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CAA Associate Actuary (China) study materials
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CAA Associate Actuary (China) practice questions

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Question 1Integrated quantitative, financial and evidence reasoning

Let $X_1, X_2, \dots, X_n$ be independent Poisson random variables representing claim counts from $n$ independent policy groups, where $X_i \sim \text{Poisson}(\lambda_i)$. What is the exact distribution of the total claim count $S = \sum_{i=1}^n X_i$?

  1. $\text{Poisson}(\sum_{i=1}^n \lambda_i)$ Correct answer.
  2. $\text{Gamma}(\sum_{i=1}^n \lambda_i, 1)$
  3. $\text{Negative Binomial}(\sum_{i=1}^n \lambda_i, 0.5)$
  4. $\text{Normal}(\sum \lambda_i, \sum \lambda_i^2)$

Explanation

The sum of independent Poisson random variables is itself a Poisson random variable whose parameter is the sum of the individual Poisson intensity parameters: $S \sim \text{Poisson}(\sum_{i=1}^n \lambda_i)$, as verified by multiplying their moment generating functions $M_S(t) = \prod M_{X_i}(t) = \exp((\sum \lambda_i)(e^t - 1))$.

Question 2Integrated quantitative, financial and evidence reasoning

A life annuity or equipment warranty model assumes time-to-failure $T$ follows an exponential distribution with mean 10 years (rate $\lambda = 0.10$). Given that the component has already survived 6 years without failure, what is the expected total lifetime $E[T | T > 6]$?

  1. 16 years Correct answer.
  2. 10 years
  3. 4 years
  4. 12 years

Explanation

Due to the memoryless property of the exponential distribution, the remaining lifetime $T - 6$ given $T > 6$ is exponentially distributed with the original mean of 10 years: $E[T - 6 | T > 6] = 10$. Therefore, the expected total lifetime is $E[T | T > 6] = 6 + 10 = 16$ years.

Question 3Integrated quantitative, financial and evidence reasoning

Let $X$ follow a Gamma distribution $\Gamma(\alpha, \beta)$ with probability density function $f(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}$ for $x > 0$. What is the moment generating function $M_X(t) = E[e^{tX}]$ for $t < \beta$?

  1. $(1 - \frac{t}{\beta})^{-\alpha}$ Correct answer.
  2. $(1 - \beta t)^{-\alpha}$
  3. $\exp(\alpha \beta t)$
  4. $(1 + \frac{t}{\beta})^\alpha$

Explanation

For $X \sim \Gamma(\alpha, \beta)$, $M_X(t) = \int_0^\infty e^{tx} \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} dx = \frac{\beta^\alpha}{(\beta - t)^\alpha} \int_0^\infty \frac{(\beta - t)^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-(\beta-t)x} dx = (1 - \frac{t}{\beta})^{-\alpha}$ for $t < \beta$.

Question 4Integrated quantitative, financial and evidence reasoning

An insurance portfolio has annual aggregate claim payouts with mean $\mu = 50$ million RMB and standard deviation $\sigma = 10$ million RMB. Using Chebyshev's Inequality, what is the minimum probability that the aggregate claim payout will fall strictly between 30 million RMB and 70 million RMB?

  1. 0.75 Correct answer.
  2. 0.50
  3. 0.89
  4. 0.95

Explanation

The interval $(30, 70)$ corresponds to $(\mu - 2\sigma, \mu + 2\sigma)$, so $k = 2$. By Chebyshev's Inequality, $P(|X - \mu| < k\sigma) \ge 1 - \frac{1}{k^2} = 1 - \frac{1}{2^2} = 1 - 0.25 = 0.75$.

Question 5Integrated quantitative, financial and evidence reasoning

An insurer covers 10,000 independent policies. Each policy produces a claim with probability $p = 0.01$ and severity exactly 50,000 RMB. Using the Central Limit Theorem with continuity correction, what is the approximate probability that total claim payouts exceed 5,500,000 RMB? (Given standard normal CDF $\Phi(1.06) \approx 0.8554$)

  1. 0.1446 Correct answer.
  2. 0.8554
  3. 0.0500
  4. 0.3085

Explanation

Let $N$ be the number of claims. $N \sim \text{Binomial}(10000, 0.01)$ with $\mu_N = 100$ and $\sigma_N = \sqrt{10000 \times 0.01 \times 0.99} = \sqrt{99} \approx 9.9499$. Total payout exceeds 5,500,000 RMB when $N > 110$. With continuity correction, $P(N \ge 111) = P(N > 110.5) \approx 1 - \Phi(\frac{110.5 - 100}{9.9499}) = 1 - \Phi(1.055) \approx 1 - \Phi(1.06) = 1 - 0.8554 = 0.1446$.

Question 6Integrated quantitative, financial and evidence reasoning

Two continuous random variables $X$ (property damage loss) and $Y$ (bodily injury loss) have joint probability density function $f(x, y) = 3(x + y)$ on the unit simplex $0 \le x \le 1, 0 \le y \le 1, x + y \le 1$. What is the marginal probability density function $f_X(x)$ for $0 \le x \le 1$?

  1. $\frac{3}{2}(1 - x^2)$ Correct answer.
  2. $3x(1 - x)$
  3. $3(1 - x)^2$
  4. $\frac{1}{2}(1 - x)$

Explanation

The coefficient 3 normalizes the joint density because the integral of $x+y$ over the unit simplex is $1/3$. Integrating over $y$ from $0$ to $1-x$ gives $f_X(x)=\int_0^{1-x}3(x+y)dy=3x(1-x)+\frac{3}{2}(1-x)^2=\frac{3}{2}(1-x^2)$ for $0\le x\le1$.

