Here are a set of exercises that guide the viewer through some of the theoretical foundations of Loss Data Analytics. Each tutorial is based on one or more questions from the professional actuarial examinations – typically the Society of Actuaries Exam C.
Tutorial Structure. Each guided tutorial has a strategy set that describes the context. When you hit the “Start quiz” button, you begin the tutorial that is comprised of a series of miniquestions designed to lead you to the target question. At each stage, hints are provided as well as feedback on the correct solution of each miniquestion.
Your Assignment. In reviewing these exercises, ideally the viewer will:
 Work the problem posed referring only to basic theory
 Even if you get the answer correct, review the strategy for this type of problem by clicking (revealing) the Strategy for … header
 If you feel comfortable with the strategy and got the problem correct, then you may choose to move on. However, you might also decide to follow the stepbystep process for solving the problem by clicking on the “Start Quiz” button. It is not really a quiz — it is a guided tutorial.
Strategy for Limited Fluctuation Credibility Problems
Let \(X_1,X_2…X_n\) be iid. Actuaries use the phrase full credibility to mean that we can use sample mean \(\bar{X}\) for pricing. To achieve this standard, it is common to require the sample size n to be large enough so that \(\bar{X}\) is within 5% of the mean (\(\mu\)) at least \(90\%\) of the time. Using \(r=0.05\) and \(p=0.9\), then we require n large enough so that
$$\Pr\left((1r)\mu\leq\overline{X}\leq(1+r)\mu\right)\geq p .$$ To find the smallest value of n, replace the inequality by equality and use the central limit to solve for n as follows:
$$\begin{array}{ll}
p
&=\Pr\left(\frac{(1r)\mu\mu}{\sigma/\sqrt{n}}\leq \frac{\overline{X}\mu}{\sigma/\sqrt{n}}\leq \frac{(1+r)\mu\mu}{\sigma/\sqrt{n}}\right) \\
&=\Pr\left(\frac{r\mu}{\sigma/\sqrt{n}}\leq \frac{\overline{X}\mu}{\sigma/\sqrt{n}}\leq \frac{r\mu}{\sigma/\sqrt{n}}\right) \\
& =\Pr\left(y_p\leq N(0,1) \leq y_p\right) .\end{array}$$ With this, \( y_p={\Phi}^{1}\left((1+p)/2\right) \) is a quantile from the standard normal distribution. For example, if p = 0.90, then \( y_{0.9}={\Phi}^{1}\left(0.95\right) = 1.645.\) From the solution, you see that $y_p=\frac{r\mu}{\sigma/\sqrt{n}}.$ Solving for n yields
$$ n_f \geq \left(\frac{y_p \sigma}{r\mu}\right)^2= \left(\frac{y_p}{r}\right)^2 CV^2 \quad \quad … (1)$$
Here, \(n_f\) is the exposure needed for full credibility and \(CV = \sigma/\mu\) is the coefficient of variation.
Note: If you are establishing the standard for full credibility for pure premium or aggregate losses, the formulae for the expected number of claims needed for full credibility is :
$$E(N)\times n_f \quad \quad … (2)$$ Where \(E(N)\) is the expected number of claims and \(n_f\) is the number of exposures required for full credibility calculated from equation (1) above.
When there is inadequate experience for full credibility , we calculate the credibility factor \(Z\) by
$$Z=
\left\{\begin{array}{ll}
\sqrt{\frac{n}{n_f}} & \text{if} \quad n\leq n_f \\
1 & \text{otherwise}\end{array} \right. … (3)$$
We use \(Z\) to calculate the credibility premium \(P_c\) given by;
$$P_c=Z\overline{X} + (1Z)M \quad \quad … (4)$$ Here, \(M\) is the manual premium which is generally part of the problem givens.
To apply these rules,
 Using the question, identify the distribution of the claims, \(p\) and \(r\).
 Using equation (1) above calculate the standard for full credibility in terms of exposure.
 If the question requires you to establish the standard for full credibility for claim sizes(severity) in terms of exposure units, the exposure unit is a claim, so the the standard expressed in exposure units is the same as the standard for number of claims and it is given by equation (1). But if you are required to establish the standard for full credibility for aggregate losses in terms of expected number of claims, its is given by equation (2) above.
 If the question requires you to calculate the credibility factor, it is given by equation (3).
 If the question requires you to calculate the credibility premium, it is given by equation (4).
CredibilityPoisson Frequency SOA #2
You are given:
(i) The number of claims has a Poisson distribution.
(ii) Claim sizes have a Pareto distribution with parameters \(\theta=0.5\) and \(\alpha=6\).
(iii) The number of claims and claim sizes are independent.
(iv) The number of claims in the first year was 1200.
(v) The aggregate loss in the first year was 6.75 million.
(vi) The manual premium for the first year was 5.00 million.
(vii) The exposure in the second year is identical to the exposure in the first year.
(vii) The observed pure premium should be within \(2\%\) of the expected pure premium \(90\%\) of the time.
(a) Calculate the expected number of claims needed for full credibility.
(b) Determine the limited fluctuation credibility net premium(in millions) for the second year.
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Try to solve the problem on your own. (The answer is (a) 16,912.65 and (b) 5.466 million)
Then, regardless of whether you get the right answer or not, you can check out this “quiz.” It is not really a quiz — it is a guided tutorial. The following provides a series of questions to lead you to the solution to this problem from the Society of Actuaries Exam C Question 2.
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Question 1 of 4
1. Question
From the question , the number of claims \(N\) follows a Poisson distribution and the Claim sizes \(X\) have a Pareto distribution with parameters \(\theta=0.5\) and \(\alpha=6\). Given that \(p=90\%\) and \(r=2\%\), in establishing the standard for full credibility for aggregate losses, which of the following expression represents the expected number of claims for full credibility.
