameter that is being estimated.
EPV is the Expected Value of the Process Variance, which is Eθ(Var(X│θ)) = E(X2) – Eθ(E(X│θ)2). The variance X is first determined for each of the possible values of θ, after which the expected value this variance is taken over the distribution of θ to find EPV.
VHM is the Variance of Hypothetical Means, which is Varθ(E(X│θ)) = Eθ(E(X│θ)2) – E(X)2.
To find VHM, first find the mean of all observations, then find the second raw moment of the means of all observations for each of the possible values of θ, then subtract the square of the former result from the latter result.
It is useful to know that EPV + VHM = Total variance.
For Bühlmann credibility, as for classical (limited fluctuation) credibility, the credibility-weighted result is ZX- + (1-Z)M, where X- is the sample mean and M is the external mean.
Moreover, Bühlmann credibility offers the least squares approximation of the Bayesian estimate. This means that the Bühlmann credibility result is one such that the weighted average of the squared differences between the Bühlmann credibility result and the Bayesian estimate for each possible case is the smallest possible.
If the prior distribution of θ is gamma and the model distribution is Poisson, the Bühlmann credibility estimate is the same as the Bayesian estimate.
An excellent free source for learning about both limited fluctuation credibility and Bühlmann credibility is “Credibility” by Howard C. Mahler and Curtis Gary Dean. Actuarial students are strongly encouraged to read this entire study guide and to solve all of the examples and practice problems provided. Solutions are offered by Mr. Mahler and Mr. Dean in the study guide, so students have an opportunity to check their work and answers.
The current study guide will focus on exam-style questions pertaining to Bühlmann credibility. If more basic practice is desired, Sections 3-5 of Mr. Mahler’s and Mr. Dean’s study guide offer a variety of practice problems that will build up students’ skills to the point of their being able to solve the problems here.
Some of the problems in this section were designed to be similar to problems from past versions of Exam 4/C, offered jointly by the Casualty Actuarial Society and the Society of Actuaries. They use original exam questions as their inspiration – and the specific inspiration is cited to give students an opportunity to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.
Exam C Sample Questions and Solutions from the Society of Actuaries.
Mahler, H.C.; and Dean, C.G., “Credibility,” Foundations of Casualty Actuarial Science (Fourth Edition), 2001, Casualty Actuarial Society, Chapter 8, Section 1 and Sections 2-5.
Original Problems and Solutions from The Actuary’s Free Study Guide
Problem S4C68-1. Similar to Question 18 of the Exam C Sample Questions from the Society of Actuaries. Event A is four times more likely to occur than Event B, and one of the events must occur, but only once a year. The following are probabilities of monetary losses L in the event that each event occurs:
Event A: Pr(L = 34) = 0.3, Pr(L = 56) = 0.4, Pr(L = 89) = 0.3.
Event B: Pr(L = 34) = 0.2, Pr(L = 56) = 0.6, Pr(L = 89) = 0.2.
In Year 1, you observe a single loss of 56. Use Bühlmann credibility to find the expected loss in Year 2.
Solution S4C68-1. First, we find EPV and VHM.
EPV: E(A) = 0.3*34 + 0.4*56 + 0.3*89 = 59.3.
E(A2) = 0.3*342 + 0.4*562 + 0.3*892 = 3977.5.
Var(A) = E(A2) – E(A)2 = 3977.5 – 59.32 = Var(A) = 461.01.
E(B) = 0.2*34 + 0.6*56 + 0.2*89 = 58.2.
E(B2) = 0.2*342 + 0.6*562 + 0.2*892 = 3697.
Var(B) = E(B2) – E(B)2 = 3697 – 58.22 = Var(B) = 309.76.
Pr(A) = 4*Pr(B) and Pr(A) + Pr(B) = 1 → Pr(A) = 4/5 and Pr(B) = 1/5.
Thus, EPV = (4/5)Var(A) + (1/5)Var(B) = (4/5)461.01 + (1/5)309.76 = EPV = 430.76.
VHM: E(Total) = (4/5)E(A) + (1/5)E(B) = (4/5)59.3 + (1/5)58.2 = 59.08.
E(Total2) = (4/5)E(A)2 + (1/5)E(B)2 = (4/5)59.32 + (1/5)58.22 = 3490.64.
VHM = E(Total2) – E(Total)2 = 3490.64 – 59.082 = VHM = 0.1936.
K = EPV/ VHM = K = 2225.
N = 1, so Z = 1/(1+K) = 1/2226 = 0.000492362983 →
Expected loss in Year 2 = 0.000492362983*56 + (1-0.000492362983)*59.08 = 59.07861635.
Problem S4C68-2. Similar to Question 32 of the Exam C Sample Questions from the Society of Actuaries. The number of elephants that a snake steps on in a year follows a Poisson distribution with mean λ. The prior distribution of λ is a gamma distribution with α = 3 and θ = 1.9. During a three-year period, you observe that the snake has stepped on 2 elephants in Year 1, 4 elephants in Year 2, and 0 elephants in Year 3. Use Bühlmann credibility to estimate the number of elephants the snake will step on in Year 4. Use the Exam 4 / C Tables as needed.
Solution S4C68-2. First, we find the prior mean E(λ), which is also the EPV, since the distribution of the relevant variable is Poisson. Since λ is gamma-distributed, E(λ) = αθ = 3*1.9 = 5.7. Now we find VHM = Var(λ) = αθ2 = 3*1.92 = 10.83.
Thus, K = EPV/ VHM = 10/19. Since N = 3, Z = 3/(3+10/19) = Z = 0.8507462687.
The mean number of observed stepped-on elephants per year is (2+4+0)/3 = 2.