Question 7Integrated quantitative, financial and evidence reasoning

A sample of $n$ independent observations $X_1, \dots, X_n$ is drawn from a continuous uniform distribution $U(0, \theta)$. What is the Method of Moments estimator $\hat{\theta}_{MM}$ and is it unbiased?

  1. $\hat{\theta}_{MM} = 2\bar{X}$, and it is an unbiased estimator of $\theta$ Correct answer.
  2. $\hat{\theta}_{MM} = \bar{X}$, and it is a biased estimator of $\theta$
  3. $\hat{\theta}_{MM} = X_{(n)}$, and it is an unbiased estimator of $\theta$
  4. $\hat{\theta}_{MM} = \frac{n+1}{n} X_{(n)}$, and it is a biased estimator of $\theta$

Explanation

For $U(0, \theta)$, the theoretical first moment is $E[X] = \frac{\theta}{2}$. Equating to the sample mean $\bar{X} = \frac{\theta}{2}$ gives $\hat{\theta}_{MM} = 2\bar{X}$. Since $E[2\bar{X}] = 2E[\bar{X}] = 2(\frac{\theta}{2}) = \theta$, $\hat{\theta}_{MM}$ is strictly unbiased.

Question 8Integrated quantitative, financial and evidence reasoning

For a random sample of size $n$ drawn from an exponential distribution with rate parameter $\lambda$ ($f(x) = \lambda e^{-\lambda x}, x > 0$), what is the Cramér-Rao Lower Bound (CRLB) for the variance of any unbiased estimator of $\lambda$?

  1. $\frac{\lambda^2}{n}$ Correct answer.
  2. $\frac{\lambda}{n}$
  3. $\frac{1}{n\lambda^2}$
  4. $\frac{\lambda^2}{n^2}$

Explanation

The log-likelihood for a single observation is $\ln f(x; \lambda) = \ln \lambda - \lambda x$. The second derivative is $\frac{\partial^2 \ln f}{\partial \lambda^2} = -\frac{1}{\lambda^2}$. The Fisher Information for a sample of size $n$ is $I_n(\lambda) = -n E[\frac{\partial^2 \ln f}{\partial \lambda^2}] = \frac{n}{\lambda^2}$. Thus, the Cramér-Rao Lower Bound is $\frac{1}{I_n(\lambda)} = \frac{\lambda^2}{n}$.

Question 9Integrated quantitative, financial and evidence reasoning

A normal sample of size $n = 21$ has sample variance $S^2 = 16$. What pivotal quantity and distribution are used to construct a two-sided $100(1-\alpha)\%$ confidence interval for the population variance $\sigma^2$?

  1. $\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{20}$ Correct answer.
  2. $\frac{(\bar{X} - \mu)}{S / \sqrt{n}} \sim t_{20}$
  3. $\frac{S^2}{\sigma^2} \sim F_{20, 20}$
  4. $\frac{(n-1)S^2}{\sigma^2} \sim N(0, 1)$

Explanation

For a normal population, $\frac{(n-1)S^2}{\sigma^2} = \frac{\sum_{i=1}^n (X_i - \bar{X})^2}{\sigma^2}$ follows a Chi-Square distribution with $n-1 = 21 - 1 = 20$ degrees of freedom ($\chi^2_{20}$), which serves as the standard pivotal quantity for variance inference.

Question 10Integrated quantitative, financial and evidence reasoning

In statistical quality control of insurance claims processing, which statement correctly defines the statistical power of a hypothesis test?

  1. The probability of correctly rejecting the null hypothesis when the alternative hypothesis is true ($1 - \beta$) Correct answer.
  2. The probability of rejecting the null hypothesis when it is actually true (Type I error $\alpha$)
  3. The probability of failing to reject the null hypothesis when the alternative is true (Type II error $\beta$)
  4. The probability that the test statistic falls strictly inside the acceptance region under $H_0$

Explanation

Statistical power is defined as $1 - \beta = P(\text{Reject } H_0 | H_1 \text{ is true})$, representing the probability that the test correctly detects a real effect and rejects a false null hypothesis.

Question 11Integrated quantitative, financial and evidence reasoning

An actuary groups claim frequency data into $k = 6$ discrete count bins ($0, 1, 2, 3, 4, \ge 5$) to conduct a Pearson Chi-Square goodness-of-fit test against a theoretical Poisson distribution. The Poisson intensity parameter $\lambda$ is estimated from the sample data using Maximum Likelihood. What are the degrees of freedom for the test statistic?

  1. 4 Correct answer.
  2. 5
  3. 6
  4. 3

Explanation

The degrees of freedom for a Chi-Square goodness-of-fit test are given by $df = k - 1 - m$, where $k$ is the number of bins and $m$ is the number of estimated parameters. Here, $k = 6$ and $m = 1$ (estimating $\lambda$), so $df = 6 - 1 - 1 = 4$.

Question 12Integrated quantitative, financial and evidence reasoning

In a One-Way Analysis of Variance (ANOVA) comparing mean claim settlement amounts across $k = 4$ regional branches with $n = 5$ claims sampled per branch (total $N = 20$), the Between-Group Sum of Squares is $SSB = 300$ and the Within-Group Sum of Squares is $SSW = 480$. What is the calculated $F$-statistic?

  1. 3.33 Correct answer.
  2. 2.50
  3. 0.625
  4. 4.17

Explanation

Mean Square Between is $MSB = \frac{SSB}{k - 1} = \frac{300}{4 - 1} = \frac{300}{3} = 100$. Mean Square Within is $MSW = \frac{SSW}{N - k} = \frac{480}{20 - 4} = \frac{480}{16} = 30$. The F-statistic is $F = \frac{MSB}{MSW} = \frac{100}{30} = 3.33$.