Correct
IncorrectIf you are establishing the standard for full credibility for pure premium or aggregate losses, the formulae for the number of exposure needed for full credibility is:
$$ n_f = \left(\frac{y_p}{r}\right)^2 CV^2 \quad \quad … (1)$$
For aggregate claims \(S\);
$$CV^2=\frac{Var(S)}{{E(S)}^2}$$Given the number of claims follows a Poisson distribution;
$$\begin{array}{ll}
E(S)&=E(N)E(X)&=\lambda E(X) \\
Var(S)&=Var(X)E(N) + Var(N){E(X)}^2 &=\lambda (Var(X)+{E(X)}^2) \end{array}$$Then;
$$\begin{array}{lll}
CV^2 &=\frac{Var(S)}{{E(S)}^2}&=\frac{\lambda (Var(X)+{E(X)}^2)}{\left(\lambda E(X)\right)^2} \\
CV^2 &=\frac{\lambda (Var(X)+{E(X)}^2)}{{\lambda}^2 {E(X)}^2}&=\frac{ (Var(X)+{E(X)}^2)}{\lambda {E(X)}^2} \\
CV^2 &=\frac{1}{\lambda}\left(1+\frac{(Var(X)}{{E(X)}^2}\right)& \end{array}$$Given \(p=90\%\);
$$y_p={\Phi}^{1}\left((1+p)/2\right)={\Phi}^{1}\left((1+0.9)/2\right)={\Phi}^{1}(0.95)=1.645$$
This implies from equation (1) the number of exposure needed for full credibility is;
$$ n_f =\frac{1}{\lambda}\left(\frac{1.645}{0.02}\right)^2\left(1+\frac{(Var(X)}{{E(X)}^2}\right)$$Then the formulae for the expected number of claims needed for full credibility is :
$$E(N)\times n_f $$$$\begin{array}{l}
=E(N)\times \frac{1}{\lambda}\left(\frac{1.645}{0.02}\right)^2\left(1+\frac{(Var(X)}{{E(X)}^2}\right) \\
=\lambda \times \frac{1}{\lambda}\left(\frac{1.645}{0.02}\right)^2\left(1+\frac{(Var(X)}{{E(X)}^2}\right) \\
=\left(\frac{1.645}{0.02}\right)^2\left(1+\frac{(Var(X)}{{E(X)}^2}\right)
\end{array}$$Therefore, the expression that represents the standard for full credibility in terms of expected number of claims given that \(p=90\%\) and \(r=2\%\) is :
$$\left(\frac{1.645}{0.02}\right)^2\left(1+\frac{(Var(X)}{{E(X)}^2}\right) $$
Hint
If you are establishing the standard for full credibility for pure premium or aggregate losses, the formulae for the number of exposure needed for full credibility is:
$$ n_f \geq \left(\frac{y_p}{r}\right)^2 CV^2 $$
Then the formulae for the expected number of claims needed for full credibility is :
$$E(N)\times n_f $$ 
Question 2 of 4
2. Question
Given claim sizes \((X)\) have a Pareto distribution with parameters \(\theta=0.5\) and \(\alpha=6\) and the expression for the expected number of claims for full credibility to be ;
$$\left(\frac{1.645}{0.02}\right)^2\left(1+\frac{Var(X)}{{E(X)}^2}\right) $$Calculate the expected number of claims needed for full credibility.
Correct
IncorrectFor claim sizes \((X)\) that follow Pareto distribution with parameters \(\theta=0.5\) and \(\alpha=6\) , form the C tables;
$$E(X)=\frac{\theta}{\alpha1}=\frac{0.5}{61}=0.1$$
$$E(X^2)=\frac{{2\theta}^2}{(\alpha1)(\alpha2)}=\frac{2({0.5}^2)}{(61)(62)}=\frac{0.5}{(5)(4)}=0.025$$
$$Var(X)=E(X^2)E(X)^2=0.025.1^2=0.015$$Then the expected number of claims needed for full credibility is;
$$\left(\frac{1.645}{0.02}\right)^2\left(1+\frac{(Var(X)}{{E(X)}^2}\right) =\left(\frac{1.645}{0.02}\right)^2\left(1+\frac{0.015}{{0.1}^2}\right)=16912.65 $$
Hint
See the Exam C tables for expressions for \(E(X)\) and \(Var(X)\) for the Pareto distribution.

Question 3 of 4
3. Question
From the question, the number of claims in the first year was 1200. calculate the credibility factor \(Z\).
Correct
IncorrectTo calculate the credibility factor \(Z\), we use the formula;
$$Z=
\left\{\begin{array}{ll}
\sqrt{\frac{n}{n_f}} & \text{if} \quad n\leq n_f \\
1 & \text{otherwise}\end{array} \right. … (3)$$Because the standard for full credibility in terms of expected number of claims is \(n_f=16912.65\), given number of claims n=1200 in the first year means there is inadequate experience for full credibility. Then ;
$$Z=\sqrt{\frac{n}{n_f}} =\sqrt{\frac{1200}{16912.65}}=0.266$$Hint
To calculate the credibility factor \(Z\), we use the formula; ;
$$Z=
\left\{\begin{array}{ll}
\sqrt{\frac{n}{n_f}} & \text{if} \quad n\leq n_f \\
1 & \text{otherwise}\end{array} \right. … (3)$$ 
Question 4 of 4
4. Question
From the question, the aggregate loss in the first year was 6.75 million and the manual premium was 5.00 million. Calculate the credibility net premium (in millions) for the second year.
Correct
IncorrectTo calculate the credibility premium \(P_c\), we use the formula;
$$P_c=Z\overline{X} + (1Z)M $$
Where , M is the Manual premium and Z=0.266 is credibility factor. From the question, the aggregate loss in the first year was 6.75 million and the manual premium was 5.00 million. Then premium for the second year is ;
$$P_c=Z\overline{X} + (1+Z)M =0.266(6.75)+(10.266)(5)=5.466 $$Hint
To calculate the credibility premium \(P_c\), we use the formula;
$$P_c=Z\overline{X} + (1Z)M $$
Where , \(M\) is the Manual premium and \(Z\) is credibility factor.
CredibilityNon Poisson Frequency SOA #65
You are given the following information about a general liability book of business comprised of 2500 insureds:
(i) \(X_i=\sum_{j=1}^{N_i}Y_{ij}\) is a random variable representing the annual loss of the \(i\)th insured.
(ii) \(N_1,N_2,…,N_{2500}\) are independent and identically distributed random variables following a negative binomial distribution with parameters \(r=2\) and \(\beta=0.2\).
(iii) \(Y_{i1},Y_{i2},…,Y_{iN_i}\), are independent and identically distributed random variables following a Pareto distribution with \(\alpha=3\) and \(\theta=1000\).
(iv) The full credibility standard is to be within \(5\%\) of the expected aggregate losses \(90\%\) of the time.
Using limited fluctuation credibility theory, calculate the partial credibility of the annual loss experience for this book of business.
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Then, regardless of whether you get the right answer or not, you can check out this “quiz.” It is not really a quiz — it is a guided tutorial. The following provides a series of questions to lead you to the solution to this problem from the Society of Actuaries Exam C Question 65.
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1. Question
From the question , the number of claims \(N\) follow a negative binomial distribution with parameters \(r=2\) and \(\beta=0.2\) and the Claim sizes \(Y\) follow a Pareto distribution with \(\alpha=3\) and \(\theta=1000\). Given that \(p=90\%\) and \(r=5\%\), in establishing the standard for full credibility for aggregate losses, which of the following expression represents the exposure needed for full credibility.