Thus, the expected number of stepped-on elephants in Year 4 is
2*0.8507462687 + 5.7(1-0.8507462687) = 2.552238806 elephants.
Problem S4C68-3. Similar to Question 32 of the Exam C Sample Questions from the Society of Actuaries. The number of elephants that a snake steps on in a year follows a Poisson distribution with mean λ. The prior distribution of λ is a gamma distribution with α = 3 and θ = 1.9. During a three-year period, you observe that the snake has stepped on 2 elephants in Year 1, 4 elephants in Year 2, and 0 elephants in Year 3. Use Bayesian estimation to find the expected number of elephants the snake will step on in Year 4. Use the Exam 4 / C Tables as needed.
Solution S4C68-3. This is an instance of Case 1 of conjugate prior distributions from Section 65:
Prior distribution of λ is gamma with parameters α and θ.
Model distribution is Poisson with parameter λ.
Posterior distribution is gamma with parameters α + Σxi and θ/(nθ+1).
The prior distribution is gamma with α = 3 and θ = 1.9. Here, n = number of years observed = 3, and Σxi = sum of observations = 2+4+0 = 6. Thus, α* = 3 + 6 = 9 and θ* = 1.9/(3*1.9+1) = 19/67. The expected number of stepped-on elephants in Year 4 is thus α*θ* = 9(19/67) = 2.552238806 elephants. Note the Bühlmann credibility and Bayesian estimation give the exact same result!
Problem S4C68-4. Similar to Question 35 of the Exam C Sample Questions from the Society of Actuaries. The unconditional probability of observing a value of 3 is ½, and the unconditional probability of observing a value of 5 is ½. If 3 is observed, the Bayesian estimate of the second observation is 3.4. If 5 is observed, the Bayesian estimate of the second observation is 4.8. Use Bühlmann credibility to estimate the second observation, given that the first observation is 3.
Solution S4C68-4. Bühlmann credibility is known as least squares credibility, because Bühlmann credibility gives a least squares approximation of the Bayesian estimate. This means that the weighted average of the squared differences between the Bühlmann credibility estimate and each of the Bayesian estimates must be minimized.
The credibility factor Z is associated with the value of the first observation, whereas the complement of credibility (1-Z) is associated with the external mean, which is the weighted average of the two Bayesian estimates: (1/2)3.4 + (1/2)4.8 = 4.1.
Hence, the quantity to be minimized in order to find Z is
q(Z) = (1/2)(3Z + 4.1(1-Z) – 3.4)2 + (1/2)(5Z + 4.1(1-Z) – 4.8)2
q(Z) = (1/2)(-1.1Z + 0.7)2 + (1/2)(0.9Z – 0.7)2
q'(Z) = -1.1(-1.1Z + 0.7) + 0.9(0.9Z – 0.7) = 0 at minimum.
Thus, 1.1(-1.1Z + 0.7) = 0.9(0.9Z – 0.7) →
-1.21Z + 0.77 = 0.81Z – 0.63 → 2.02Z = 1.4 → Z = 70/101 = 0.6930693069.
Hence, our Bühlmann credibility estimate is (70/101)3 + (31/101)4.1 = 3.337623762.
Problem S4C68-5. Similar to Question 48 of the Exam C Sample Questions from the Society of Actuaries. There are 15 observations in a sample, and the sum of the observations is 78. The magnitude of each observation X depends on the parameter Θ via the following joint distribution:
Pr(X = 4, Θ = 3) = 0.3
Pr(X = 6, Θ = 3) = 0.4
Pr(X = 4, Θ = 4) = 0.2
Pr(X = 6, Θ = 4) = 0.1
Use this information to find the Bühlmann credibility premium, i.e., the expected value of the next observation.
Solution S4C68-5. First, we find EPV and VHM. We can separate the joint distribution by values of Θ. Pr(Θ = 3) = 0.3+0.4 = 0.7, so Pr(X = 4│ Θ = 3) = 0.3/0.7 = 3/7 and Pr(X = 6│ Θ = 3) = 4/7. Thus, E(X │ Θ = 3) = 4(3/7) + 6(4/7) = 36/7. E(X2 │ Θ = 3) = 42(3/7) + 62(4/7) = 192/7. Hence, Var(X│ Θ = 3) = 192/7 – (36/7)2 = 0.9795918367.
Pr(Θ = 4) = 1 – 0.7 = 0.3, so Pr(X = 4│ Θ = 4) = 0.2/0.3 = 2/3, and Pr(X = 6│ Θ = 4) = 1/3.
Thus, E(X │ Θ = 4) = 4(2/3) + 6(1/3) = 14/3. E(X2 │ Θ = 4) = 42(2/3) + 62(1/3) = 68/3.
Hence, Var(X│ Θ = 4) = 68/3 – (14/3)2 = 8/9.
EPV = 0.7*Var(X│ Θ = 3) + 0.3*Var(X│ Θ = 4) = 0.7*0.9795918367 + 0.3(8/9) = 0.9523809524.
E(X) = 0.7*(36/7) + 0.3*(14/3) = 5.
E(X2) = 0.7*(36/7)2 + 0.3*(14/3)2 = 25.04761905
VHM = E(X2) – E(X)2 = 25.04761905 – 52 = 0.0476190476.
K = EPV/VHM = 0.9523809524/0.0476190476 = 20.
The number of observations, N, is 15 so Z = 15/(15 + 20) = (3/7).
The observed mean is 78/15 = 5.2. Hence, the Bühlmann credibility premium is (3/7)5.2 + (4/7)5 = 5.085714286 = 178/35.
See other sections of The Actuary’s Free Study Guide for Exam 4 / Exam C.