Question 13Integrated quantitative, financial and evidence reasoning

In Bayesian actuarial credibility, the claim probability $\theta$ for a policyholder follows a Beta prior distribution $\text{Beta}(a, b)$. Over $n$ policy years, the policyholder experiences $k$ claim events modeled as $\text{Binomial}(n, \theta)$. What is the posterior distribution of $\theta$?

  1. $\text{Beta}(a + k, b + n - k)$ Correct answer.
  2. $\text{Beta}(a + k, b + n)$
  3. $\text{Gamma}(a + k, b + n)$
  4. $\text{Normal}(a + k, b + n - k)$

Explanation

The Beta distribution is the conjugate prior for the Binomial likelihood. Multiplying the prior $\theta^{a-1}(1-\theta)^{b-1}$ by the likelihood $\theta^k (1-\theta)^{n-k}$ produces a kernel of $\theta^{(a+k)-1} (1-\theta)^{(b+n-k)-1}$, which is exactly $\text{Beta}(a + k, b + n - k)$.

Question 14Integrated quantitative, financial and evidence reasoning

Let $(X, Y)$ have a bivariate normal distribution with means $\mu_X = 100, \mu_Y = 200$, standard deviations $\sigma_X = 15, \sigma_Y = 25$, and correlation coefficient $\rho = 0.60$. If an observed loss $X = 120$, what is the conditional expectation $E[Y | X = 120]$ and conditional variance $\mathrm{Var}(Y | X = 120)$?

  1. $E[Y|X=120] = 220$, $\mathrm{Var}(Y|X=120) = 400$ Correct answer.
  2. $E[Y|X=120] = 212$, $\mathrm{Var}(Y|X=120) = 625$
  3. $E[Y|X=120] = 220$, $\mathrm{Var}(Y|X=120) = 250$
  4. $E[Y|X=120] = 232$, $\mathrm{Var}(Y|X=120) = 400$

Explanation

For bivariate normal variables: $E[Y | X = x] = \mu_Y + \rho \frac{\sigma_Y}{\sigma_X}(x - \mu_X) = 200 + 0.60 \times \frac{25}{15} \times (120 - 100) = 200 + 1.0 \times 20 = 220$. The conditional variance is $\mathrm{Var}(Y | X = x) = \sigma_Y^2 (1 - \rho^2) = 25^2 (1 - 0.60^2) = 625 \times 0.64 = 400$.

Question 15Integrated quantitative, financial and evidence reasoning

Let $X_1, X_2, \dots, X_n$ be independent and identically distributed exponential random variables with parameter $\lambda$ ($f(x) = \lambda e^{-\lambda x}, x > 0$). What is the exact probability density function of the minimum order statistic $X_{(1)} = \min(X_1, \dots, X_n)$?

  1. $f_{X_{(1)}}(x) = n\lambda e^{-n\lambda x}$ for $x > 0$ Correct answer.
  2. $f_{X_{(1)}}(x) = \lambda e^{-\lambda x / n}$ for $x > 0$
  3. $f_{X_{(1)}}(x) = (1 - e^{-\lambda x})^n$ for $x > 0$
  4. $f_{X_{(1)}}(x) = n(1 - e^{-\lambda x})^{n-1} \lambda e^{-\lambda x}$ for $x > 0$

Explanation

For the minimum order statistic, $P(X_{(1)} > x) = P(X_1 > x, \dots, X_n > x) = \prod_{i=1}^n P(X_i > x) = (e^{-\lambda x})^n = e^{-n\lambda x}$. The CDF is $F_{X_{(1)}}(x) = 1 - e^{-n\lambda x}$, so the density is $f_{X_{(1)}}(x) = n\lambda e^{-n\lambda x}$, which is exponentially distributed with rate $n\lambda$.

Question 16Integrated quantitative, financial and evidence reasoning

For independent observations $X_1,\dots,X_n$ from a Poisson distribution with unknown mean $\lambda$, which statistic is sufficient for $\lambda$ by the Neyman-Fisher factorization criterion?

  1. $T=\sum_{i=1}^n X_i$ Correct answer.
  2. $T=\max_i X_i$
  3. $T=\sum_{i=1}^n (X_i-\bar X)^2$
  4. $T=X_1$ alone

Explanation

The joint likelihood is $L(\lambda;x)=e^{-n\lambda}\lambda^{\sum x_i}/\prod x_i!$. Its dependence on $\lambda$ and the sample occurs only through $\sum x_i$, while $1/\prod x_i!$ is free of $\lambda$. The factorization criterion therefore makes $T=\sum X_i$ sufficient for $\lambda$.

Question 17Integrated quantitative, financial and evidence reasoning

What is the fundamental mathematical distinction between the Weak Law of Large Numbers (WLLN) and the Strong Law of Large Numbers (SLLN) regarding the sample mean $\bar{X}_n$ of i.i.d. claims?

  1. WLLN guarantees convergence in probability ($\bar{X}_n \xrightarrow{P} \mu$), while SLLN guarantees almost sure convergence ($P(\lim_{n\to\infty} \bar{X}_n = \mu) = 1$) Correct answer.
  2. WLLN applies only when variance is finite, while SLLN applies when first moments do not exist
  3. WLLN guarantees almost sure convergence, while SLLN guarantees convergence in distribution
  4. WLLN describes the limiting distribution shape, while SLLN describes parameter consistency

Explanation

The Weak Law of Large Numbers (Khinchin) establishes convergence in probability: $\lim_{n\to\infty} P(|\bar{X}_n - \mu| > \epsilon) = 0$. The Strong Law of Large Numbers (Kolmogorov) establishes almost sure (with probability 1) convergence: $P(\lim_{n\to\infty} \bar{X}_n = \mu) = 1$, which is a strictly stronger mode of stochastic convergence.