Correct
IncorrectIf you are establishing the standard for full credibility for pure premium or aggregate losses, the formulae for the number of exposure needed for full credibility is:
$$ n_f = \left(\frac{y_p}{r}\right)^2 CV^2 \quad \quad … (1)$$
For aggregate claims \(X\);
$$CV_X^2=\frac{Var(X)}{{E(X)}^2}$$Given the number of claims \(N\) follows a negative binomial distribution with parameters \(r=2\) and \(\beta=0.2\); and the Claim sizes\(Y\) follow a Pareto distribution with \(\alpha=3\) and \(\theta=1000\):
$$\begin{array}{ll}
E(X)&=E(N)E(Y) \\
Var(X)&=Var(Y)E(N) + Var(N){E(Y)}^2
\end{array}$$Then;
$$\begin{array}{ll}
CV_X^2
&=\frac{Var(X)}{{E(X)}^2}\\
&=\frac{Var(Y)E(N) + Var(N){E(Y)}^2}{\left(E(N)E(Y)\right)^2}\\
&=\frac{Var(Y)E(N) + Var(N){E(Y)}^2}{{E(N)}^2 {E(Y)}^2}& \\
&=\left(\frac{Var(Y)}{{E(Y)}^2E(N)}+\frac{Var(N)}{{E(N)}^2}\right)
\end{array}$$Given \(p=90\%\);
$$y_p={\Phi}^{1}\left((1+p)/2\right)={\Phi}^{1}\left((1+0.9)/2\right)={\Phi}^{1}(0.95)=1.645$$
This implies from equation (1) he number of exposure needed for full credibility is;
$$ n_f =\left(\frac{1.645}{0.05}\right)^2\left(\frac{Var(Y)}{{E(Y)}^2E(N)}+\frac{Var(N)}{{E(N)}^2}\right)$$Therefore, the expression that represents the standard for full credibility in terms of exposure given that \(p=90\%\) and \(r=5\%\) is :
$$\left(\frac{1.645}{0.05}\right)^2\left(\frac{Var(Y)}{{E(Y)}^2E(N)}+\frac{Var(N)}{{E(N)}^2}\right) $$
Hint
If you are establishing the standard for full credibility for pure premium or aggregate losses, the formulae for the number of exposure needed for full credibility is:
$$ n_f \geq \left(\frac{y_p}{r}\right)^2 CV^2 $$

Question 2 of 3
2. Question
From the question , the number of claims \(N\) follow a negative binomial distribution with parameters \(r=2\) and \(\beta=0.2\) and the Claim sizes \(Y\) follow a Pareto distribution with \(\alpha=3\) and \(\theta=1000\). Given the expression for the exposure needed for full credibility to be ;
$$\left(\frac{1.645}{0.05}\right)^2\left(\frac{Var(Y)}{{E(Y)}^2E(N)}+\frac{Var(N)}{{E(N)}^2}\right) $$
Calculate the exposure needed for full credibility.Correct
IncorrectFor claim sizes \((Y)\) that follow Pareto distribution with parameters \(\alpha=3\) and \(\theta=1000\) , form the C tables;
$$E(Y)=\frac{\theta}{\alpha1}=\frac{1000}{31}=500$$
$$E(Y^2)=\frac{{2\theta}^2}{(\alpha1)(\alpha2)}=\frac{2({1000}^2)}{(31)(32)}=\frac{2000000}{(2)(1)}=1000000$$
$$Var(Y)=E(Y^2)E(Y)^2=1000000500^2=750,000$$For the number of claims \(N\) that follow a negative binomial distribution with parameters \(r=2\) and \(\beta=0.2\), form the C tables;
$$E(N)=r\beta=2(0.2)=0.4$$
$$Var(N)=r\beta(1+\beta)=0.4(1+0.2)=0.48$$Then the exposure needed for full credibility is;
$$\begin{array}{ll}
n_f
&=\left(\frac{1.645}{0.05}\right)^2\left(\frac{Var(Y)}{{E(Y)}^2E(N)}+\frac{Var(N)}{{E(N)}^2}\right)\\
&=\left(\frac{1.645}{0.05}\right)^2\left(\frac{750000}{{500}^2(0.4)}+\frac{0.48}{{0.4}^2}\right)\\
&=11,365.31
\end{array}$$Hint
See the Exam C tables for expressions for \(E(Y)\) and \(Var(Y)\) for Pareto distribution; \(E(N)\) and \(Var(N)\) for Negative binomial distribution.

Question 3 of 3
3. Question
From the question, the general liability book of business comprised of 2500 insureds. calculate the credibility factor \(Z\).
Correct
IncorrectTo calculate the credibility factor \(Z\), we use the formula;
$$Z=
\left\{\begin{array}{ll}
\sqrt{\frac{n}{n_f}} & \text{if} \quad n\leq n_f \\
1 & \text{otherwise}\end{array} \right. … (3)$$Because the standard for full credibility in terms of exposure is \(n_f=11,365.31\) , given number of exposure n=2500 means there is inadequate experience for full credibility. Then ;
$$Z=\sqrt{\frac{n}{n_f}} =\sqrt{\frac{2500}{11,365.31}}=0.47$$Hint
To calculate the credibility factor \(Z\), we use the formula; ;
$$Z=
\left\{\begin{array}{ll}
\sqrt{\frac{n}{n_f}} & \text{if} \quad n\leq n_f \\
1 & \text{otherwise}\end{array} \right. … (3)$$
Strategy for Bühlmann Credibility
The Bühlmann credibility is sometimes known as greatest accuracy credibility. As before, \(X_1,X_2, \ldots, X_n\) are from a policyholder. Let \(\theta\) be a latent random variable associated with a policyholder and \(\{X_i\theta\}\) are iid, Then:
$$\begin{array}{lll}
\mu(\theta) & =E(X_i \theta) &=\text{“hypothetical mean”} \\
v(\theta) & =Var(X_i \theta) &=\text{“process variance”} \end{array}$$
The credibility premium is :
$$P_c=Z\overline{X} + (1Z)M \quad \quad (1)$$As before, \(M\) is the manual premium which is often given to be \(\mu=E(X)\). In this case, the credibility factor \(Z\) is
$$\begin{array}{ll}
Z & =\frac{n}{n+k} & & \quad \quad (2)\\
\end{array}$$ where k is the ratio of “expectation of the process variance” to the “variance of the hypothetical mean”
$$ k = \frac{\mathrm{E~} v(\theta)}{\mathrm{Var~}\mu(\theta)}
= \frac{\mathrm{E}\left(\mathrm{Var}(X_i \theta)\right)}{\mathrm{Var}\left(\mathrm{E}(X_i \theta)\right)} \quad \quad (3)
$$ As before, n is the number of observations.
To apply these rules,
 Use the problem to identify the conditional model distribution of each observation given \(\theta\), \(f_{X\theta}(x\theta)\) and the prior distribution of \(\theta\), \(\pi(\theta)\).
 Using the conditional model distribution \(f_{X\theta}(x\theta)\), find the hypothetical mean \(E(X\theta)\) and process variance \(Var(X\theta)\).
 Using the results above and the prior distribution \(\pi(\theta)\), Calculate \(k\) using equation (3) above.
 To get credibility factor \(Z\) use equation (2) above.
 If the question requires you to calculate the credibility premium, it is given by equation (1).
Bühlmann Credibility SOA #219
For a portfolio of policies, you are given:
(i) The annual claim amount on a policy has probability density function:
$$f(x\theta)=\frac{2x}{{\theta}^2}, \quad 0\lt x \lt \theta $$
(ii) The prior distribution of \(\theta\) has density function:
$$\pi(\theta)=4{\theta}^3, \quad \quad 0\lt\theta\lt 1 $$
(iii) A randomly selected policy had claim amount 0.1 in Year 1.
Calculate the Bühlmann credibility estimate of the claim amount for the selected policy in Year 2
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1. Question
From the question, the conditional model distribution given \(\theta\), \(f_{X\theta}(x\theta)\) is given by:
$$f(x\theta)=\frac{2x}{{\theta}^2}, \quad 0\lt x \lt \theta $$
Which of the following expressions below represent the hypothetical mean \(E(X\theta)\) and process variance \(Var(X\theta)\).Correct
Incorrect$$E(X\theta)=\int xf_{X\theta}(x\theta) dx$$
Given:
$$f(x\theta)=\frac{2x}{{\theta}^2}, \quad 0\lt x \lt \theta $$
Then
$$E(X\theta)=\int_0^{\theta} x\frac{2x}{{\theta}^2} dx=\int_0^{\theta} \frac{2x^2}{{\theta}^2} dx=\left(\frac{2}{{\theta}^2}\right)\left(\frac{{\theta}^3}{3}\right)=\frac{2\theta}{3}$$
and
$$E(X^2\theta)=\int_0^{\theta} x^2\frac{2x}{{\theta}^2} dx=\int_0^{\theta} \frac{2x^3}{{\theta}^2} dx=\left(\frac{2}{{\theta}^2}\right)\left(\frac{{\theta}^4}{4}\right)=\frac{{\theta}^2}{2}$$
This implies:
$$Var(X\theta)=E(X^2\theta)E(X\theta)^2=\frac{{\theta}^2}{2}\left(\frac{2\theta}{3}\right)^2=\frac{{\theta}^2}{18}$$Therefore , the hypothetical mean \(E(X\theta)=\frac{2\theta}{3}\) and the process variance \(Var(X\theta)=\frac{{\theta}^2}{18}\)
Hint
$$E(X^k\theta)=\int x^kf_{X\theta}(x\theta) dx$$

Question 2 of 4
2. Question
From the question, the prior distribution of \(\theta\) has density function:
$$\pi(\theta)=4{\theta}^3, \quad \quad 0\lt\theta\lt 1 $$
Given that the hypothetical mean \(E(X\theta)=\frac{2\theta}{3}\) and the process variance \(Var(X\theta)=\frac{{\theta}^2}{18}\) find Bühlmann’s \(k\).Correct
IncorrectFrom equation (3)
$$k = \frac{v}{a} = \frac{E\left(Var(X_i \theta)\right)}{Var\left(E(X_i \theta)\right)}$$
Given that the hypothetical mean \(E(X\theta)=\frac{2\theta}{3}\) and the process variance \(Var(X\theta)=\frac{{\theta}^2}{18}\) and$$\pi(\theta)=4{\theta}^3, \quad \quad 0\lt\theta\lt 1 $$
it follows that :
$$E\left(Var(X_i \theta)\right)=E\left(\frac{{\theta}^2}{18}\right)=\frac{1}{18}E({\theta}^2)=\frac{1}{18}\int_0^14{\theta}^5=\frac{1}{27}$$and
$$Var\left(E(X_i \theta)\right)=Var\left(\frac{2\theta}{3}\right)=\frac{4}{9}Var(\theta)$$
but;
$$
\begin{array}{ll}
Var(\theta)&=E({\theta}^2)E(\theta)^2=\int_0^1 4{\theta}^5\left(\int_0^1 4{\theta}^4\right)^2=\frac{4}{6}\left(\frac{4}{5}\right)^2\\
& =\frac{2}{75}
\end{array}$$$$k = \frac{v}{a} = \frac{E\left(Var(X_i \theta)\right)}{Var\left(E(X_i \theta)\right)}=\frac{\frac{1}{27}}{\frac{4}{9}\frac{2}{75}}=\frac{25}{8}$$
Hint
$$k = \frac{v}{a} = \frac{E\left(Var(X_i \theta)\right)}{Var\left(E(X_i \theta)\right)}$$

Question 3 of 4
3. Question
Given Bühlmann’s \(k=\frac{25}{8}\), calculate the credibility factor.
Correct
Incorrect$$Z =\frac{n}{n+k}$$
Given , \(k=\frac{25}{8}\) and number of observations \(n=1\)
$$Z =\frac{1}{1+\frac{25}{8}}=\frac{8}{33}$$Hint
$$Z =\frac{n}{n+k}$$
where n is number of observation and \(k\) is Bühlmann’s \(k\)

Question 4 of 4
4. Question
Calculate the Bühlmann credibility estimate of the claim amount for the selected policy in Year 2
Correct
IncorrectThe credibility premium is :
$$P_c=Z\overline{X} + (1Z)M $$Where , \(M\) is the Manual premium which is often \(\mu=E(X)\) and \(Z\) is the credibility factor.
$$E(X)=E\left(\frac{2\theta}{3}\right)=\left(\frac{2}{3}\right)E(\theta)=\left(\frac{2}{3}\right)\left(\frac{4}{5}\right)=\frac{8}{15}$$
Given \(Z=\frac{8}{33}\) and \(\overline{X}=0.1\)
$$P_c=Z\overline{X} + (1Z)M =0.1\left(\frac{8}{33}\right)+ \left(1\frac{8}{33}\right)\frac{8}{15}=0.4283$$Hint
The credibility estimate is :
$$P_c=Z\overline{X} + (1Z)M $$Where , \(M\) is the Manual premium which is often \(\mu=E(X)\) and \(Z\) is the credibility factor.
Bühlmann Credibility Aggregate losses SOA #8
You are given:
(i) Claim counts follow a Poisson distribution with mean \(\theta\).
(ii) Claim sizes follow an exponential distribution with mean \(10\theta\).
(iii) Claim counts and claim sizes are independent, given \(\theta\).
(iv) The prior distribution has probability density function:
$$\pi(\theta)=\frac{5}{{\theta}^6}, \quad\quad \theta\gt 1$$
Calculate Bühlmann’s \(k\) for aggregate losses.