Question 18Integrated quantitative, financial and evidence reasoning

An insurance product has an estimated price elasticity of demand equal to $\varepsilon = -0.60$. If the insurer raises the gross premium rate by 10% and uses the elasticity estimate to project a 6% fall in quantity demanded, what is the projected change in total premium revenue?

  1. Total revenue will increase by approximately 3.4% Correct answer.
  2. Total revenue will decrease by approximately 6.0%
  3. Total revenue will decrease by approximately 3.4%
  4. Total revenue will remain unchanged because elasticity is negative

Explanation

Since demand is price inelastic ($|\varepsilon| = 0.60 < 1$), the percentage decrease in quantity demanded ($%\Delta Q = \varepsilon \times %\Delta P = -0.60 \times 10\% = -6\%$) is smaller than the percentage increase in price ($+10\%$). Total revenue changes by $(1 + 0.10)(1 - 0.06) - 1 = 1.034 - 1 = +3.4\%$.

Question 19Integrated quantitative, financial and evidence reasoning

An investor exhibits Constant Absolute Risk Aversion (CARA) with utility function $u(w) = -e^{-a w}$, where $a > 0$ is a constant. What is the Arrow-Pratt measure of absolute risk aversion $A(w) = -\frac{u''(w)}{u'(w)}$?

  1. $a$ Correct answer.
  2. $\frac{a}{w}$
  3. $a w$
  4. $\frac{1}{a}$

Explanation

First derivative: $u'(w) = a e^{-aw}$. Second derivative: $u''(w) = -a^2 e^{-aw}$. The Arrow-Pratt measure of absolute risk aversion is $A(w) = -\frac{-a^2 e^{-aw}}{a e^{-aw}} = a$, which is constant and independent of wealth level $w$.

Question 20Integrated quantitative, financial and evidence reasoning

Two insurance companies compete in a Cournot duopoly market where the inverse market demand for standardized term insurance policies is $P = 100 - Q$, where $Q = q_1 + q_2$. Both firms have identical and constant marginal costs $MC = 10$. In Cournot equilibrium, what is the equilibrium market price $P^*$?

  1. 40 Correct answer.
  2. 30
  3. 55
  4. 10

Explanation

For Firm 1, profit is $\pi_1 = (100 - q_1 - q_2 - 10)q_1 = (90 - q_1 - q_2)q_1$. First-order condition: $\frac{\partial \pi_1}{\partial q_1} = 90 - 2q_1 - q_2 = 0 \implies q_1 = \frac{90 - q_2}{2}$. By symmetry, $q_1 = q_2 = q^* \implies q^* = \frac{90 - q^*}{2} \implies 3q^* = 90 \implies q^* = 30$. Total output $Q^* = 60$. Equilibrium price is $P^* = 100 - 60 = 40$.

Question 21Integrated quantitative, financial and evidence reasoning

In the standard macroeconomic IS-LM model for a closed economy, how does an open market purchase of government securities by the central bank (People's Bank of China) affect equilibrium interest rates and national output?

  1. Shifts the LM curve rightward, lowering interest rates and increasing output Correct answer.
  2. Shifts the IS curve rightward, raising interest rates and increasing output
  3. Shifts the LM curve leftward, raising interest rates and decreasing output
  4. Shifts both IS and LM curves leftward, leaving interest rates unchanged

Explanation

An open market purchase increases the real money supply $M/P$, which shifts the LM curve to the right. In the money market, excess money supply drives bond prices up and interest rates down ($r \downarrow$), stimulating investment spending and expanding equilibrium national output ($Y \uparrow$).

Question 22Integrated quantitative, financial and evidence reasoning

An actuary is constructing a minimum variance portfolio from two risky assets $A$ and $B$ with standard deviations $\sigma_A = 10\%, \sigma_B = 20\%$, and correlation coefficient $\rho_{AB} = 0$. What is the optimal weight $w_A$ allocated to Asset A?

  1. 0.80 Correct answer.
  2. 0.50
  3. 0.67
  4. 0.20

Explanation

The minimum variance portfolio weight for Asset A is $w_A^* = \frac{\sigma_B^2 - \text{Cov}(A,B)}{\sigma_A^2 + \sigma_B^2 - 2\text{Cov}(A,B)}$. With $\rho = 0 \implies \text{Cov}(A,B) = 0$, $w_A^* = \frac{\sigma_B^2}{\sigma_A^2 + \sigma_B^2} = \frac{20^2}{10^2 + 20^2} = \frac{400}{100 + 400} = \frac{400}{500} = 0.80$.

Question 23Integrated quantitative, financial and evidence reasoning

In a simple open-economy Keynesian model, planned expenditure is defined by $C=100+0.75(Y-T)$, $T=0.20Y$, $I=200$, $G=300$, and $NX=50-0.10Y$. With no unintended inventory change, what is equilibrium output $Y$?

  1. 1,300 Correct answer.
  2. 1,625
  3. 1,083
  4. 650

Explanation

Taxes are $0.20Y$, so disposable income is $0.80Y$ and consumption is $100+0.75(0.80Y)=100+0.60Y$. Planned expenditure is therefore $100+0.60Y+200+300+50-0.10Y=650+0.50Y$. Equilibrium requires $Y=650+0.50Y$, so $Y=1,300$.

Question 24Integrated quantitative, financial and evidence reasoning

A property insurer has a 72% incurred loss ratio and a 25% underwriting expense ratio for the year. Ignoring investment income and income taxes, what do the combined ratio and underwriting result indicate?

  1. A 97% combined ratio and a 3% underwriting profit relative to earned premium Correct answer.
  2. A 97% combined ratio and a 3% underwriting loss relative to earned premium
  3. A 47% combined ratio and a 53% underwriting profit relative to earned premium
  4. A 103% combined ratio and a 3% underwriting loss relative to earned premium

Explanation

The combined ratio is the loss ratio plus the underwriting expense ratio: $72\%+25\%=97\%$. A combined ratio below 100% means premiums exceeded losses and underwriting expenses, so the underwriting profit is approximately $100\%-97\%=3\%$ of earned premium.