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Question 1 of 4
1. Question
Let \(N\) be the Poisson claim count variable, let \(X\) be the claim size variable, and let \(S\) be the aggregate loss variable. Which of the following expressions below represent the hypothetical mean \(E(X\theta)\) and process variance \(Var(X\theta)\).
Correct
IncorrectGiven \(N\) to be the Poisson claim count variable, \(X\) to be the claim size variable, and \(S\) to be the aggregate loss variable. The hypothetical mean:
$$\mu(\theta)=E(S \theta) =E(N \theta)E(X \theta)=\theta(10\theta)=10{\theta}^2$$Process variance :
$$\begin{array}{ll}
v(\theta) &=Var(S \theta)=E(N\theta)Var(X\theta) +Var(N\theta)E(X\theta)^2\\
&=E(N\theta)E(X^2\theta)=\theta(200{\theta}^2)=200{\theta}^3
\end{array}$$Therefore , the hypothetical mean \(E(S \theta)=10{\theta}^2\) and the process variance \(Var(S \theta)=200{\theta}^3\)
Hint
Hypothetical mean:
$$\mu(\theta)=E(S \theta)=E(N \theta)E(X \theta) $$
Process variance :
$$v(\theta) =Var(S \theta)=E(N\theta)Var(X\theta) +Var(N\theta)E(X\theta)^2$$ 
Question 2 of 4
2. Question
From the question, the prior distribution of \(\theta\) has density function:
$$\pi(\theta)=\frac{5}{{\theta}^6}, \quad\quad \theta\gt 1$$
Given that the hypothetical mean \(E(S \theta)=10{\theta}^2\) find \(a\), the variance of the hypothetical mean \(Var\left(E(S \theta)\right)\).Correct
IncorrectGiven, \(E(S \theta)=10{\theta}^2\) and \(\pi(\theta)=\frac{5}{{\theta}^6}, \quad\quad \theta\gt 1\)
$$Var\left(E(S \theta)\right)=Var(10{\theta}^2)=100Var({\theta}^2)$$
but;
$$
\begin{array}{ll}
Var({\theta}^2)&=E({\theta}^4)E({\theta}^2)^2=5\int_1^{\infty} \frac{{\theta}^4}{{\theta}^6}d\theta\left(5\int_1^{\infty} \frac{{\theta}^2}{{\theta}^6}d\theta\right)^2\\
& =5\left(\frac{5}{3}\right)^2=\frac{20}{9}
\end{array}$$Therefore:
$$Var\left(E(S \theta)\right)=100Var({\theta}^2)=100\frac{20}{9}=\frac{2000}{9}$$Hint
\(Var\left(E(S \theta)\right)=E\left(E(S^2 \theta)\right)E\left(E(S \theta)\right)^2\)

Question 3 of 4
3. Question
From the question, the prior distribution of \(\theta\) has density function:
$$\pi(\theta)=\frac{5}{{\theta}^6}, \quad\quad \theta\gt 1$$
Given that the process variance \(Var(S \theta)=200{\theta}^3\) find \(v\), the expected value of the process variance \(E\left(Var(S \theta)\right)\).Correct
IncorrectGiven, \(Var(S \theta)=200{\theta}^3\) and \(\pi(\theta)=\frac{5}{{\theta}^6}, \quad\quad \theta\gt 1\)
It follows that:
$$E\left(Var(S \theta)\right)=E(200{\theta}^3)=200E({\theta}^3)=1000\int_1^{\infty} \frac{{\theta}^3}{{\theta}^6}d\theta =500$$ 
Question 4 of 4
4. Question
Given, \(v=E\left(Var(S \theta)\right)=500\) and \(a=Var\left(E(S \theta)\right)=\frac{2000}{9}\) calculate Bühlmann’s \(k\).
Correct
IncorrectGiven, \(v=E\left(Var(S \theta)\right)=500\) and \(a=Var\left(E(S \theta)\right)=\frac{2000}{9}\)
$$k = \frac{v}{a} = \frac{E\left(Var(S \theta)\right)}{Var\left(E(S \theta)\right)}=\frac{500}{\frac{2000}{9}}=2.25 $$
Hint
$$k = \frac{v}{a} = \frac{E\left(Var(S \theta)\right)}{Var\left(E(S \theta)\right)}$$
Strategy for BühlmannStraub Credibility
The Bühlmann model assumes one exposure in every period but the BühlmannStraub model generalizes Bühlmann model to a case where there are $m_j$ exposures in period $j$. Then for the BühlmannStraub model:
$$
Var(X_j \theta)= \frac{Var(X \theta)}{m_j} ;\quad \quad m_j=\text{exposure} \quad \quad (1)
$$
This is because the variance of the sample mean is the distribution variance divided by the number of observations. It follows that the credibility factor \(Z\) is given by ;
$$\begin{array}{ll}
Z & =\frac{\sum m_j}{\sum m_j+k} & & \quad \quad (2)\\
\end{array}$$
When \(m_j=1\) it implies that \(\sum m_j =n\) and the formulae is the same as that of the Bühlmann model. As before k is the ratio of “expectation of the process variance” to the “variance of the hypothetical mean”.
$$ k = \frac{\mathrm{E~} v(\theta)}{\mathrm{Var~}\mu(\theta)}
= \frac{\mathrm{E}\left(\mathrm{Var}(X_i \theta)\right)}{\mathrm{Var}\left(\mathrm{E}(X_i \theta)\right)} \quad \quad (3)
$$
and the
To apply these rules,
 Use the problem to identify the conditional model distribution of each observation given \(\theta\), \(f_{X\theta}(x\theta)\) and the prior distribution of \(\theta\), \(\pi(\theta)\).
 Using the conditional model distribution \(f_{X\theta}(x\theta)\), find the hypothetical mean \(E(X\theta)\) and process variance \(Var(X\theta)\).
 Using the results above and the prior distribution \(\pi(\theta)\), Calculate \(k\) using equation (3) above.
 To get credibility factor \(Z\) use equation (2) above.
 If the question requires you to calculate the credibility premium, it is the same as that for the Bühlmann model.
BühlmannStraub Credibility SOA #21
You are given:
(i) The number of claims incurred in a month by any insured has a Poisson distribution with mean \(\lambda\)
(ii) The claim frequencies of different insureds are independent.
(iii) The prior distribution is gamma with probability density function:
$$ f(\lambda)=\frac{(100\lambda)^6 e^{100\lambda}}{120\lambda}$$
(iv)
$$
{\scriptsize
\begin{matrix}
\begin{array}{ccc}
\hline
\text{Month} & \text{Number of Insureds} & \text{Number of Claims} \\
\hline
1 & 100 & 6 \\
2 & 150 & 8 \\
3 & 200 & 11 \\
4 & 300 & ? \\
\hline \\
\end{array}
\end{matrix}
}
$$
Calculate the BühlmannStraub credibility estimate of the number of claims in Month 4.