Question 25Integrated quantitative, financial and evidence reasoning

In a closed economy with no taxes, households consume 80% of each additional unit of disposable income. If planned investment falls by 25 billion RMB and the simple Keynesian multiplier model applies, what is the resulting change in equilibrium output?

  1. A decrease of 125 billion RMB Correct answer.
  2. A decrease of 25 billion RMB
  3. A decrease of 100 billion RMB
  4. An increase of 125 billion RMB

Explanation

With a marginal propensity to consume of 0.80, the simple expenditure multiplier is $1/(1-0.80)=5$. Therefore $\Delta Y=5\times(-25)=-125$ billion RMB.

Question 26Integrated quantitative, financial and evidence reasoning

A borrower and lender agree to a one-year loan at a fixed nominal interest rate of 5%. Inflation was expected to be 2% but turns out to be 4%, with no default. Relative to their expectations at origination, who benefits from the unexpected inflation?

  1. The borrower benefits because the realized real interest rate is lower than expected Correct answer.
  2. The lender benefits because the realized real interest rate is higher than expected
  3. Neither benefits because the nominal interest rate is fixed
  4. Both benefit because unexpected inflation raises the loan's real value

Explanation

The ex post real return is approximately the nominal rate minus actual inflation, or $5\%-4\%=1\%$, rather than the roughly 3% real return expected at origination. The borrower repays in money with less purchasing power than expected, so the borrower benefits at the lender's expense.

Question 27Integrated quantitative, financial and evidence reasoning

Which statement best describes the principle of indemnity in property insurance?

  1. The insured should be restored approximately to the financial position held immediately before the covered loss, subject to policy terms Correct answer.
  2. The insured must receive the policy limit after every covered event regardless of the amount of loss
  3. The insurer must return every premium whenever no claim occurs
  4. The insured may recover the same loss in full from several insurers and retain all proceeds

Explanation

Indemnity aims to compensate the insured for the covered financial loss without allowing a profit from the event. Deductibles, limits, valuation clauses, and other policy terms can constrain the amount paid.

Question 28Integrated quantitative, financial and evidence reasoning

In insurance economics, what is the Rothschild-Stiglitz model's key finding regarding equilibrium in a competitive insurance market with asymmetric information and adverse selection?

  1. A pooling equilibrium is not stable; if a competitive equilibrium exists, it is separating, with full coverage for high-risk types and partial coverage for low-risk types Correct answer.
  2. A pooling equilibrium always exists where all individuals pay the average community rate
  3. High-risk individuals are offered partial insurance while low-risk individuals receive full insurance
  4. Adverse selection eliminates moral hazard and leads to Pareto optimal pricing for all buyers

Explanation

In the Rothschild-Stiglitz screening model, a pooling allocation is vulnerable to cream-skimming. If a competitive equilibrium exists, it is separating: high-risk types receive full coverage at their actuarially fair premium, while low-risk types accept partial coverage to make their contract unattractive to high-risk types. For some population proportions and preferences, no pure-strategy competitive equilibrium exists.

Question 29Integrated quantitative, financial and evidence reasoning

Which contractual mechanism is primarily designed to mitigate ex-ante moral hazard in property and casualty insurance?

  1. Deductibles and coinsurance that leave a portion of the financial loss with the insured Correct answer.
  2. Guaranteed renewable clauses with fixed lifetime premium schedules
  3. Full first-dollar reimbursement with zero copayments
  4. Mandatory automatic inflation adjustments on policy coverage limits

Explanation

Ex-ante moral hazard occurs when policyholders take fewer preventive precautions to avoid loss after purchasing insurance. Cost-sharing provisions such as deductibles (免赔额), coinsurance (共保比例), and copayments force the insured to retain financial risk, aligning incentives to maintain loss prevention efforts.

Question 30Integrated quantitative, financial and evidence reasoning

For a portfolio of independent, identically distributed insurance losses, what is the main effect of increasing the number of exposure units while examining the average loss per exposure?

  1. The expected average loss is unchanged, while the variance of the average loss decreases Correct answer.
  2. Both the expected average loss and its variance increase in direct proportion to the number of exposures
  3. The expected average loss falls to zero even when each exposure has a positive expected loss
  4. The variance of the average loss is unchanged because the losses are identically distributed

Explanation

If each independent loss has mean $\mu$ and variance $\sigma^2$, the average of $n$ losses has mean $\mu$ and variance $\sigma^2/n$. Pooling therefore makes the average loss more predictable without changing its expected value.

Question 31Integrated quantitative, financial and evidence reasoning

Using the indirect method, an insurer reports net income of 120 million RMB, depreciation of 20 million, a 15 million gain on sale of investments, a 10 million increase in receivables, and a 5 million decrease in payables. What is net cash flow from operating activities, assuming no other adjustments?

  1. 110 million RMB Correct answer.
  2. 150 million RMB
  3. 140 million RMB
  4. 90 million RMB

Explanation

Start with net income and reverse noncash or nonoperating items and working-capital changes: $120+20-15-10-5=110$ million RMB. Depreciation is added back, the investment-sale gain is removed, and increases in receivables and decreases in payables both reduce operating cash flow.

Question 32Integrated quantitative, financial and evidence reasoning

An insurance investment fund earns a nominal annual interest rate of $i^{(4)} = 8.0\%$ compounded quarterly. What is the equivalent constant force of interest $\delta$ per annum?

  1. 7.92% Correct answer.
  2. 8.00%
  3. 8.24%
  4. 7.70%

Explanation

The effective annual rate is $1 + i = (1 + \frac{i^{(4)}}{4})^4 = (1 + 0.02)^4 = 1.082432$. The continuous force of interest is $\delta = \ln(1 + i) = \ln(1.082432) = 4 \times \ln(1.02) = 4 \times 0.0198026 = 0.07921$ or $7.92\%$.