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Then, regardless of whether you get the right answer, you can check out this “quiz.” It is not really a quiz — it is a guided tutorial. The following provides a series of questions to lead you to the solution to this problem from the Society of Actuaries Exam C Question 21.
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Question 1 of 4
1. Question
From the question, given \(\lambda\), the number of claims follows a Poisson distribution and the prior distribution of \(\lambda\) has density function:
$$ f(\lambda)=\frac{(100\lambda)^6 e^{100\lambda}}{120\lambda}$$Calculate the variance of the hypothetical mean \(\mathrm{Var~}\mu(\theta)\) and expectation of the process variance \(\mathrm{E~} v(\theta)\).
Correct
IncorrectFrom the question, the number of claims incurred in a month by any insured has a Poisson distribution with mean \(\lambda\). This implies that
$$\begin{array}{llll}
\mu(\theta) & =E(N_i \theta) &=\text{“hypothetical mean”} &=\lambda\\
v(\theta) & =Var(N_i \theta) &=\text{“process variance”}&=\lambda \end{array}$$Also, the prior distribution of \(\lambda\) has density function:
$$ f(\lambda)=\frac{(100\lambda)^6 e^{100\lambda}}{120\lambda}$$The prior distribution of \(\lambda\) can be identified to be from a gamma distribution with parameters \(\alpha=6\) and \(\theta=1/100\).
Using the Exam C tables:
$$\begin{array}{lllll}
\mathrm{E~} v(\theta) &=E(\lambda) &=\alpha\theta &=\frac{6}{100}&=0.06\\
\mathrm{Var~}\mu(\theta) &=Var(\lambda) &=\alpha\theta^2 &=\frac{6}{100^2} &= 0.0006 \end{array}$$Therefore, the variance of the hypothetical mean \(\mathrm{Var~}\mu(\theta)=0.0006 \) and expectation of the process variance \(\mathrm{E~} v(\theta)=0.06 \).
$$\begin{array}{ll}
\mathrm{E~} v(\theta) &=0.06\\
\mathrm{Var~}\mu(\theta) &= 0.0006 \end{array}$$Hint
Given that the number of claims incurred in a month by any insured has a Poisson distribution with mean \(\lambda\), then;
$$\begin{array}{llll}
\mu(\theta) & =E(N_i \theta) &=\text{“hypothetical mean”} &=\lambda\\
v(\theta) & =Var(N_i \theta) &=\text{“process variance”}&=\lambda \end{array}$$ 
Question 2 of 4
2. Question
Given that;
$$\begin{array}{ll}
\mathrm{E~} v(\theta) &=0.06\\
\mathrm{Var~}\mu(\theta) &= 0.0006 \end{array}$$find \(k\) to calculate the BühlmannStraub credibility factor.
Correct
IncorrectFrom equation (3)
$$ k = \frac{\mathrm{E~} v(\theta)}{\mathrm{Var~}\mu(\theta)}
= \frac{\mathrm{E}\left(\mathrm{Var}(X_i \theta)\right)}{\mathrm{Var}\left(\mathrm{E}(X_i \theta)\right)}
$$Given that;
$$\begin{array}{ll}
\mathrm{E~} v(\theta) &=0.06\\
\mathrm{Var~}\mu(\theta) &= 0.0006 \end{array}$$Then;
$$ k = \frac{\mathrm{E~} v(\theta)}{\mathrm{Var~}\mu(\theta)}
= \frac{0.06}{0.0006} =100
$$Hint
$$ k = \frac{\mathrm{E~} v(\theta)}{\mathrm{Var~}\mu(\theta)}
= \frac{\mathrm{E}\left(\mathrm{Var}(X_i \theta)\right)}{\mathrm{Var}\left(\mathrm{E}(X_i \theta)\right)}
$$ 
Question 3 of 4
3. Question
Calculate BühlmannStraub credibility factor given that \(k=100\).
Correct
IncorrectFrom equation (2) , the BühlmannStraub credibility factor is given by:
$$\begin{array}{ll}
Z & =\frac{\sum m_j}{\sum m_j+k} & & \quad \quad (2)\\
\end{array}$$Given \(k=100\), and the total number of insureds contributing experience or sum of exposures \(\sum m_j=450\)
$$Z =\frac{450}{450+100}=\frac{9}{11}$$Hint
The BühlmannStraub credibility factor is given by;
$$\begin{array}{ll}
Z & =\frac{\sum m_j}{\sum m_j+k} & & \\
\end{array}$$ 
Question 4 of 4
4. Question
Calculate the BühlmannStraub credibility estimate of the number of claims in Month 4.
Correct
IncorrectThe credibility premium is :
$$P_c=Z\overline{X} + (1Z)M $$Where , \(M\) is the Manual premium which is often \(\mu=E(X)\) and \(Z\) is the credibility factor.
$$E(X)=\mathrm{E~}\left( E(N_i \theta)\right)=E(\lambda)=\alpha\theta=\frac{6}{100}=0.06$$Given \(Z=\frac{9}{11}\) and \(\overline{X}=\frac{\sum N_j}{\sum m_j}=\frac{25}{450} \) , The credibility estimate of the expected number of claims for one insured in month 4:
$$P_c=Z\overline{X} + (1Z)M =\left(\frac{25}{450}\right)\left(\frac{9}{11}\right)+ \left(1\frac{9}{11}\right)\frac{6}{100}=0.056364$$Then for 300 insureds the expected number of claims is 300(0.056364)=16.9.
Hint
The credibility estimate is :
$$P_c=Z\overline{X} + (1Z)M $$Where , \(M\) is the Manual premium which is often \(\mu=E(X)\) and \(Z\) is the credibility factor.
Strategy for Solving Bayesian Premium Problems
For Bayesian Premium problems:
 Use the problem to identify the conditional model distribution of each observation given \(\theta\), \(f_{X\theta}(x\theta)\). Then get the likelihood of the model distribution, which is typically the product of the model distributions for all observations,
$$f_{X\theta}(x_1…x_n\theta)=\prod_{i=1}^N f_{X\theta}(x_i\theta) .$$  From the question identify the prior distribution of \(\theta\), \(\pi(\theta)\).
 Now obtain the posterior distribution using the Bayes theorem;
$$\pi(\thetax_1…x_n)=\frac{f_{X\theta}(x_1…x_n\theta)\pi(\theta)}{\int f_{X\theta}(x_1…x_n\theta)\pi(\theta) d\theta}$$Note that the denominator of this expression \( f(x) = \int f_{X\theta}(x_1…x_n\theta)\pi(\theta) d\theta\) is the marginal distribution of \(x\) and does not involve \( \theta \).  With this, to calculate the posterior probability that \(a \lt \theta \lt b\);
$$\pi(a \lt \theta \lt bx_1…x_n)=\int_{a}^{b} \pi(\thetax_1…x_n) d\theta$$  For a new data \(y\) , the predictive distribution is
$$f(yx_1…x_n)=\int f(y\theta)\pi(\thetax_1…x_n)d\theta .$$Here, \(f(y\theta)\) is obtained from the conditional model distribution
There are several ways that one can use the predictive distribution. For some problems, you may wish to calculate Bayesian prediction for a new data \(y\). Specifically, given the past observations \(x_1…x_n\), you can use
$$E(yx_1…x_n)=\int E(y\theta)\pi(\thetax_1…x_n)d\theta. $$Here, \(E(y\theta)\) is the expected value of the conditional model distribution.
Bayesian Premium SOA #45
You are given:
(i) The amount of a claim, \(X\), is uniformly distributed on the interval \([0,\theta]\).
(ii) The prior density of \(\theta\) is \(\pi(\theta)=\frac{500}{\theta^2}, \quad \quad \theta \gt 500 \).
Two claims, \(x_1=400\) and \(x_2=600\) are observed. You calculate the posterior distribution as:
$$f(\thetax_1, x_2)=3\left(\frac{600^3}{\theta^4}\right), \theta \gt 600$$
Calculate the Bayesian premium , \(E(X_3x_1,x_2)\).
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Question 1 of 3
1. Question
From the question the amount of a claim, \(X\), is uniformly distributed on the interval \([0,\theta]\). Which of the following represents the expectation of the conditional model distribution?
Correct
IncorrectFrom the question the model distribution is a uniform distribution with unknown limit \(\theta\). Hence the conditional density function is
$$f_{X\theta}(x\theta)=\frac{1}{\theta}, 0\lt x \lt \theta$$
It follows that;$$E(x\theta)=\int_0^{\theta}x\frac{1}{\theta} dx =\frac{\theta}{2} $$
Hint
The model distribution is a uniform distribution with unknown limit \(\theta\). Hence the conditional density function is
$$f_{X\theta}(x\theta)=\frac{1}{\theta}, 0\lt x \lt \theta$$ 
Question 2 of 3
2. Question
The posterior distribution after the two claims, \(x_1=400\) and \(x_2=600\) are observed is given by
$$f(\thetax_1, x_2)=3\left(\frac{600^3}{\theta^4}\right), \theta \gt 600$$
For a new data \(X_3\), which of the following represents the expectation of the predictive distribution?
Correct
IncorrectUsing our notation, the posterior distribution was given as
$$\pi(\thetax_1, x_2)=f(\thetax_1, x_2)=3\left(\frac{600^3}{\theta^4}\right), \theta \gt 600$$
In the first part, we determined that mean of the model distribution to be
\( E(x\theta)= \theta / 2\). A general formula for the expectation of the predictive distribution is
$$E(X_{n+1}x_1…x_n)=\int E(X_{n+1}\theta)\pi(\thetax_1…x_n)d\theta.$$
For this problem, with \(n=2\), this is
$$E(X_3x_1, x_2)=\int_{600}^{\infty} \frac{3\theta}{2} \left(\frac{600^3}{\theta^4}\right) d\theta. $$Alternatively, for a new data \(X_{n+1}\) , one can determine the predictive distribution
$$f(X_{n+1}x_1…x_n)=\int f(X_{n+1}\theta)\pi(\thetax_1…x_n)d\theta .$$
From this, the expectation of the predictive distribution is
$$E(X_{n+1}x_1…x_n)=\int x f(xx_1…x_n) dx.$$Hint
For a new data \(X_{n+1}\) , the predictive distribution is
$$f(X_{n+1}x_1…x_n)=\int f(X_{n+1}\theta)\pi(\thetax_1…x_n)d\theta .$$ 
Question 3 of 3
3. Question
Calculate the Bayesian premium , \(E(X_3x_1,x_2)\).
Correct
IncorrectFrom the prior part, the Bayesian premium for a new data \(X_3\) is the expected value from the predictive distribution and is given by
$$E(X_3x_1, x_2)=\int_{600}^{\infty} \frac{3\theta}{2} \left(\frac{600^3}{\theta^4}\right) d\theta .$$Integrating gives
$$E(X_3x_1, x_2)=\frac{3(600^3)}{2}\frac{\theta^{2}}{2}\bigg{}_{600}^{\infty}=\frac{3(600^3)(600^{3})}{4}=450 .$$
Bayesian Estimation SOA #64/11
For a group of insureds, you are given;
(i) The amount of a claim in uniformly distributed but will not exceed a certain unknown limit \(\theta\).
(ii) The prior distribution of \(\theta\) is \(\pi(\theta)=\frac{500}{{\theta}^2}, \theta>500\).
(iii) Two claims of 400 and 600 are observed.
(a) Calculate the posterior probability that \(700 \lt \theta \lt 900\) .
(b) Calculate the probability that the next claim will exceed 550.
(c) Calculate the Bayesian premium, that is, the expected value of the next claim \(y\) given two claims of 400 and 600 are observed, \(E(yx_1=400,x_2=600)\).
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Question 1 of 5
1. Question
From the question, the model distribution follows a uniform distribution with unknown limit \(\theta\). Using the two claims observed, obtain the likelihood of the model distribution \(f_{X\theta}(x_1…x_n\theta)\).
Correct
IncorrectUnder conditional independence, the likelihood of the model distribution is the product of the model distributions for all observations
$$f_{X\theta}(x_1, x_2\theta) =\prod_{i=1}^2 f_{X\theta}(x_i\theta).$$
From the question, the model distribution is a uniform distribution with unknown limit \(\theta\). This has conditional probability density function
$$f_{X\theta}(x\theta)=\frac{1}{\theta}, 0\lt x \lt \theta .$$
Thus, the likelihood of the model distribution is
$$f_{X\theta}(x_1=400,x_2=600\theta)=\prod_{i=1}^2 \frac{1}{\theta}=\left(\frac{1}{\theta}\right)\left(\frac{1}{\theta}\right)=\frac{1}{{\theta}^2}.$$Hint
Under conditional independence, the likelihood of the model distribution is the product of the model distributions for all observations
$$f_{X\theta}(x_1…x_n\theta ) =\prod_{i=1}^n f_{X\theta}(x_i\theta).$$ 
Question 2 of 5
2. Question
The prior distribution of \(\theta\) is \(\pi(\theta)=\frac{500}{{\theta}^2}, \theta>500\). Derive the posterior density, which is the revised density based on the two claims observed, \(\pi(\thetax_1=400,x_2=600)\).