Question 33Integrated quantitative, financial and evidence reasoning

A structured settlement provides an annuity due paying 100,000 RMB at the beginning of each year for $n = 10$ years. If the effective annual interest rate is $i = 5.0\%$, what is the present value of this annuity due $\ddot{a}_{\overline{10}|}$? (Given $v^{10} = (1.05)^{-10} \approx 0.613913$)

  1. 810,782 RMB Correct answer.
  2. 772,173 RMB
  3. 1,000,000 RMB
  4. 851,321 RMB

Explanation

The present value of an ordinary annuity immediate is $a_{\overline{10}|} = \frac{1 - v^{10}}{i} = \frac{1 - 0.613913}{0.05} = \frac{0.386087}{0.05} = 7.721735$. For an annuity due, payments occur at the beginning of each period: $\ddot{a}_{\overline{10}|} = (1 + i) a_{\overline{10}|} = 1.05 \times 7.721735 = 8.107822$. Multiplying by 100,000 RMB yields 810,782 RMB.

Question 34Integrated quantitative, financial and evidence reasoning

A commercial property loan of $L = 1,000,000$ RMB is amortized by 20 equal annual end-of-year payments $R$ at an annual effective interest rate $i = 6.0\%$. What is the principal repaid in the 5th payment $P_5$?

  1. 34,320 RMB Correct answer.
  2. 87,184 RMB
  3. 52,865 RMB
  4. 29,155 RMB

Explanation

The annual payment is $R=L/a_{\overline{20}|}=1,000,000/11.4699\approx87,184.56$ RMB. The first payment's principal is $P_1=R-iL=87,184.56-60,000=27,184.56$ RMB. Principal portions grow at the loan rate, so $P_5=P_1(1.06)^4\approx27,184.56\times1.262477=34,320$ RMB.

Question 35Integrated quantitative, financial and evidence reasoning

A 3-year zero-coupon bond has a Macaulay Duration of exactly 3.0 years. If the yield to maturity is $y = 5.0\%$ per annum (annual compounding), what is the Modified Duration $D_{mod}$ of the bond?

  1. 2.857 years Correct answer.
  2. 3.000 years
  3. 3.150 years
  4. 2.721 years

Explanation

Modified Duration is defined as $D_{mod} = \frac{D_{mac}}{1 + y}$. For $D_{mac} = 3.0$ years and $y = 0.05$, $D_{mod} = \frac{3.0}{1 + 0.05} = \frac{3.0}{1.05} \approx 2.857$ years.

Question 36Integrated quantitative, financial and evidence reasoning

In life contingency mathematics, how is the force of mortality $\mu_{x+t}$ formally defined in terms of the survival function $S(x)$ and conditional survival probability ${}_t p_x$?

  1. $\mu_{x+t} = -\frac{1}{{}_t p_x} \frac{d}{dt}({}_t p_x) = -\frac{d}{dt} \ln({}_t p_x)$ Correct answer.
  2. $\mu_{x+t} = \frac{d}{dt}({}_t q_x)$
  3. $\mu_{x+t} = \frac{{}_t q_x}{{}_t p_x}$
  4. $\mu_{x+t} = -\ln(1 - {}_t p_x)$

Explanation

The force of mortality (hazard rate of death) represents the instantaneous rate of mortality at age $x+t$: $\mu_{x+t} = \lim_{\Delta t \to 0^+} \frac{P(T_x \le t + \Delta t | T_x > t)}{\Delta t} = -\frac{1}{{}_t p_x} \frac{d}{dt}({}_t p_x) = -\frac{d}{dt} \ln({}_t p_x)$.

Question 37Integrated quantitative, financial and evidence reasoning

For a life aged $x$, a whole life insurance pays 1 RMB at the end of the year of death. If the effective annual discount factor is $v = 0.95$ ($d = 0.05$) and the actuarial present value of a whole life annuity due is $\ddot{a}_x = 12.50$, what is the actuarial present value of the whole life insurance $A_x$?

  1. 0.375 Correct answer.
  2. 0.625
  3. 0.500
  4. 0.250

Explanation

The fundamental identity linking discrete whole life insurance and life annuities due is $A_x = 1 - d \ddot{a}_x$. Given $d = 0.05$ and $\ddot{a}_x = 12.50$, $A_x = 1 - (0.05 \times 12.50) = 1 - 0.625 = 0.375$.

Question 38Integrated quantitative, financial and evidence reasoning

For a fully discrete whole life insurance of 1,000,000 RMB issued to $(x)$ with net annual premium $P_x = 15,000$ RMB, at duration $t = 10$, the actuarial functions are $A_{x+10} = 0.35$ and $\ddot{a}_{x+10} = 13.0$. What is the 10th year prospective net premium policy reserve ${}_{10} V_x$?

  1. 155,000 RMB Correct answer.
  2. 350,000 RMB
  3. 195,000 RMB
  4. 135,000 RMB

Explanation

The prospective benefit reserve is defined as: ${}_t V_x = S \cdot A_{x+t} - P_x \cdot \ddot{a}_{x+t} = (1,000,000 \times 0.35) - (15,000 \times 13.0) = 350,000 - 195,000 = 155,000$ RMB.