Correct
IncorrectFrom Bayes theorem
$$\pi(\thetax_1, x_2)=\frac{f_{X\theta}(x_1, x_2\theta)\pi(\theta)}{\int f_{X\theta}(x_1,x_2\theta)\pi(\theta) d\theta}.$$ From the question, two observed claims are 400 and 600, and the prior distribution is
$$\pi(\theta)=\frac{500}{{\theta}^2}, \theta>500.$$
With Bayes theorem
$$\begin{array}{ll}
\pi(\thetax_1=400,x_2=600)
&=\frac{f_{X\theta}(x_1=400,x_2=600\theta)\pi(\theta)}{\int f_{X\theta}(x_1=400,x_2=600\theta)\pi(\theta) d\theta}\\
&=\frac{\left(\frac{1}{{\theta}^2}\right)\left(\frac{500}{{\theta}^2}\right)}{\int_{600}^{\infty} \left(\frac{1}{{\theta}^2}\right)\left(\frac{500}{{\theta}^2}\right)d\theta}
\end{array}$$
Integrating the denominator of \(\pi(\thetax_1=400,x_2=600)\) gives
$$\int_{600}^{\infty} \left(\frac{1}{{\theta}^2}\right)\left(\frac{500}{{\theta}^2}\right) d\theta=\int_{600}^{\infty}\frac{500}{{\theta}^4} d\theta=\frac{500}{3(600)^3}.$$Note that the range of \(\theta\) is now \(\theta\gt 600\) . This is because having observed the value of 600 we know a parameter value less than or equal to 600 is not possible.
Then ;
$$\pi(\thetax_1=400,x_2=600)=\frac{\frac{500}{{\theta}^4}}{\frac{500}{3(600)^3}}=\frac{3(600)^3}{{\theta}^4}$$
Finally, adding the domain of \(\theta\) gives
$$\pi(\thetax_1=400,x_2=600)=\frac{3(600)^3}{{\theta}^4}, \theta \gt 600$$Hint
Use Bayes theorem
$$\pi(\thetax_1…x_n)=\frac{f_{X\theta}(x_1…x_n\theta)\pi(\theta)}{\int f_{X\theta}(x_1…x_n\theta)\pi(\theta) d\theta}.$$ 
Question 3 of 5
3. Question
In the prior step, we derived the posterior density, \(\pi(\thetax_1=400,x_2=600)=\frac{3(600)^3}{{\theta}^4}, \theta \gt 600\). Now, calculate the posterior probability that \(700 \lt \theta \lt 900\) .
Correct
IncorrectIn the prior step, we calculated the posterior density \(\pi(\thetax_1=400,x_2=600) \).
To calculate the posterior probability that \(700 \lt \theta \lt 900\), integrate to get
$$\begin{array}{ll}
\Pr(700 \lt \theta \lt 900400,600)
&=\int_{700}^{900} \frac{3(600)^3}{{\theta}^4} d\theta\\
&=3(600)^3\int_{700}^{900}\frac{1}{{\theta}^4} d\theta\\
&=3(600)^3 \left(\frac{1}{3(700)^3}\frac{1}{3(900)^3} \right)\\
&=0.333 .
\end{array}$$Hint
A “posterior probability” can be computed by integrating a posterior density over an appropriate range. For example, to calculate the posterior probability that \(a \lt \theta \lt b\), you might use
$$\Pr(a \lt \theta \lt bx_1, \ldots, x_n )=\int_{a}^{b} \pi(\thetax_1, \ldots, x_n ) d\theta .$$ 
Question 4 of 5
4. Question
Calculate the probability that the next claim will exceed 550.
Correct
IncorrectFor a new data \(y\) , the predictive density is
$$f(yx_1,x_2)=\int f(y\theta)\pi(\thetax_1,x_2)d\theta .$$ Here, \(f(y\theta)\) is obtained from the conditional model distribution. With this, we can write a predictive survival function as
$$
\begin{array}{ll}
\Pr(y > cx_1,x_2)
&= \int_c^{\infty} f(y x_1,x_2) dy \\
&=\int \Pr(y >c\theta) \pi(\thetax_1,x_2)d\theta .
\end{array}
$$
Recall that posterior density is
$$\pi(\thetax_1=400,x_2=600)=\frac{3(600)^3}{{\theta}^4}, \theta \gt 600$$
The model distribution follows a uniform distribution with unknown limit \(\theta\), therefore the posterior probability of an observation \(y\) exceeding 550 is;
$$\Pr(y>550\theta)=\frac{\theta550}{\theta}$$
With this, the predictive probability of an observation \(y\) exceeding 550 is
$$\begin{array}{ll}
\Pr(y>550400,600)
&= \int_{600}^{\infty} \Pr(y>550\theta)\pi(\theta400,600)d\theta \\
&=\int_{600}^{\infty} \left(\frac{\theta550}{\theta}\right) \left(\frac{3(600)^3}{{\theta}^4}\right) d\theta\\
&=3(600)^3\int_{600}^{\infty}\frac{1}{{\theta}^4}\frac{550}{{\theta}^5 }d\theta\\
&=0.3125
\end{array}$$Hint
For a new data \(y\) , the predictive density is
$$f(yx_1…x_n)=\int f(y\theta)\pi(\thetax_1…x_n)d\theta .$$ Here, \(f(y\theta)\) is obtained from the conditional model distribution. With this, we can write a predictive survival function as
$$
\begin{array}{ll}
\Pr(y > cx_1…x_n)
&= \int_c^{\infty} f(y x_1…x_n) dy \\
&=\int \Pr(y >c\theta) \pi(\thetax_1…x_n)d\theta .
\end{array}
$$ 
Question 5 of 5
5. Question
Calculate the Bayesian premium, that is expected value of the next claim \(y\) given two claims of 400 and 600 are observed, \(E(yx_1=400,x_2=600)\).
Correct
IncorrectTo calculate Bayesian prediction for a new data \(y\), given the past observations \(x_1=400,x_2=600\), we use
$$E(y400,600)=\int_{600}^{\infty} E(y\theta)\pi(\theta400,600)d\theta.$$The model distribution follows a uniform distribution with unknown limit \(\theta\). Therefore the expected value of a new observation \(y\) is
$$E(y\theta)=\frac{\theta}{2}.$$
From prior work, the posterior density is
$$\pi(\theta400,600)=\frac{3(600)^3}{{\theta}^4}, \theta \gt 600$$
Putting these together, we have
$$\begin{array}{ll}
E(y400,600)
&=\int_{600}^{\infty} \left(\frac{\theta}{2}\right)\left(\frac{3(600)^3}{{\theta}^4}\right) d\theta\\
&=\frac{3(600)^3}{2} \int_{600}^{\infty} \frac{1}{{\theta}^3} d\theta\\
&=450.
\end{array}$$Hint
To calculate Bayesian prediction for a new data \(y\), given the past observations \(x_1…x_n\), you can use
$$E(yx_1…x_n)=\int E(y\theta)\pi(\thetax_1…x_n)d\theta. $$ Here, \(E(y\theta)\) is the expected value of the conditional model distribution.