Question 39Integrated quantitative, financial and evidence reasoning

In continuous life insurance theory, Thiele's Differential Equation for the policy reserve ${}_t V$ of a policy with death benefit $S_t$, continuous net premium rate $P_t$, force of mortality $\mu_{x+t}$, and force of interest $\delta$ is expressed as:

  1. $\frac{d}{dt} {}_t V = \delta {}_t V + P_t - \mu_{x+t}(S_t - {}_t V)$ Correct answer.
  2. $\frac{d}{dt} {}_t V = \delta {}_t V - P_t + \mu_{x+t} S_t$
  3. $\frac{d}{dt} {}_t V = P_t - \mu_{x+t} S_t - \delta {}_t V$
  4. $\frac{d}{dt} {}_t V = (\delta + \mu_{x+t}){}_t V - S_t$

Explanation

Thiele's Differential Equation states that the instantaneous change in policy reserve equals interest earned on the reserve plus net premium income minus the expected cost of net amount at risk: $\frac{d}{dt} {}_t V = \delta {}_t V + P_t - \mu_{x+t}(S_t - {}_t V)$, where $(S_t - {}_t V)$ is the net amount at risk.

Question 40Integrated quantitative, financial and evidence reasoning

A non-life claim severity $X$ is exponentially distributed with mean 10,000 RMB. A policy pays $(X-d)_+$ per loss after an ordinary deductible of $d=5,000$ RMB. What is the expected insurer payment per loss?

  1. Approximately 6,065 RMB Correct answer.
  2. Approximately 3,935 RMB
  3. 10,000 RMB
  4. 5,000 RMB

Explanation

For an exponential severity with mean $\theta=10,000$, $E[(X-d)_+]=\int_d^\infty P(X>x)dx=\theta e^{-d/\theta}$. Thus the expected payment is $10,000e^{-0.5}\approx6,065$ RMB.

Question 41Integrated quantitative, financial and evidence reasoning

For two independent lives $(x)$ and $(y)$, which formula correctly represents the actuarial present value of a last survivor life annuity due $\ddot{a}_{\overline{xy}}$ in terms of single-life and joint-life annuities due?

  1. $\ddot{a}_{\overline{xy}} = \ddot{a}_x + \ddot{a}_y - \ddot{a}_{xy}$ Correct answer.
  2. $\ddot{a}_{\overline{xy}} = \ddot{a}_x + \ddot{a}_y + \ddot{a}_{xy}$
  3. $\ddot{a}_{\overline{xy}} = \frac{1}{2}(\ddot{a}_x + \ddot{a}_y)$
  4. $\ddot{a}_{\overline{xy}} = \sqrt{\ddot{a}_x \cdot \ddot{a}_y}$

Explanation

By the inclusion-exclusion principle of survival status, the last-survivor survival probability is ${}_t p_{\overline{xy}} = {}_t p_x + {}_t p_y - {}_t p_{xy}$. Multiplying by discount factor $v^t$ and summing yields $\ddot{a}_{\overline{xy}} = \ddot{a}_x + \ddot{a}_y - \ddot{a}_{xy}$.

Question 42Integrated quantitative, financial and evidence reasoning

In auto physical damage ratemaking, a vehicle fleet has an expected claim frequency of 0.08 claims per vehicle-year and an expected average claim severity of 12,500 RMB. If fixed administrative expenses are 200 RMB per vehicle and variable expenses plus target profit margin are 20% of the gross premium, what is the indicated gross premium per vehicle?

  1. 1,500 RMB Correct answer.
  2. 1,200 RMB
  3. 1,000 RMB
  4. 1,440 RMB

Explanation

Pure premium = $\text{Frequency} \times \text{Severity} = 0.08 \times 12,500 = 1,000$ RMB. Gross premium formula: $G = \frac{\text{Pure Premium} + \text{Fixed Expense}}{1 - \text{Variable Expense Ratio}} = \frac{1,000 + 200}{1 - 0.20} = \frac{1,200}{0.80} = 1,500$ RMB.

Question 43Integrated quantitative, financial and evidence reasoning

In a non-life loss reserving triangle, cumulative paid claims for Accident Year 2024 are 40 million RMB at 12 months development and 60 million RMB at 24 months development. The historical volume-weighted link ratio from 24 to 36 months is $f_{24-36} = 1.25$, and the cumulative development factor to ultimate from 36 months is $CDF_{36-\text{ult}} = 1.10$. What are the projected ultimate losses for Accident Year 2024?

  1. 82.5 million RMB Correct answer.
  2. 75.0 million RMB
  3. 66.0 million RMB
  4. 90.0 million RMB

Explanation

Current evaluation for AY 2024 at 24 months is $C_{2024, 24} = 60$ million RMB. The cumulative development factor from 24 months to ultimate is $CDF_{24-\text{ult}} = f_{24-36} \times CDF_{36-\text{ult}} = 1.25 \times 1.10 = 1.375$. Projected ultimate losses $= C_{2024, 24} \times CDF_{24-\text{ult}} = 60 \times 1.375 = 82.5$ million RMB.

Question 44Integrated quantitative, financial and evidence reasoning

For Accident Year 2025, an actuary establishes an initial expected loss ratio of $70\%$ on earned premium of 100 million RMB (Initial Expected Ultimate Losses $= 70$ million RMB). At 12 months development, cumulative reported claims are 28 million RMB, and the cumulative development factor to ultimate is $CDF = 2.00$. Using the Bornhuetter-Ferguson (BF) method, what is the estimated Incurred But Not Reported (IBNR) reserve?

  1. 35.0 million RMB Correct answer.
  2. 28.0 million RMB
  3. 42.0 million RMB
  4. 14.0 million RMB

Explanation

Under the Bornhuetter-Ferguson method, the percentage of losses unreported at development age $t$ is $1 - \frac{1}{CDF} = 1 - \frac{1}{2.00} = 1 - 0.50 = 0.50 (50\%)$. The IBNR reserve is calculated by multiplying the initial expected ultimate losses by the percentage unreported: $\text{IBNR}_{BF} = \text{Expected Losses} \times (1 - \frac{1}{CDF}) = 70 \times 0.50 = 35.0$ million RMB.

Question 45Integrated quantitative, financial and evidence reasoning

What is the primary conceptual distinction between the Cape Cod (Stanard-Bühlmann) loss reserving method and the standard Bornhuetter-Ferguson method?

  1. Cape Cod estimates the expected loss ratio internally from historical triangle experience and on-level premium, whereas BF requires an externally selected a priori loss ratio Correct answer.
  2. Cape Cod is applied only to paid claims, while BF is restricted exclusively to claim counts
  3. Cape Cod assumes zero development after 24 months
  4. Cape Cod ignores on-level earned premium adjustments

Explanation

The Cape Cod method is an enhancement of the Bornhuetter-Ferguson method where the expected loss ratio is not selected subjectively/externally, but is estimated directly from the aggregated reported losses divided by the sum of 'used' on-level earned premium across all accident years: $\hat{ELR} = \frac{\sum C_i}{\sum P_i / CDF_i}$.

Question 46Integrated quantitative, financial and evidence reasoning

A ceding insurer enters into a 4-line surplus reinsurance treaty with a net retention line of 2 million RMB per risk. The maximum underwriting capacity of the treaty per risk is 10 million RMB ($2M \text{ retention} + 4 \times 2M \text{ surplus}$). If the insurer writes a policy with sum insured 8 million RMB, what proportion of any claim is retained by the primary insurer?

  1. 25% Correct answer.
  2. 20%
  3. 40%
  4. 50%

Explanation

For an 8 million RMB policy with a 2 million RMB retention line, the retention proportion is: $\text{Retention Ratio} = \frac{\text{Net Retention}}{\text{Sum Insured}} = \frac{2,000,000}{8,000,000} = 0.25$ or $25\%$. The remaining $75\%$ (6 million RMB) is ceded to the surplus reinsurer.

Question 47Integrated quantitative, financial and evidence reasoning

An insurer purchases an Excess of Loss (XoL) reinsurance treaty structured as '30 million RMB excess of 20 million RMB' per occurrence. If a major catastrophe produces a gross claim of 45 million RMB, what amounts are paid by the primary insurer and the reinsurer respectively?

  1. Primary Insurer pays 20 million RMB; Reinsurer pays 25 million RMB Correct answer.
  2. Primary Insurer pays 20 million RMB; Reinsurer pays 30 million RMB
  3. Primary Insurer pays 15 million RMB; Reinsurer pays 30 million RMB
  4. Primary Insurer pays 25 million RMB; Reinsurer pays 20 million RMB

Explanation

The treaty covers losses exceeding the attachment point of 20 million RMB up to a layer limit of 30 million RMB (covering losses between 20M and 50M RMB). For a 45 million RMB gross loss, the primary insurer retains the initial attachment of 20 million RMB, and the reinsurer pays the entire excess portion: $45 - 20 = 25$ million RMB.

Question 48Integrated quantitative, financial and evidence reasoning

In a homogeneous Poisson process $\{N(t), t \ge 0\}$ with constant claim arrival intensity $\lambda = 4$ claims per day, what is the probability that exactly 2 claims occur during a given 12-hour period ($t = 0.5$ days)?

  1. 0.2707 Correct answer.
  2. 0.1465
  3. 0.1839
  4. 0.5413

Explanation

For a time interval of length $t = 0.5$ days, the expected number of claims is $\mu = \lambda t = 4 \times 0.5 = 2.0$. The claim count follows $N(0.5) \sim \text{Poisson}(2.0)$. Therefore, $P(N(0.5) = 2) = \frac{e^{-2.0} (2.0)^2}{2!} = \frac{e^{-2} \times 4}{2} = 2 e^{-2} \approx 2 \times 0.135335 = 0.27067$.

Question 49Integrated quantitative, financial and evidence reasoning

A non-homogeneous Poisson process has time-varying intensity function $\lambda(t) = 2 + 3t$ for $t \ge 0$ (measured in hours). What is the probability of observing zero claims during the 2-hour interval from $t = 1$ to $t = 3$?

  1. $\exp(-16)$ Correct answer.
  2. $\exp(-8)$
  3. $\exp(-12)$
  4. $\exp(-20)$

Explanation

The expected number of claims in interval $[1, 3]$ is $\mu = \int_1^3 \lambda(t) dt = \int_1^3 (2 + 3t) dt = [2t + \frac{3}{2}t^2]_1^3 = (2(3) + 1.5(9)) - (2(1) + 1.5(1)) = (6 + 13.5) - (2 + 1.5) = 19.5 - 3.5 = 16$. For a Poisson distribution, $P(N = 0) = \frac{e^{-16} 16^0}{0!} = e^{-16}$.

Question 50Integrated quantitative, financial and evidence reasoning

In an aggregate risk model, total claims $S = \sum_{i=1}^N X_i$ follow a Compound Poisson process where annual claim frequency is $N \sim \text{Poisson}(\lambda = 100)$ and individual claim severity $X$ has mean $E[X] = 5,000$ RMB and variance $\mathrm{Var}(X) = 4,000,000 \text{ RMB}^2$. What is the standard deviation of annual aggregate claims $\sigma_S$?

  1. 53,852 RMB Correct answer.
  2. 50,000 RMB
  3. 29,000 RMB
  4. 20,000 RMB

Explanation

For a compound Poisson distribution, the aggregate variance is given by $\mathrm{Var}(S) = \lambda E[X^2] = \lambda (\mathrm{Var}(X) + (E[X])^2) = 100 \times (4,000,000 + 5,000^2) = 100 \times (4,000,000 + 25,000,000) = 100 \times 29,000,000 = 2,900,000,000 \text{ RMB}^2$. The standard deviation is $\sigma_S = \sqrt{2,900,000,000} \approx 53,851.65$ RMB.